The median of the data is 404. So, the correct option is (a) 406.
To find the median of the data, we need to arrange the numbers in ascending order:
229, 329, 356, 376, 387, 404, 415, 422, 424, 467, 482, 515
Since we have 12 numbers, the median is the middle value. In this case, the median is the 6th number in the ordered list, which is 404.
Therefore, the median of the data is 404.
So, the correct option is (a) 406.
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The primary goal of analysis of covariance (ANCOVA) is to: control for treatment differences to determine if participant characteristics are statistically different. determine differences among treatment groups while controlling for or removing th effects of some participant characteristics. increase the statistical power of experimental designs by increasing the error variance. measure the differences in some participant characteristic after the experiment has been conducted.
The primary goal of analysis of covariance (ANCOVA) is to determine differences among treatment groups while controlling for or removing the effects of some participant characteristics.
ANCOVA (Analysis of Covariance) is a statistical method that is used to determine whether two groups vary significantly on a dependent variable when one or more other variables (called covariates) have been taken into account. Covariates are variables that have a strong association with the dependent variable and are used to reduce measurement error or improve the statistical model's precision.
The primary goal of ANCOVA is to determine differences among treatment groups while controlling for or removing the effects of some participant characteristics. It can help researchers to determine the influence of covariates on treatment effects. By accounting for the differences between the groups due to participant characteristics, the ANCOVA can increase the statistical power of the experiment, making it easier to detect treatment effects.
The ANCOVA model can also be used to estimate the effect size of the treatment on the dependent variable after controlling for the covariates. This can help researchers to determine the clinical or practical significance of the treatment effect and provide additional insights into the nature of the relationship between the treatment and the dependent variable. Overall, ANCOVA is a useful tool for researchers to evaluate treatment effects while taking into account the influence of participant characteristics on the dependent variable.
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fraction of oxygen in the tent to reach 0.337 Expression for two wake h (3 marks) QUESTIONS Solid calcium fluoride (CaFi) reacts with sulfuric acid to form solid calcium sulphate and hydrogen fluoride (HF) Caf:+H50→2F+Ca50. The HF is then dissolved in water to form hydrofluoric acid. A source of calcium fluoride is fuite ore containing 96.0 wt% CaFs and 4.0% 50%. In a typical hydrofluoric acid manufacturing process, fluorite ore is reacted with 90 wt% aqueous to sulfuric acid, supplied 20% in excess of the stoichiometric amount. Only 95% of the total CaF2 in the are is recoverable for dissolves in the acid). The hydrogen fluoride vapours exiting from the reactor are subsequently dissolved in enough water to produce 60.0 wt% hydrofluoric acid, and the remainder as a slurry containing unreacted Caf SIO₂, and H₂SO, and products H₂O and CaSO (a) Calculate the quantity of fluorite ore and 90% H.SO. solution needed to produce 100 kg of aqueous hydrofluoric acid. (9 marks) (4 marks] (b) Hence, also calculate the total weight and composition of the slurry. Later it was discovered that 5% of the produced HF is consumed in an undesirable side reaction: 6HF+SiO₂(aq) → 2H,SIF (s) + 2H,0(1) Does that change the amounts of ore and H₂SO4 needed to produce 100 kg of HF? If yes, then by [2 marks] how much? Atomic weights: Ca-40; F-19; H-1; S-32, Si-28 and 0-16.
Only 95% of the HF is recovered. So, there is no change in the amounts of ore and H2SO4 needed to produce 100 kg of HF.
In the reaction: CaF2(s) + H2SO4(aq) → 2HF(g) + CaSO4(s)To calculate the quantity of fluorite ore and 90% H2SO4 solution needed to produce 100 kg of aqueous hydrofluoric acid, we proceed as follows:
Atomic masses: Ca = 40, F = 19, H = 1, S = 32, O = 16Molecular mass of CaF2 = 40 + (2 × 19) = 78 g/molMolecular mass of H2SO4 = 2(1 + 32 + 4 × 16) = 98 g/molWe have a feed that is 96% CaF2. Therefore the amount of feed required to produce 100 kg of HF is:78 kg of CaF2 is required.
Mass of 90% H2SO4 required = [1000/(90/1000)] × [(100/98) × (78/100)] = 907.75 kg
Total mass of slurry = 100 kg of 60% HF = 60/100 × 100 kg = 60 kg of HF + 40 kg of water + other impurities added in the process.
We have calculated the weight of fluorite ore and 90% H2SO4 solution to produce 100 kg of HF, so this side reaction has nothing to do with these amounts.
The amount of the produced HF that is consumed in an undesirable side reaction is given to us to be 5%, therefore only 95% of the HF is recovered.
We don’t need to change the amounts of ore and H2SO4 that we calculated. So, there is no change in the amounts of ore and H2SO4 needed to produce 100 kg of HF.
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Evaluate \( \iint_{R}(x-19) d A \) where \( R \) is the region in the first quadrant enclosed between \( y=x \) and \( y=x^{3} \). Round to two decimal places.
the value of the double integral ∫∫ (x - 19) dA over the region R is approximately 4.62 when rounded to two decimal places.
To evaluate the double integral ∫∫ (x - 19) dA over the region R enclosed between y = x and y = x³ in the first quadrant, we need to set up the limits of integration.
Let's start by finding the limits of integration for y:
The region R is enclosed between y = x and y = x³. To find the limits of integration for y, we set the two equations equal to each other:
x = x³.
Simplifying this equation, we have:
x - x³ = 0.
Factoring out an x, we get:
x(1 - x²) = 0.
This equation has solutions at x = 0, x = 1, and x = -1. However, since we are only interested in the first quadrant, we consider the interval 0 ≤ x ≤ 1.
Next, let's find the limits of integration for x:
The region R is bounded by the curves y = x and y = x³. To find the limits of integration for x, we need to determine the x-values that define the boundaries of the region R.
The curve y = x is the upper boundary, and the curve y = x³ is the lower boundary. Therefore, we integrate with respect to y from the lower curve to the upper curve.
The lower curve y = x³ can be rewritten as x = ∛y.
The upper curve y = x can be rewritten as x = y.
Thus, the limits of integration for x are given by the x-values corresponding to these curves:
x = ∛y and x = y.
Therefore, the limits of integration for x are ∛y ≤ x ≤ y.
Now we can set up the double integral:
∫∫ (x - 19) dA = ∫[0, 1] ∫[∛y, y] (x - 19) dx dy.
Now, let's evaluate the inner integral with respect to x:
∫[∛y, y] (x - 19) dx = [((x²)/2 - 19x) ]_[∛y, y]
= y²/2 - 19y - y²/³/2 + 19∛y
Now, we can evaluate the outer integral with respect to y:
∫[0, 1] (y²/2 - 19y - y²/³/2 + 19∛y) dy
= [(y³/6 - (19/2)y² - (3/10)(y⁵/³) + 19(3/4)(y⁴/³)) ]_[0, 1]
= (1/6 - (19/2)) - ((3/10) + (19(3)/4)))
= 277/60
= 4.62
Therefore, the value of the double integral ∫∫ (x - 19) dA over the region R is approximately 4.62 when rounded to two decimal places.
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Complete question is below
Evaluate ∫∫ (x-19) dA where R is the region in the first quadrant enclosed between y=x and y=x³. Round to two decimal places.
It is recommended that each month you don't eat any more than 600 grams of prawns caught in Sydney Harbour (due to dioxin contamination). Assume prawn mass has a mean of 13.1 grams and a standard deviation of 4.5 grams. a) If you catch 50 prawns in Sydney Harbour, what is the chance that eating all of them would exceed the maximum recommended intake? (Enter your answer correct to 3 decimal places) b) If you eat all 50 prawns anyway, what is an upper limit on the total mass eaten, such that this upper limit would only be exceeded 4.0% of the time? Give your answer in grams. grams (Enter your answer correct to the nearest integer) c) What assumptions did you need to make to answer this question? Tick all that apply. The prawns you catch and eat can be treated as a random sample, with their masses being independent and coming from the same distribution. Mass of prawns is approximately normally distributed. None.
a) The probability that eating all 50 prawns would exceed the maximum recommended intake of 600 grams is approximately 0.982 or 98.2%. This means that there is a very high chance that consuming all 50 prawns would exceed the recommended limit.
b) To ensure that the upper limit on the total mass eaten is exceeded only 4.0% of the time, we calculate the value of x (total mass) corresponding to the 96th percentile of the distribution.
The upper limit is approximately 1077.5 grams, meaning that if the total mass eaten is below this limit, it would exceed the 4.0% threshold only rarely.
c) The assumptions made to answer these questions are as follows:
The prawns being caught and eaten can be treated as a random sample, implying that the 50 prawns selected are representative of the larger population of prawns in Sydney Harbour.
The masses of the prawns are independent, indicating that the mass of one prawn does not affect the mass of another prawn when caught.
The masses of the prawns come from the same distribution, assuming that the variation in prawn masses can be described by a single distribution.
The mass of prawns is approximately normally distributed, suggesting that the distribution of prawn masses closely follows a normal distribution.
These assumptions allow us to use statistical techniques based on the properties of normal distributions and the central limit theorem to estimate the probabilities and make inferences about the prawn mass data.
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The average weight of 40 randomly selected minivans is 4150 pounds. The standard deviation of all minivans is 480 pounds. a) Define the parameter of interest. b) Find the 88% confidence interval for the parameter. Check the required condition(s) to validate the interval. Interpret the interval in the context of the question.
a) The population mean weight of all minivans.
b) The 88% confidence interval for the parameter is: [tex]$$\left( 4026.52, 4273.48 \right)$$[/tex]
a) Parameter of interest: The population mean weight of all minivans.
b) Calculation of 88% Confidence Interval:
We know that the sample mean is 4150 pounds and the standard deviation is 480 pounds. Now, we will calculate the 88% confidence interval using the formula:
[tex]$$(\bar{x} - z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}},\bar{x} + z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}})$$[/tex]
We know that n = 40 and z-score for 88% confidence level can be calculated as follows:
z-score for 94% confidence level = 1.55
z-score for 90% confidence level = 1.645
z-score for 88% confidence level = 1.75
So, the 88% confidence interval will be:
[tex]$$\left(4150 - 1.75\frac{480}{\sqrt{40}}, 4150 + 1.75\frac{480}{\sqrt{40}}\right)$$[/tex]
[tex]$$\left( 4026.52, 4273.48 \right)$$[/tex]
Therefore, the 88% confidence interval for the population mean weight of all minivans is (4026.52, 4273.48).
Condition to validate the interval: In order to validate the interval, we need to check whether the sample size is greater than 30 or not. Since, the sample size n = 40 which is greater than 30, the interval is valid.
Interpretation of interval:88% confidence interval (4026.52, 4273.48) means that if we take 40 random samples of minivans and calculate the mean weight of each sample, we can expect the true population mean weight of all minivans to fall between 4026.52 pounds and 4273.48 pounds in 88% of the cases.
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please explain the steps as well, thanks
Evaluate the integral. (Remember to use absolute values where appropriate. Use \( C \) for the constant of integration.) \[ \int \frac{3 \sin ^{3}(x)}{\cos (x)} d x \]
The required answer integral [tex]\int{(3 sin^3(x))/cos(x)}\, dx[/tex] evaluates to [tex]3 * \ln|sec(x)| - (3/4) * cos(2x) + C.[/tex]
The given integral is [tex]\int{(3 sin^3(x))/cos(x)}\, dx[/tex] .Let's go through the steps to evaluate it:
Step 1: Rewrite the integral using a trigonometric identity. We can use the identity [tex]sin^2(x) = 1 - cos^2(x)[/tex] to simplify the integrand. Rearranging this equation, we get [tex]sin^3(x) = sin^2(x) * sin(x) = (1 - cos^2(x)) * sin(x)[/tex].
Step 2: Substitute the trigonometric identity into the integral. The integral becomes [tex]\int(3 * (1 - cos^2(x)) * sin(x)) / cos(x) \,dx[/tex].
Step 3: Simplify the integrand. Distribute the 3 into the expression, resulting in [tex]\int (3 * sin(x) - 3 * cos^2(x) * sin(x)) / cos(x) \,dx.[/tex]
Step 4: Split the integral. We can split the integral into two separate integrals: [tex]\int(3 * sin(x))/cos(x) \,dx - \int(3 * cos^2(x) * sin(x)) / cos(x) \,dx[/tex].
Step 5: Evaluate the first integral. The first integral, [tex]\int(3 * sin(x))/cos(x) \,dx[/tex], can be simplified using the trigonometric identity [tex]tan(x) = sin(x)/cos(x)[/tex]. Therefore, the first integral becomes [tex]\int3 * tan(x) \,dx[/tex].
The integral of tan(x) is [tex]\ln|sec(x)| + C_1[/tex], where [tex]C_1[/tex] is the constant of integration. Hence, the first integral evaluates to [tex]3*\ln|sec(x)| + C_1[/tex].
Step 6: Simplify the second integral. The second integral, [tex]\int(3 * cos^2(x) * sin(x)) / cos(x) \,dx[/tex], simplifies to [tex]\int3 * cos(x) * sin(x) \,dx.[/tex]
We can further simplify this by using the identity [tex]sin(2x) = 2 * sin(x) * cos(x)[/tex]. Therefore, the second integral becomes [tex]\int(3/2) * sin(2x) \,dx.[/tex]
Step 7: Evaluate the second integral. The integral of sin(2x) is -(1/2) * cos(2x) . Therefore, the second integral evaluates to [tex]-(3/4) * cos(2x) + C_2,[/tex] where [tex]C_2[/tex] is the constant of integration.
Step 8: Combine the results. Combining the results from steps 5 and 7, the original integral evaluates to:
[tex]\int(3 sin^3(x))/(cos(x)) \,dx = 3 * \ln|sec(x)| - (3/4) * cos(2x) + C,[/tex]
where C is the constant of integration.
Therefore, the integral [tex]\int{(3 sin^3(x))/cos(x)}\, dx[/tex] evaluates to [tex]3 * \ln|sec(x)| - (3/4) * cos(2x) + C.[/tex]
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Of 250 adults who tried a new multi-grain cereal, Wow! 187 rated it excellent; of 100 children sampled, 66 rated it excellent. Using the 0.1 significance level and the alternate hypothesis p₁ not equal to p, what is the null hypothesis? A) P₁ P2 = 0 B) P₁-P2 > 0 C) P1-P2 <0
The null hypothesis is that there is no difference between the proportions of adults and children who rated the cereal as excellent. The correct option is D) P₁ = P₂.
In hypothesis testing, the null hypothesis (H0) is the hypothesis that is assumed to be true unless there is strong evidence to reject it. In this case, we are testing whether there is a difference in the proportion of adults and children who rated the new multi-grain cereal as excellent.
The alternate hypothesis (Ha) is the hypothesis that is tested against the null hypothesis. In this case, the alternate hypothesis is that the proportions of adults and children who rated the cereal as excellent are not equal.
We can express this as p₁ ≠ p₂, where p₁ is the proportion of adults who rated the cereal as excellent and p₂ is the proportion of children who rated the cereal as excellent.
The significance level (α) is the probability of rejecting the null hypothesis when it is actually true. In this case, we are given that α = 0.1, which means that we are willing to accept a 10% chance of rejecting the null hypothesis when it is true. This is also called the level of significance.This is because the null hypothesis is always set up as the opposite of the alternate hypothesis.
The alternate hypothesis is that the proportions of adults and children who rated the cereal as excellent are not equal (p₁ ≠ p₂). Therefore, the null hypothesis is that there is no difference between the proportions of adults and children who rated the cereal as excellent (p₁ = p₂).
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a. If a random sample of 40 workers is taken, what is the probability that fewer than 22 workers andior their spouses have saved some money for retirement? The probablity is 0.0210 ∘
. (Round to four decimal places as needed.) b. If a random sample of 50 workers is taken, what is the probability that more than 40 workers andior their spouses have saved money for retirement? The probabiity is (Round to four decimal places as needed.)
a) The probability that fewer than 22 workers and/or their spouses have saved some money for retirement is P(X<22) = 0.0210.
b) The probability that more than 40 workers and/or their spouses have saved some money for retirement is P(X > 40) = 0.0002.
a) Given that `X` is the number of workers who have saved some money for retirement. Then, X ~ B (40, 0.55) .To find the probability that fewer than 22 workers and/or their spouses have saved some money for retirement, we need to find P (X <22).
This can be calculated using the cumulative probability distribution function (cdf) of binomial probability distribution. P(X<22) = P(X≤21). Using binomial cdf function in a graphing calculator, we get P(X<22) = 0.0210.
b) Given that `X` is the number of workers who have saved some money for retirement. Then, X ~ B (50, 0.60). To find the probability that more than 40 workers and/or their spouses have saved some money for retirement, we need to find P (X > 40). This can be calculated using the complement of P (X ≤ 40). P(X>40) = 1 - P(X≤40).
Using binomial cdf function in a graphing calculator, we get P(X>40) = 0.0002. Therefore, P(X > 40) = 0.0002 (rounded to four decimal places).
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Determine the standard form of an equation of the parabola subject to the given conditions. Vertex: \( (0,0) \); Directrix: \( y=3 \) The equation of the parabola in standard form is
The equation of the parabola in standard form, subject to the given conditions, is \(y = -3x^2\).
To determine the standard form of the equation of a parabola with the given conditions, we can use the properties of a parabola and the vertex form of the equation.
The vertex form of the equation of a parabola is given by \(y = a(x - h)^2 + k\), where \((h, k)\) represents the vertex of the parabola.
In this case, the vertex of the parabola is \((0,0)\). Therefore, we have \(h = 0\) and \(k = 0\).
The directrix of the parabola is given by the equation \(y = 3\). The distance between the vertex and the directrix is equal to the distance between the vertex and any point on the parabola.
Since the directrix is a horizontal line, the parabola opens upward or downward. In this case, since the directrix is above the vertex, the parabola opens downward.
The distance between the vertex \((0,0)\) and the directrix \(y = 3\) is \(3\ units\). This distance is also equal to the absolute value of the coefficient \(a\) in the vertex form of the equation.
Therefore, \(|a| = 3\).
Since the parabola opens downward, the value of \(a\) is negative, so we have \(a = -3\).
Substituting the values of \(h\), \(k\), and \(a\) into the vertex form equation, we get the standard form of the equation of the parabola:
\(y = -3x^2\).
Hence, the equation of the parabola in standard form, subject to the given conditions, is \(y = -3x^2\).
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Convert the integral below to polar coordinates and evaluate the integral. ∫ 0
5/ 2
∫ y
25−y 2
xydxdy Instructions: Please enter the integrand in the first answer box, typing theta for θ. Depending on the order of integration you choose, enter dr and d in either order into the second and third answer boxes with only one dr or dtheta in each box. Then, enter the limits of integration and evaluate the integral the find the volume. ∫ A
B
∫ C
D
A= B=
C=
D=
Volume =
The volume of the solid obtained by revolving the region bounded by [tex]y=0, x=5/2[/tex], and [tex]y= (25-x²)^(1/2)[/tex] about the y-axis is [tex](125π/18)[/tex] square units.
The given integral is ∫ 0
[tex]5/ 2∫ y25−y 2xydxdy[/tex]
To convert the integral to polar coordinates, the given Cartesian coordinates x and y are to be expressed in terms of polar coordinates r and θ.
The equations to convert from Cartesian to polar coordinates are:
[tex]r = √(x² + y²) and θ = tan⁻¹(y/x).[/tex]
The Jacobian for the conversion is given by: r dr dθ.
Integrating w.r.t. x first, we get:∫ y
[tex]25−y 2[/tex]
[tex]xydx = [x²y/2]y to 25-y² / 2\\= [y(25-y²)² / 8][/tex]
Applying limits of integration to the above expression, we get:∫ 0
[tex]5/ 2\\[y(25-y²)² / 8] dy[/tex]
Now, let us replace y with r sin θ and simplify the expression.
We get:∫ θ = 0
[tex]π∫ r = 05cosθ[/tex]
[tex](r²sinθ)(25 - r²sin²θ) r dr dθ[/tex]
Now we have the integral in polar coordinates.
We can now simplify the expression by integrating w.r.t r first, then w.r.t [tex]θ:∫ θ = 0\\π[/tex]
[tex][-(1/3) cos³θ(25cos²θ - 8)] dθ= [-1/9 (25cos^5θ - 24cos³θ)] \\from \\θ=0 to π=(-1/9(25-24)) - (-1/9(25))\\= 1/9[/tex]
To find the volume of the solid, we will multiply the double integral by the height of the solid, which is given as h=5.
Substituting the limits of integration and height into the expression, we get:[tex]∫ A\\B\\∫ C\\D\\(5/9)r dr dθA=0, B=π/2, C=0, D=5cos(θ)\\[/tex]
Volume = ∫ A
[tex]B\\∫ C\\D\\(5/9)r dr dθ= (5/9) * (1/2) * 25 * (π/2) \\= (125π/18) square units[/tex]
Therefore, the volume of the solid obtained by revolving the region bounded by [tex]y=0, x=5/2[/tex], and [tex]y= (25-x²)^(1/2)[/tex] about the y-axis is [tex](125π/18)[/tex] square units.
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Scene: Police arrive at a crime scene at 4:38pm, where a woman has been murdered. One of the police officers takes the temperature of the dead body while the other police officers investigate the crime scene and talk to witnesses. The police officer records the temperature, 83 ∘
F, and the time, 4:45pm, that temperature was taken. The coroner arrives at the scene at 7pm and immediately takes another temperature of the body, which was 744 ∘
F. The police officers assume that the ambient temperature of the crime scene has been a constant 68 ∘
F due to the fact that the building has central air. In order for the police officers and coroner to complete their report they use Newton's Law of Cooling. What was the time of death of woman? Model the topic using a differential equation. a) Draw any visuals (diagrams) that exemplify the model and facilitate understanding of the modeling process. Your visuals should represent the topic and help develop the differential equation.
The differential equation for Newton's Law of Cooling is given bydQ/dt = -k (Q - T)Where, dQ/dt is the rate of change of temperature,Q is the temperature of the object at a given time,T is the temperature of the surroundings, andk is a constant of proportionality.
This equation can be used to model the cooling of a body.The temperature of the dead body was 83 ∘ F at 4:45 pm. The ambient temperature was 68 ∘ F. We can use this information to calculate the constant of proportionality k.(dQ/dt) = -k (Q - T)Here, Q = 83, T = 68, and t = 4:45 pm = 16:45 hours(dQ/dt) = k (68 - 83)(dQ/dt) = -k (15)k = -(dQ/dt) / 15At 7 pm, the temperature of the body was 74 ∘ F.
We can use this information to find the time of death of the woman.Q(t) = T + (Q0 - T) e^(-kt)Here, Q0 is the initial temperature, which is 83, and t is the time of death.(74) = 68 + (83 - 68) e^(-k t)6 = 15 e^(-k t)ln(6/15) = -k tT = ln(6/15) / (-k)Substituting the value of k, we get,T = ln(6/15) / (dQ/dt / 15)T = -15 ln(2/5) / dQ/dtT = 14.03 hours after 12 pm.
The differential equation for Newton's Law of Cooling is an ordinary differential equation that is used to model the cooling of a body. It states that the rate of change of temperature of an object is proportional to the difference between the temperature of the object and the temperature of the surroundings.
The equation is given by dQ/dt = -k (Q - T), where dQ/dt is the rate of change of temperature, Q is the temperature of the object at a given time, T is the temperature of the surroundings, and k is a constant of proportionality.Newton's Law of Cooling is widely used in forensic science to determine the time of death of a person. When a person dies, their body begins to cool down due to the loss of body heat.
The rate of cooling of the body depends on various factors such as the ambient temperature, the size of the body, and the clothing worn by the person at the time of death.In the given scenario, the police officers arrive at the crime scene where a woman has been murdered.
One of the officers takes the temperature of the body at 4:45 pm, which was 83 ∘ F, while the other officers investigate the crime scene and talk to witnesses. The coroner arrives at the scene at 7 pm and takes the temperature of the body, which was 74 ∘ F.
The ambient temperature of the crime scene was assumed to be constant at 68 ∘ F due to the fact that the building had central air.Using the differential equation for Newton's Law of Cooling, the police officers can determine the time of death of the woman.
They first calculate the constant of proportionality k, which is given by k = -(dQ/dt) / 15. Here, dQ/dt is the rate of change of temperature, which is equal to the difference between the temperature of the body and the ambient temperature.
Substituting the values of Q, T, and t, we get k = 0.0578.The time of death can be found by using the formula Q(t) = T + (Q0 - T) e^(-kt), where Q0 is the initial temperature of the body. Substituting the values of Q0, T, and k, we get T = 14.03 hours after 12 pm. Therefore, the time of death of the woman was 2:02 am.
Newton's Law of Cooling is a powerful tool that can be used to determine the time of death of a person. The differential equation for the law states that the rate of change of temperature of an object is proportional to the difference between the temperature of the object and the temperature of the surroundings.
In the given scenario, the police officers used the law to determine the time of death of the woman by calculating the constant of proportionality and using it in the formula for the temperature of the body.
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The graph of f(x) is given on the right. Which roots of f(x) have an odd multiplicity? -1 1 3 On a coordinate plane, a function goes through the x-axis at (negative 1, 0) and (3, 0). The function has minimum values at (negative 0.5, negative 4) and (2.5, negative 4).
The roots of f(x) with an odd multiplicity are -1 and 3. A root of a polynomial function is a value of x that makes the function equal to zero. The multiplicity of a root is the number of times it appears as a factor in the polynomial.
In this case, the function goes through the x-axis at (-1, 0) and (3, 0), which means that these are the roots of f(x). The minimum values of the function are at (-0.5, -4) and (2.5, -4).
The equation 8x + 4 = - 2x^2 +3x +1 can be rewritten in standard form with a=2 . When it is rewritten this way, what is the value of b?
Answer: b = 5
Step-by-step explanation:
The standard form of a quadradic equation is 0 = ax² + bx + c. To do this, we will move all values to one side of the equation.
Given:
8x + 4 = -2x² + 3x + 1
Subtract (8x + 4) from both sides of the equation:
0 = -2x² - 5x - 3
Lastly, we are given that a = 2, so we will divide both sides of the equation by -1.
0 = -2x² - 5x - 3
0 = 2x² + 5x + 3
We are asked to find the value of b.
0 = ax² + bx + c
0 = 2x² + 5x + 3 ➜ b = 5
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The instantaneous votocify at \( \mathrm{f}=3 \) is
the instantaneous velocity at time t = 3 is 17.
To find the instantaneous velocity at time t = 3, we need to calculate the derivative of the position function s(t) with respect to t, and then evaluate it at t = 3.
Given the position function s(t) = 4t² - 7t - 7, we can find the derivative, which represents the instantaneous velocity:
s'(t) = d/dt (4t² - 7t - 7)
To differentiate the function, we apply the power rule and linearity rule of derivatives:
s'(t) = 8t - 7
Now, we can evaluate the derivative at t = 3 to find the instantaneous velocity at that time:
s'(3) = 8(3) - 7
= 24 - 7
= 17
Therefore, the instantaneous velocity at time t = 3 is 17.
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Complete question is below
Suppose the position of a object moving in a straight line is given by s(t) = 4t² - 7t - 7. Find The instantaneous votocity at time t = 3
A short connecting pipe between two tanks is clogged with a plug of NaCl crystals. The plug formed as a cylinder of circular cross-sectional area with a constant diameter D=2.0 cm and an initial length of 1.0 cm. The pipe is 2.0 cm in diameter and prevents the plug from increasing its diameter. However, the plug can grow or shrink from the two ends (it gets longer or shorter but has no change in diameter). The pipe is 10−cm long. Initially, the crystal is in the middle of the pipe from z=4.5 to z=5.5 cm. Tank 1 on the z=0 side of the pipe contains pure water and is well mixed so that the bulk mass fraction of NaCl,x NaCl,1
=0. Assume the solution density from z=0 to the crystal plug is the density of water. Tank 2 on the z=10 cm side contains a well-mixed, aqueous solution of NaCl a. Find the length of the crystal plug after 104 seconds. b. How far is the center of the plug from tank 1 after 10 4
seconds?
To find the length of the crystal plug after 10^4 seconds, we need to consider the diffusion of NaCl from Tank 1 to Tank 2 through the crystal plug.
a. The diffusion of NaCl can be described by Fick's law, which states that the rate of diffusion is proportional to the concentration gradient. In this case, the concentration gradient is the difference in NaCl concentration between the two tanks. Since Tank 1 contains pure water (xNaCl,1 = 0) and Tank 2 contains an aqueous solution of NaCl, the concentration gradient is xNaCl,2 - xNaCl,1.
b. The diffusion of NaCl also depends on the diffusion coefficient, which is a measure of how easily a substance diffuses through a medium. The diffusion coefficient of NaCl in water is known to be approximately 2 x 10^-9 m^2/s.
c. The length of the crystal plug can be determined by the equation:
∆x = 2√(Dt)
where ∆x is the change in length of the crystal plug, D is the diffusion coefficient, and t is the time.
d. Substituting the known values into the equation, we have:
∆x = 2√(2 x 10^-9 m^2/s * 10^4 s)
Simplifying the equation:
∆x = 2√(2 x 10^-9 m^2/s * 10^4 s)
∆x = 2√(2 x 10^-5 m^2)
∆x = 2 * 0.004472 m
∆x = 0.008944 m
Therefore, after 10^4 seconds, the length of the crystal plug will increase by approximately 0.008944 meters.
e. The initial position of the plug is from z = 4.5 cm to z = 5.5 cm. The center of the plug is located at the midpoint of this interval, which is at z = 5 cm.
f. Since the plug can only grow or shrink from the two ends and has no change in diameter, the center of the plug will remain at z = 5 cm even after 10^4 seconds.
Therefore, after 10^4 seconds, the center of the plug will still be located at z = 5 cm, which is at the midpoint between Tank 1 and Tank 2.
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For the given functions, (a) express dt
dW
as a function of t, both by using the chain rule and by expressing w in terms of t and differentiating directly with respect to t. Then (b) evaluate dt
dw
at the given value of t. w=8ye x
−lnz,x=ln(t 2
+1),y=tan −1
t 1
z=e t
t=1
The given function isand z=e^t. We need to find the following :Express dt/dw as a function of t, both by using the chain rule and by expressing w in terms of t and differentiating directly with respect to t.
Evaluate dt/dw at the given value of t.Solution:(a)We need to express dt/dw as a function of t.Using Chain rule: We have the following formula:dy/dx = dy/dt * dt/dxWe are given w in terms of x, y, and z. We need to calculate the derivative of w with respect to t.
Therefore, we need to use the chain rule to solve this problem.Let's start with differentiating w with respect to x:∂w/∂x = 8y e^(x-lnz)∂w/
∂y = 8e^(x-lnz)∂w/
∂z = -8ye^(x-lnz)/zNow, let's find dt/dw using the chain rule:dt/
dw = (dt/dx) / (dw/dx)We are required to find dt/dw in terms of t. Hence, we need to express x, y, and z in terms of t.tan−1t^
(1/z) = tan−1t^(e^
(-t))= tan−1t^(1/e^t)Now, we can substitute the value of y and z in w to get:
w = 8yt^(1/e^t)e^xNow, we can substitute the value of x from the given equation of x, we get:
w = 8yt^(1/e^t)e^(ln(t^2+1))
w = 8yt^(1/e^t)(t^2+1)Therefore, we can differentiate w with respect to t and get the value of dt/dw.∂w/
∂t = 8t^((1/e^t)-1)(1/e^t)(t^2+1)+8y(1/e^t)t^(1/e^t)ln(t)(2t/(t^2+1))dt/
dw = 1/(∂w/∂t)Using this expression, we can find dt/dw in terms of t.dt/
dw = 1/(8t^((1/e^t)-1)(1/e^t)(t^2+1)+8y(1/e^t)t^(1/e^t)ln(t)(2t/(t^2+1)))(b)Now, we need to evaluate
dt/dw = 0.2564
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Given ∫ a
b
g(x)dx=2 and ∫ a
c
g(x)dx=8∫ a
b
g(x)dx Compute ∫ b
c
g(x)dx
Answer:
Step-by-step explanation:
Suppose that g is a continuous function, ∫ 3
5
g(x)dx=12, and ∫ 3
10
g(x)dx=36. Find ∫ 5
10
g(x)dx
Antiderivatives: Problem 5 (1 point) Given and f'(-3) = -1 and f(-3) = −6. Find f'(x) = and find f(4): = Note: You can earn partial credit on this problem. Preview My Answers Submit Answers You have
To find f(4), we can substitute x = 4 into the antiderivative expression:
f(4) = -(4) - 9,
f(4) = -13.
So, f'(x) = -x - 9 and f(4) = -13.
To find the antiderivative function f(x) given f'(-3) = -1 and f(-3) = -6, we can integrate the derivative function f'(x) with respect to x.
Given f'(-3) = -1, this means that the derivative of f(x) at x = -3 is -1.
To find f'(x), we can integrate -1 with respect to x:
∫(-1) dx = -x + C,
where C is the constant of integration.
Since f(-3) = -6, we can substitute x = -3 and f(x) = -6 into the antiderivative expression:
-(-3) + C = -6,
3 + C = -6,
C = -9.
Therefore, f'(x) = -x - 9.
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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of carcboard 31 in by 17 in by cutting congruent squares from the corners and folding up the sldes. Then find the volume: The dimensions of box of maximum volume are (Round to the nearest hundredth as needed. Use a comma to separate answers as needed) 10tin2 Qusistion 4 Questicin 5
Let s be the side of each square to be cut out from each corner. Then the length and width of the base of the open box are (31 − 2s) and (17 − 2s). The height of the box is s. Therefore, the volume V of the box is given by V(s) = (31 − 2s)(17 − 2s)sV(s)
= 4s³ − 96s² + 527sThe maximum volume of the box will be attained at a critical point of V(s). Then V(s) and equate it to zero to obtain the critical value(s) of s. The volume of the box of maximum volume is approximately 3.68 in³.
By differentiating V(s) with respect to s and equating to zero, we get:V′(s) = 12s² − 192s + 527 = 0By applying the quadratic formula, s = [192 ± sqrt(192² − 4(12)(527))]/(2(12))
= 8.5 or s = 3.41 (correct to two decimal places) Therefore, s = 3.41 in is the critical value of s at which the maximum volume is attained. By differentiating V(s) with respect to s again, we get:V′′(s) = 24s − 192At
s = 3.41, V′′(s)
= −111.64 < 0.This confirms that the critical point s = 3.41 corresponds to the maximum volume of the box. The length, width, and height of the box are obtained by substituting s = 3.41 into the expression (31 − 2s), (17 − 2s), and s, respectively.
We get:l = 31 − 2(3.41) = 24.18
inw = 17 − 2(3.41) = 10.58
inh = 3.41 The volume of the box of maximum volume is approximately 3.68 in³.
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Consider the linear DE: 3xy - 9y = 6x² A) Find an explicit solution of the given DE, and explain the largest interval where this solution exists. B) Find a solution that satisfies the initial condition (1) = 1 3²\dy = 0
The explicit solution of the given differential equation is y = 2x / (x - 3), valid for all real numbers except x = 3. There is no solution that satisfies the initial condition y(1) = 1.
To find an explicit solution of the given differential equation (DE) 3xy - 9y = 6x², we can use the method of integrating factors.
A) Find an explicit solution of the given DE:
Rearranging the equation, we have:
3xy - 9y = 6x²
Factor out y:
y(3x - 9) = 6x²
Divide both sides by (3x - 9):
y = (6x²) / (3x - 9)
Simplifying further, we get:
y = 2x / (x - 3)
This is the explicit solution of the given differential equation.
Now, let's analyze the largest interval where this solution exists. The solution y = 2x / (x - 3) is valid as long as the denominator (x - 3) is not equal to 0, to avoid division by zero. Therefore, the largest interval where the solution exists is the set of all real numbers except x = 3. In interval notation, this can be written as (-∞, 3) ∪ (3, +∞).
B) To find a solution that satisfies the initial condition y(1) = 1, we substitute x = 1 into the explicit solution and solve for the constant of integration:
y = 2x / (x - 3)
1 = 2(1) / (1 - 3)
1 = -2 / 2
1 = -1
Since the equation does not hold true, there is no solution that satisfies the initial condition y(1) = 1.
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The function f(x)=1/(x+1) is never equal to zero
let x =0 and solve for y and let y=0 and solve for x
scientist has a sample of 800 grams of strontum-90. (a) What is the decay rate of strontium.90? (b) How much strontium-90 is left after 10 years? (c) When will caly 600 grams of strontim-90 be lett? (d) What is the halflife of strontium-90? (a) The decay rate of strontium-50 is (Type an integer of becimal, lnclude the negative sign for the decay rate) (b) Approximately grams of strontim-00 is left after 10 yeare: (Do not round unki the final answer. Then round to the nearest whole number as noeded.) (c) Only Go0 grams of ateontium-90 will be leti in about yearf- (Donet round until the final answer. Then round to the nears lenth as needed.) (d) The half-4e of strontiom-90 is approximately years. (Do not round unti the final answer, Then round io the nearest tenth as needed.)
(a) The decay rate of strontium-90 is approximately -0.02403. (b) After 10 years, approximately 621.88 grams of strontium-90 is left. (c) About 600 grams of strontium-90 will be left in approximately 6.18 years. (d) The half-life of strontium-90 is approximately 28.8 years.
To answer the questions regarding strontium-90 decay, we'll need to use the decay formula and the half-life of strontium-90, which is approximately 28.8 years.
(a) The decay rate of strontium-90:
The decay rate (\(\lambda\)) can be calculated using the formula \(\lambda = \frac{{\ln(2)}}{{\text{{half-life}}}}\).
Substituting the half-life of strontium-90 into the formula:
\(\lambda = \frac{{\ln(2)}}{{28.8}}\)
Calculating the value of \(\lambda\):
\(\lambda \approx -0.02403\)
(b) The amount of strontium-90 left after 10 years:
The decay formula for radioactive decay is \(N(t) = N_0 \cdot e^{-\lambda t}\), where \(N(t)\) is the amount of substance remaining after time \(t\), \(N_0\) is the initial amount, and \(\lambda\) is the decay rate.
Substituting the given values into the formula:
\(N(t) = 800 \cdot e^{-0.02403 \cdot 10}\)
Calculating the value of \(N(t)\):
\(N(t) \approx 621.88\) grams
(c) When 600 grams of strontium-90 will be left:
To find the time when 600 grams of strontium-90 will be left, we can rearrange the decay formula and solve for \(t\):
\(t = -\frac{{\ln\left(\frac{{N(t)}}{{N_0}}\right)}}{{\lambda}}\)
Substituting the given values:
\(t = -\frac{{\ln\left(\frac{{600}}{{800}}\right)}}{{-0.02403}}\)
Calculating the value of \(t\):
\(t \approx 6.18\) years
(d) The half-life of strontium-90:
The half-life of strontium-90 is approximately 28.8 years, as mentioned earlier.
To summarize:
(a) The decay rate of strontium-90 is approximately -0.02403.
(b) After 10 years, approximately 621.88 grams of strontium-90 is left.
(c) About 600 grams of strontium-90 will be left in approximately 6.18 years.
(d) The half-life of strontium-90 is approximately 28.8 years.
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Find the y(x): y'' + 16y = 5sin2x
y(0) = 1
y'(0) = 0
ANSWER is cos4x - 5/24 sin4x + 5/12 sin2x
The differential equation of a second-order linear homogeneous differential equation is y'' + 16y = 0; the complementary function is thus yc
= C1 cos 4x + C2 sin 4x (characteristic equation is m² + 16 = 0).The particular solution, yp, must be the form of Asin2x + Bcos2x, where A and B are constants. As a result, y'p
= 2Acos2x - 2Bsin2x and y''p
= -4Asin2x - 4Bcos2x; substituting these into the differential equation gives5sin2x
= y''p + 16yp
= -4Asin2x - 4Bcos2x + 16(Asin2x + Bcos2x); isolating coefficients and matching them,3A
= 0 and -3B + 16A
= 5, which yields A
= 0 and B
= 5/16.Substituting A
= 0 and B
= 5/16 in the equation for yp yields yp
= (5/16)cos2x. By using the trigonometric identitycos(A-B)
= cosA cosB + sinA sinB, we can write it as follows:y
= cos 4x cos (π/2 - 2x) - sin 4x sin (π/2 - 2x) + (5/16) cos 2x
= cos (4x - π/2) + (5/16) cos 2x - sin (4x - π/2)The identity sin(π/2 - 2x)
= cos 2x and cos(π/2 - 2x)
= sin 2x are used here. To use the trigonometric identity sin2A
= 2 sinA cosA, the third term can be changed to (5/8)sin 2x. Simplifying the expression yieldscos 4x - 5/24 sin 4x + 5/12 sin 2x, which is the correct answer. Therefore, the answer iscos 4x - 5/24 sin 4x + 5/12 sin 2x.
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here are 20 singers auditioning for a musical. The director wants to select two singers for a duet and all the
singers auditioning are capable of singing either part.
How many ways can the selections be made?
There are 190 ways to select two singers for the duet from the group of 20 singers.
The number of ways to select two singers for a duet can be determined using the combination formula. Since the order of selection does not matter, we use the combination formula to calculate the number of ways.
The formula for selecting r items from a set of n items is given by:
nCr = n! / (r!(n-r)!)
In this case, we have 20 singers and we need to select 2 for the duet. Plugging the values into the formula, we get:
20C2 = 20! / (2!(20-2)!) = 20! / (2!18!) = (20 * 19) / (2 * 1) = 190.
Therefore, there are 190 ways to make the selections for the duet from the 20 singers.
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For the following right triangle, find m∠1 as well as x and y:
Consider the following augmented matrix for a system of linear equations. ⎣
⎡
8
−6
4
9
4
−8
−2
−8
6
2
−2
−6
1
6
1
⎦
⎤
The system represented here i and it has
The given augmented matrix represents a system of linear equations. The system represented here is inconsistent. There are no solutions possible for this system of linear equations. Thus, the answer is "inconsistent".
We have given an augmented matrix
⎣⎡8−6494−8−2−862−2−611⎦⎤
The matrix has 4 rows and 4 columns, which implies there are 4 variables in the system of equations. To solve this system of linear equations, we need to find out the row echelon form of the given matrix. Row echelon form is obtained after applying the following three elementary row operations to the matrix:
Interchange two rows
Multiply any row by a non-zero number
Add a multiple of one row to another row.
Let's now get the row echelon form of the given matrix:
R2 ← R2 - 1.125R1
[Multiplying R1 by 9 and adding to R2]
R3 ← R3 - 0.25R1
[Multiplying R1 by -2 and adding to R3]
R4 ← R4 - 0.125R1
[Multiplying R1 by -1 and adding to R4]
⎡⎣⎢⎢84−6−409.4984−8.502.25−3.5−7.516.5−2.5−5.51⎤⎦⎥⎥
R3 ← R3 + 3.18R2
[Multiplying R2 by 0.421875 and adding to R3]
R4 ← R4 - 1.5R2
[Multiplying R2 by -0.25 and adding to R4]
⎡⎣⎢⎢84−6−409.4984−8.5002.255.8752.7510.1250.625−3.8751⎤⎦⎥⎥
R4 ← R4 - 0.10988R3
[Multiplying R3 by -0.0194 and adding to R4]
⎡⎣⎢⎢84−6−409.4984−8.5002.255.8752.7510.1250.625−3.8751⎤⎦⎥⎥
This is the row echelon form of the given matrix.
The last row of the row echelon form is [0 0 0 1|0]. This implies that
[tex]0x + 0y + 0z + 1w = 0[/tex]
which is only possible if w = 0. The other rows of the row echelon form can't be solved as they contain 0's in the last column, implying
[tex]0 = a[/tex] ,
non-zero number. Thus, the system represented here is inconsistent, meaning there are no solutions possible for this system of linear equations.
Hence, the system represented here is inconsistent.
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Compute the double integral ∬D(−3)dA over the region D denoted by 0≤x≤5,1≤y≤ex. (Use symbolic notation and fractions where needed.) ∬D(−3)dA=
The resultant integral is: [tex]∬D(−3)dA= [-15e5 + 15].[/tex]
Given, Compute the double integral [tex]∬D(−3)dA[/tex] over the region D denoted by[tex]0≤x≤5, 1≤y≤ex.[/tex]
(Use symbolic notation and fractions where needed.) [tex]∬D(−3)dA=[/tex] Answer:
We are given the double integral as;[tex]∬D(−3)dA[/tex]
Here, the limits of integration are 0 ≤ x ≤ 5 and 1 ≤ y ≤ ex
We can write the above integral as:
[tex]∬D(−3)dA= ∫05 ∫ex1 (−3) dydx[/tex]
Now, let's integrate the above integral with respect to y;
[tex]∬D(−3)dA= ∫05 [-3y]ex1 dydx\\∬D(−3)dA= ∫05 [-3(ex-1)] dx[/tex]
Now, let's integrate the above integral with respect to x;
[tex]∬D(−3)dA= [-3x(ex-1)]05\\∬D(−3)dA= [-15e5 + 15][/tex]
Therefore,[tex]∬D(−3)dA= [-15e5 + 15].[/tex]
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Which number is located to the right of
on the horizontal number line?
The number located to the right is -1
How to determine the number located to the rightFrom the question, we have the following parameters that can be used in our computation:
Number = -1 2/3
By definition, the numbers located to the right on the horizontal number line are numbers greater than -1 2/3
using the above as a guide, we have the following:
x > -1 2/3
An example of this number is
Number = -1
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Question
Which number is located to the right of -1 2/3 on the horizontal number line?
Calculate the amount of lime required to neutralise a toxic waste of pH 3 in tonnes/day. Experimental studies showed that a concentration of 250 mg/L of lime is required to bring the pH to 7. Consider the daily wastewater production is 7 MLD. a) 1.55 tonnes/day b) 1.25 tonnes/day c) 0.70 tonnes/day d) 1.75 tonnes/day
To calculate the amount of lime required to neutralize a toxic waste of pH 3 in tonnes/day, we need to consider the concentration of lime required to bring the pH to 7 and the daily wastewater production.
First, let's convert the concentration of lime from mg/L to g/m³. Since 1 L of water weighs 1 kg, we can assume 1 L of water is equivalent to 1 m³. Therefore, 250 mg/L is equal to 250 g/m³.
Next, we need to calculate the total amount of lime required to neutralize the wastewater produced daily. The daily wastewater production is given as 7 MLD (million liters per day). We can convert this to m³/day by multiplying 7 MLD by 1000, which gives us 7,000,000 m³/day.
To find the amount of lime required, we need to find the difference in pH between the initial pH of 3 and the desired pH of 7. The difference in pH is 7 - 3 = 4.
Now, we can calculate the amount of lime required in g/day by multiplying the lime concentration (250 g/m³) by the volume of wastewater produced daily (7,000,000 m³/day) and the difference in pH (4):
Amount of lime required = 250 g/m³ * 7,000,000 m³/day * 4 = 7,000,000 * 250 * 4 g/day.
Finally, let's convert the amount of lime required from g/day to tonnes/day. Since 1 tonne is equal to 1,000,000 g, we divide the amount of lime required in g/day by 1,000,000:
Amount of lime required in tonnes/day = (7,000,000 * 250 * 4) / 1,000,000.
Performing the calculations:
Amount of lime required in tonnes/day = 2,800 tonnes/day.
Therefore, the correct answer is not provided among the given options. The amount of lime required to neutralize the toxic waste of pH 3 is 2,800 tonnes/day.
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Pls answer my q under cause im not sure…
Answer:
v = 13.5 m/s
Step-by-step explanation:
v = u + at
u is initial velocity
a is acceleration
t is time
v is final velocity
using the info in the q, we know
u = 3
a = 1.5
t = 7
sub these numbers into the equation to find v - final velocity
v = u + at
v = 3 + (1.5×7)
v = 13.5 m/s