the probability that the mean weight of these 400 bags is less than 14.6 ounces is practically zero.
Given that weight of corn chips dispensed into a 14-ounce bag follows a normal distribution with mean (μ) = 14.5 ounces and standard deviation (σ) = 0.1 ounce.
We need to find the probability that the mean weight of these 400 bags is less than 14.6 ounces.
Since the sample size (n) is large (n > 30), we can use the central limit theorem, which states that the sample mean follows a normal distribution with a mean of the population mean (μ) and a standard deviation of the population standard deviation divided by the square root of the sample size (σ/√n).
So, the mean weight of 400 bags of chips follows a normal distribution with mean μ = 14.5 ounces and standard deviation σ/√n = 0.1/√400 = 0.005 ounce.
Let X be the weight of corn chips dispensed into a single bag. Then, we need to find the probability P(x(bar )< 14.6), where x(bar ) is the sample mean weight of 400 bags of chips.
Using the standard normal distribution, we can standardize the sample mean as:
Z = (x(bar ) - μ) / (σ/√n)
Z = (14.6 - 14.5) / (0.005)
Z = 20
Now, we need to find the probability that Z is less than 20, which is practically zero. Therefore, the probability that the mean weight of these 400 bags is less than 14.6 ounces is almost zero.
In symbols, P(x(bar ) < 14.6) ≈ 0.
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A livestock company reports that the mean weight of a group of young steers is 1146 pounds with a standard deviation of 86 pounds. Based on the model ​N(1146​,86​) for the weights of​ steers, what percent of steers weigh a) over 1200 ​pounds? ​b) under 1100 ​pounds? ​c) between 1250 and 1300 ​pounds?
a) 26.43% of steers weigh over 1200 pounds.
b) 29.46% of steers weigh under 1100 pounds.
c) 7.26% of steers weigh between 1250 and 1300 pounds.
The proportion of steers that weigh over 1200 pounds the area to the right of 1200 under the normal curve with mean 1146 and standard deviation 86.
A z-score and the standard normal distribution to find this area.
The z-score is:
z = (1200 - 1146) / 86 = 0.63
A standard normal distribution table or calculator the area to the right of z = 0.63 is 0.2643.
The proportion of steers that weigh under 1100 pounds, the area to the left of 1100 under the normal curve with mean 1146 and standard deviation 86.
Again, we can use a z-score and the standard normal distribution to find this area.
The z-score is:
z = (1100 - 1146) / 86 = -0.54
A standard normal distribution table or calculator the area to the left of z = -0.54 is 0.2946.
The proportion of steers that weigh between 1250 and 1300 pounds The area between the z-scores corresponding to these weights.
The z-score for 1250 pounds is:
z1 = (1250 - 1146) / 86 = 1.23
The z-score for 1300 pounds is:
z2 = (1300 - 1146) / 86 = 1.79
A standard normal distribution table or calculator the area to the left of z1 is 0.8907, and the area to the left of z2 is 0.9633.
The area between z1 and z2 is:
0.9633 - 0.8907 = 0.0726
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Question 2 of 10
Which situation is most likely to have a constant rate of change?
OA. Number of flowers in a flower bed compared with the area planted
B. The total amount paid for gas compared with the number of
gallons purchased.
C. Distance a delivery truck travels compared with the number of
deliveries made
D. Points scored in a basketball game compared with the number of
quarters played
Answer: The situation that is most likely to have a constant rate of change is option B: "The total amount paid for gas compared with the number of gallons purchased."
This is because the price of gas per gallon is usually constant, so the rate of change of the total amount paid for gas should be constant with respect to the number of gallons purchased. In other words, if you plot the total amount paid for gas against the number of gallons purchased, you would expect a straight line with a constant slope.
In contrast, the number of flowers in a flower bed compared with the area planted (option A), the distance a delivery truck travels compared with the number of deliveries made (option C), and points scored in a basketball game compared with the number of quarters played (option D) are less likely to have a constant rate of change because they can be affected by various factors such as weather, traffic, player performance, and so on.
Answer:
B.
Step-by-step explanation:
suppose a packaging system fills boxes such that the weights are normally distributed with a mean of 16.3 ounces and a standard deviation of 0.21 ounces. what is the probability that a box weighs between 16.4 and 16.5 ounces? report your answer to 2 decimal places.
The probability that a box weighs between 16.4 and 16.5 ounces is approximately 14.45% (rounded to 2 decimal places). To solve this problem, we need to use the z-score formula:
z = (x - μ) / σ
where x is the weight of the box, μ is the mean weight of all boxes, σ is the standard deviation of weights, and z is the number of standard deviations away from the mean.
In this case, we want to find the probability that a box weighs between 16.4 and 16.5 ounces. We can convert these weights to z-scores as follows:
z1 = (16.4 - 16.3) / 0.21 = 0.48
z2 = (16.5 - 16.3) / 0.21 = 0.95
Using a z-score table or calculator, we can find the area under the standard normal curve between these two z-scores:
P(0.48 ≤ z ≤ 0.95) = 0.1736
Therefore, the probability that a box weighs between 16.4 and 16.5 ounces is 0.17 or 17% (rounded to 2 decimal places).
Hi! To find the probability that a box weighs between 16.4 and 16.5 ounces, we can use the z-score formula and the standard normal table.
First, let's calculate the z-scores for 16.4 and 16.5 ounces using the formula: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.
For 16.4 ounces:
z1 = (16.4 - 16.3) / 0.21 ≈ 0.48
For 16.5 ounces:
z2 = (16.5 - 16.3) / 0.21 ≈ 0.95
Now, use the standard normal table to find the area between these z-scores:
P(0.48 < z < 0.95) = P(z < 0.95) - P(z < 0.48) ≈ 0.8289 - 0.6844 = 0.1445
The probability that a box weighs between 16.4 and 16.5 ounces is approximately 14.45% (rounded to 2 decimal places).
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If 2x + 4 = 36 what is the value of X
Answer:
x = 16
Step-by-step explanation:
to solve for x, you need to isolate x on one side of the equation and move everything else to the other side. You can do this by using inverse operations, such as subtraction, addition, multiplication, and division. Here are the steps to follow:
Start with the original equation: 2x + 4 = 36Subtract 4 from both sides to eliminate the constant term on the left side: 2x + 4 - 4 = 36 - 4Simplify both sides: 2x = 32Divide both sides by 2 to eliminate the coefficient of x on the left side: 2x / 2 = 32 / 2Simplify both sides: x = 16So the value of x is 16.
Wanda’s Widgets used market surveys and linear regression to develop a demand function based on the wholesale price. The demand function is q = –140p + 9,000. The expense function is E = 2.00q + 16,000. At a price of $10.00, how many widgets are demanded?
With the help of demand function, when the wholesale price is $10.00, Wanda's Widgets will demand 7,600 widgets.
What is function?
n mathematics, a function is a rule that assigns a unique output value to every input value in a specified set. In other words, it is a relationship between two sets of values, where each input value in the first set is associated with a unique output value in the second set.
The demand function is given by q = –140p + 9,000, where q is the quantity demanded and p is the wholesale price.
To find the quantity demanded when the price is $10.00, we can substitute p = 10 in the demand function and solve for q:
q = –140(10) + 9,000
q = –1,400 + 9,000
q = 7,600
Therefore, when the wholesale price is $10.00, Wanda's Widgets will demand 7,600 widgets.
Note that the expense function E = 2.00q + 16,000 is not used to find the quantity demanded in this problem. It is used to calculate the total expenses based on the quantity demanded.
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Data were collected from a random sample of 390 home sales from a community in 2003. Let Price denote the selling price (in $1,000), BDR denote the number of bedrooms, Bath denote the number of bathrooms, Hsize denote the size of the house (in square feet), Lsize denote the lot size (in square feet), Age denote the age of the house (in years), and Poor denote a binary variable that is equal to 1 if the condition of the house is reported as "poor." An estimated regression yields
Price = 120.4 + 0.490 * BDR + 23.6 * Bath + 0.158 * Hsize + 0.004 * Lsize + error
Suppose that a homeowner adds a new bathroom to her house, which increases the size of the house by 101 square feet. What is the expected increase in the value of the house?
The expected increase in the value of the house is
Approximately 23.6 thousand dollars
Approximately 15.8 thousand dollars
Approximately 39.6 thousand dollars
none of the above
The expected increase in the value of the house is approximately $39.6 thousand dollars.
The expected increase in the value of the house, resulting from adding a new bathroom and increasing the size of the house by 101 square feet, is approximately $39.6 thousand dollars.
This estimate is based on the given regression equation, where the coefficient for the number of bathrooms is 23.6 and the coefficient for the house size is 0.158. These coefficients indicate the expected change in the selling price associated with a one-unit increase in the respective variable.
Therefore, by multiplying the coefficient for bathrooms by the increase in bathrooms (23.6 * 1) and the coefficient for house size by the increase in size (0.158 * 101), we can estimate the expected increase in the value of the house.
To calculate the expected increase, we need to consider the coefficients associated with the bathroom variable and the house size variable in the regression equation. The coefficient for bathrooms is 23.6, indicating that for every additional bathroom, the selling price is expected to increase by $23.6 thousand. In this case, the homeowner added one bathroom, so the expected increase due to the additional bathroom is 23.6 * 1 = $23.6 thousand.
Similarly, the coefficient for the house size variable is 0.158, indicating that for every additional square foot of house size, the selling price is expected to increase by $0.158 thousand (or $158). Since the homeowner increased the house size by 101 square feet, the expected increase due to the increase in size is 0.158 * 101 = $15.958 thousand (approximately $15.8 thousand).
To find the total expected increase, we add the expected increases due to the additional bathroom and the increase in house size: $23.6 thousand + $15.8 thousand = $39.6 thousand (approximately). Therefore, the expected increase in the value of the house is approximately $39.6 thousand dollars.
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we desire the residuals in our model to have which probability distribution? select answer from the options below normal binomial poisson
The distribution that the residuals in our model to follow is equals to the normal probability distribution. So, option(a).
Because residuals are defined as the difference between any data point and the regression line, they are sometimes called "errors". An error in this context does not mean that there is anything wrong with the analysis. In other words, the residual is the error that is not described by the regression line. The residue(s) can also be expressed by "e". The formula is written as, Residual = Observed value – predicted value or
[tex]e = y – \hat y [/tex].
In order to draw valid conclusions from your regression, the regression residuals should follow a normal distribution. The residuals are simply the error terms or differences between the observed value of the dependent variable and the predicted value. Therefore, the residuals should have a normal distribution.
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Complete question:
we desire the residuals in our model to have which probability distribution? select answer from the options below
a) normal
b) binomial
c) poisson
Question 3 (1 point) The table shows y as a function of x. Suppose a point is added to this table. Which choice gives a point that preserves the function? a (9, −5) b (−1, −5) c (−8, −6) d (−5, 7)
If a point is added in the table, then the point which preserves the function is (d) (-5, 7).
The relation given in the table is a function, which means that every value of "x" in the domain must have exactly one corresponding value of "y" in the range.
The inputs , x = 9, x = -8, and x = -1 already have defined values in the table, so any other value assigned to these inputs would create a situation where an input has more than one output.
So, the only choice that would preserve the function is (d) (-5, 7), which assigns a "new-value" to an input that doesn't have a defined value in the table.
This new input-output pair is consistent with the existing function rule and ensures that every input in the domain has exactly one output in the range, preserving the function.
Therefore, the correct option is (d).
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The given question is incomplete, the complete question is
The table shows y as a function of x. Suppose a point is added to this table.
x y
6 -9
-8 9
-1 -4
9 -6
8 -8
Which choice gives a point that preserves the function?
(a) (9, -5)
(b) (-1, -5)
(c) (-8, -6)
(d) (-5, 7)
Find the value of M.
Side question: How do I make somebody brainlist or whatever?
m = 133°
Step-by-step explanation:
You want the value of m in the given polygon.
HeptagonThe sum of interior angles in a heptagon is (7 -2)(180°) = 900°.
This fact is used to find the value of m:
138 +106 +(m -9) +m + 133 +120 +(m +13) = 900
3m = 399 . . . . . . . subtract 501
m = 133 . . . . . . . . divide by 3
The value of m is 133°.
__
Additional comments
We suspect your answer will be just the numerical value.
An n-sided polygon has a sum of angles equal to (n -2)(180°).
You can award Brainliest when the "crown" shows up. In apps where the "crown" shows, you can click on the crown for the answer you choose. You will need to wait 24 hours unless there is a second answer to the question. (The Brainliest symbol does not appear on all platforms, so you may not see it at all.)
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Suppose we want to assess the effect of a one-day SAT prep class at a 5% level of significance. Scores on the SAT writing exam can range from 200 to 800. A random sample of 50 students takes the SAT writing test before and after a prep class. We test the hypotheses: LaTeX: H_0 H 0 : LaTeX: \mu=0 μ = 0 LaTeX: H_a H a : LaTeX: \mu>0 μ > 0 where LaTeX: \mu μ is the mean of the difference in SAT writing scores (after minus before) for all students who take the SAT prep class. The sample mean is 5 with a standard deviation of 18. Since the sample size is large, we are able to conduct the T-Test. The T-test statistic is approximately 1.96 with a P-value of approximately 0.028. What can we conclude?
The SAT prep class has no influence on the mean difference in SAT writing scores, hence the null-hypothesis (H0) states that the mean difference is zero.
The alternative theory (Ha) states that the SAT prep course has a positive impact on the mean difference in SAT writing scores, resulting in a mean difference that is greater than zero.
The sample size of 50 is sufficient for us to do the hypothesis test using the t-distribution.
The estimated t-test statistic is 1.96, and at the 5% level of significance, it is significant only if it is in the rejection zone of the null hypothesis (1.677 is the crucial value for a one-tailed test with 49 degrees of freedom).
The calculated p-value of 0.028 is less than the threshold of 0.05, the null hypothesis is also rejected in favour of the alternative hypothesis.
To draw the conclusion that the SAT prep course has a favourable impact on the mean difference in SAT writing scores. Particularly, at the 5% level of significance, the sample-mean difference of 5 is statistically significantly greater than zero.
Therefore, it is reasonable.
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A bee colony produced 0.7 pounds of honey, but bears ate 0.2 pounds of it. How much honey remains?
The amount of honey remain left after bears ate 0.2 pounds of honey is equal to 0.5 pounds.
Amount of Honey produced by bee colony is equal to 0.7 pounds
Amount of honey consumed by bears is equal to 0.2 pounds
let 'x' be the amount of honey that remain left after bears consumed some amount of honey.
If a bee colony produced 0.7 pounds of honey, and bears ate 0.2 pounds of it,
Then the amount of honey that remains is represented by an equation,
x + 0.2pounds = 0.7 pounds
This implies,
⇒ x = 0.7 - 0.2
⇒ x = 0.5 pounds
Therefore, there are 0.5 pounds of honey remaining after the bears ate 0.2 pounds of it.
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Two schedules for giving rest were compared--the massed schedule and the spaced schedule. Twenty observations of the spaced schedule produced a mean of 26 errors. On the massed schedule 14 observations resulted in a mean of 36 errors. An a level of .05 was adopted and an F = 4.21 was obtained. What conclusion is appropriate?
Based on the given information, we can conclude that the spaced schedule for giving rest is more effective in reducing errors compared to the massed schedule.
This is supported by the mean of 26 errors in the spaced schedule, which is lower than the mean of 36 errors in the massed schedule. Additionally, the obtained F value of 4.21 is greater than the critical F value at the 0.05 level of significance, indicating that there is a significant difference between the two schedules. Therefore, we reject the null hypothesis and accept the alternative hypothesis that the spaced schedule is more effective in reducing errors.
Based on the given information, you conducted a study comparing two rest schedules: massed schedule and spaced schedule. You obtained the following results:
- Spaced schedule: 20 observations, mean of 26 errors
- Massed schedule: 14 observations, mean of 36 errors
You performed an F-test with an alpha level of 0.05 and obtained an F-value of 4.21. To determine the appropriate conclusion, you would need to compare the F-value with the critical F-value for the given degrees of freedom and alpha level. Unfortunately, the critical F-value is not provided in your question.
However, if your obtained F-value (4.21) is greater than the critical F-value at α = 0.05, then you would reject the null hypothesis and conclude that there is a significant difference between the massed and spaced rest schedules in terms of the number of errors made. If the obtained F-value is smaller than the critical F-value, then you would fail to reject the null hypothesis and not conclude a significant difference between the two schedules.
Please check the critical F-value for your specific test and degrees of freedom, and compare it to your obtained F-value (4.21) to draw an appropriate conclusion.
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The maximum amounts of lead and copper allowed in drinking water are 0. 015 mg/kg for lead and 1. 3 mg/kg for copper. Express these values in parts per million.
The answer is that the maximum amount of lead allowed in drinking water is 0.015 mg/kg and the maximum amount of copper allowed is 1.3 mg/kg.
To express these values in parts per million (ppm), we need to convert the mass of the substance to the mass of the water.
To convert mg/kg to ppm, we need to multiply by 1,000,000 (1 million) and divide by the density of the water. The density of water is 1 gram per milliliter (g/mL), which is equivalent to 1,000,000 mg/L.
For lead:
0.015 mg/kg x 1,000,000 / 1,000,000 mg/L = 15 ppb (parts per billion)
For copper:
1.3 mg/kg x 1,000,000 / 1,000,000 mg/L = 1,300 ppb
Therefore, the maximum allowed levels of lead and copper in drinking water are 15 ppb and 1,300 ppb, respectively.
The maximum amounts of lead and copper allowed in drinking water, when expressed in parts per million (ppm), are 15 ppm for lead and 1,300 ppm for copper.
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Consider a set of data in which the sample mean is 33.7 and the sample standard deviation is 7.2. Calculate the z-score given that x = 30.2. Round your answer to two decimal places
The z-score for x = 30.2 is approximately -0.49.
What is z-score measures?
The z-score, also known as the standard score, is a measure used in statistics to quantify the number of standard deviations that a given data point is from the mean of a dataset.
To calculate the z-score for x = 30.2, we use the formula:
z = (x - μ) / σ
where x is the observed value, μ is the population mean, and σ is the population standard deviation. In this case, we are given the sample mean and sample standard deviation, so we will use them as estimates for the population parameters.
Substituting the given values, we have:
z = (30.2 - 33.7) / 7.2
Simplifying, we get:
z = -0.49
Rounding to two decimal places, we have:
z ≈ -0.49
Therefore, the z-score for x = 30.2 is approximately -0.49.
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The distribution of this approximate sampling distribution will be closer to approximately normal than the distribution of the population due to the Central Limit Theorem, will have the same mean as the distribution of the population ($150), and the standard deviation will be \($50/\sqrt{25}=$10\).
The standard deviation of the sampling distribution of the sample means will be equal to the standard deviation of the population divided by the square root of the sample size. Therefore, if the population standard deviation is 50 and the sample size is 25, then the standard deviation of the sampling distribution of the sample means will be 10.
The Central Limit Theorem (CLT) states that the sampling distribution of the sample means will approach a normal distribution as the sample size increases, regardless of the shape of the population distribution. This means that even if the population distribution is not normal, the distribution of the sample means will still be approximately normal as long as the sample size is sufficiently large (usually, a sample size greater than or equal to 30 is considered large enough).
Additionally, according to the CLT, the mean of the sampling distribution of the sample means will be equal to the mean of the population from which the samples are drawn. In this case, since the population mean is 150, the mean of the sampling distribution of the sample means will also be 150.
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there exists a continuous function defined for all real numbers that is concave up and always negative.T/F
it is not possible for a continuous function to be concave up and always negative.
To see why, note that a concave up function is one whose second derivative is positive. So we need to find a function whose second derivative is positive and is always negative.
However, if a function is always negative, then its values are always less than or equal to zero. This means that its second derivative must be non-positive, since the second derivative measures the rate at which the function's slope is changing.
Now suppose that we have a function f(x) that is concave up and always negative. Since f(x) is always negative, we have f(x) < 0 for all x. But since f(x) is concave up, its second derivative f''(x) is positive. This means that f(x) is increasing, and in particular, as x goes to infinity, f(x) must approach a limit. But since f(x) is always negative, its limit as x goes to infinity must be nonpositive. This is a contradiction, since a concave up function that is always negative cannot have a nonpositive limit at infinity.
Therefore, it is not possible for a continuous function to be concave up and always negative.
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What is the surface area of this right rectangular prism?
Enter your answer as a mixed number in simplest form by filling in the boxes.
yd²
Rectangular prism with length 3 yards, width 1 and 1 third yard, and height 2 and 2 thirds yards.
Answer:
31 and 4 ninths square yards.
Step-by-step explanation:
The surface area of a right rectangular prism is the amount of wrapping paper you need to cover it up. To find it, you need to measure the length, width, and height of the prism. Then you can use this magic spell:
Surface area = 2 (lw + wh + lh) square units
where l is the length, w is the width, and h is the height.
For example, you have a right rectangular prism with length 3 yards, width 1 and 1 third yard, and height 2 and 2 thirds yards. That's a big gift! Using the magic spell, we get:
Surface area = 2 ((3 x 1 1/3) + (1 1/3 x 2 2/3) + (3 x 2 2/3)) yd² Surface area = 2 ((4 + 3 7/9 + 8)) yd² Surface area = 2 (15 7/9) yd² Surface area = 31 4/9 yd²
So the surface area of the right rectangular prism is 31 and 4 ninths square yards. That's a lot of wrapping paper! I hope you have enough tape!
two codominant alleles, lm and ln, determine the human mn blood type. suppose that the lm allele occurs with a frequency of 0.80 in a population of eskimos on a small arctic island. match the expected frequencies to the m, mn, and n blood types in the population on the island for two mating scenarios: if random mating occurs, and if the inbreeding coefficient for this population is 0.05.
The expected frequencies of the M, MN, and N blood types are then:
The frequency of the M blood type is f(AA) = 0.632.
The frequency of the MN blood type is f(Aa) = 0.304.
The frequency of the N blood type is f(aa) = 0.064.
What is the frequency?
The number of periods or cycles per second is called frequency. The SI unit for frequency is the hertz (Hz). One hertz is the same as one cycle per second.
We can use the Hardy-Weinberg equilibrium to calculate the expected frequencies of the M, MN, and N blood types in the population of Eskimos on the Arctic island.
The Hardy-Weinberg equilibrium is a principle that states that the frequencies of alleles and genotypes in a population remain constant from generation to generation in the absence of evolutionary factors (mutation, migration, genetic drift, selection).
Let p be the frequency of the LM allele, and q be the frequency of the LN allele in the population. Since there are only two alleles, p + q = 1.
The frequency of the MM genotype is p², the frequency of the MN genotype is 2pq, and the frequency of the NN genotype is q².
The frequencies of the M, MN, and N blood types can be calculated from the frequencies of the genotypes:
The frequency of the M blood type is p².
The frequency of the MN blood type is 2pq.
The frequency of the N blood type is q².
Given that the frequency of the LM allele is 0.80, we have p = 0.80 and q = 0.20.
If random mating occurs, the expected frequencies of the M, MN, and N blood types are:
The frequency of the M blood type is p² = (0.80)² = 0.64.
The frequency of the MN blood type is 2pq = 2 x 0.80 x 0.20 = 0.32.
The frequency of the N blood type is q² = (0.20)² = 0.04.
If the inbreeding coefficient for this population is 0.05, we can use the following equation to calculate the expected frequencies of the genotypes:
f(AA) = (1 - F) p² + F p,
f(Aa) = (1 - F) 2pq,
f(aa) = (1 - F) q² + F q,
where F is the inbreeding coefficient.
Substituting p = 0.80, q = 0.20, and F = 0.05, we obtain:
The frequency of the MM genotype is f(AA) = (1 - 0.05) (0.80)² + 0.05 x 0.80 = 0.632.
The frequency of the MN genotype is f(Aa) = (1 - 0.05) 2 x 0.80 x 0.20 = 0.304.
The frequency of the NN genotype is f(aa) = (1 - 0.05) (0.20)² + 0.05 x 0.20 = 0.064.
Hence, The expected frequencies of the M, MN, and N blood types are then:
The frequency of the M blood type is f(AA) = 0.632.
The frequency of the MN blood type is f(Aa) = 0.304.
The frequency of the N blood type is f(aa) = 0.064.
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Billy is creating a rectangular patio in his backyard using square cement tiles. The length of the patio, in feet, is represented by the function I(x) = X + 5, and the width of the patio is represented by the function w(x) = X + 3.
Write the standard from of the function which describes the total area of the patio, a(x) in terms of x, the side length of each tile.
The area in terms of x, can be written as:
A(x) = x² + 8x + 15
How to find the equation for the area of the rectangle?Remember that the area of a rectangle is given by the product between the dimensions.
Here we know that the length is:
L(x) = x + 5
And the width is:
W(x) = x + 3
Then the formula for the area is:
A(x) = (x + 5)*(x + 3)
A(x) = x² + 5x + 3x + 15
A(x) = x² + 8x + 15
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The graph below describes the height (in feet) of a flare signal in terms of t, the time (in seconds) since the flare went off. Find the y-intercept. What is the correct interpretation of the y-intercept?
A-The flare signal had a maximum height of 163 feet.
B-The flare signal was set off at a height of 3 feet.
C-The flare signal had a maximum height of 3 feet.
D-The flare signal was set off at a height of 163 feet.
Answer:
B-The flare signal was set off at a height of 3 feet.
10. A car that is always traveling at the same speed
travels 30 miles every 0. 5 hours. How many miles
does it travel in 4. 5 hours?
The car will travel 270 miles in 4.5 hours.
How many miles does the car travel?To find out how many miles the car travels in 4.5 hours, we can use the formula of distance which is "distance = speed * time".
The car travels 30 miles every 0.5 hours. This means its speed is:
= distance / time
= 30 miles / 0.5 hours
= 60 miles per hour
In 4.5 hours, the car will travel (distance):
= Speed x Time
= 60 miles per hour x 4.5 hours
= 270 miles.
Full question "A car that is always traveling at the same speed travels 30 miles every 0. 5 hours. How many miles does it travel in 4. 5 hours?"
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What is the fundamental difference in the graphs of polynomial functions and rational functions.
Polynomial functions and rational functions are both types of functions that are commonly studied in mathematics. However, there are fundamental differences in the graphs of these two types of functions.
A polynomial function is a function of the form f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0, where n is a non-negative integer, and the a_i's are coefficients. The graph of a polynomial function is a smooth curve that can have any number of turns, but does not have any breaks or holes.
Polynomial functions can have degree 0 (a constant function), degree 1 (a linear function), degree 2 (a quadratic function), and so on.
On the other hand, a rational function is a function of the form f(x) = p(x)/q(x), where p(x) and q(x) are both polynomial functions. The graph of a rational function can have breaks or holes where the denominator is zero. The degree of the numerator and denominator can be the same, but it is not a requirement.
One fundamental difference in the graphs of polynomial functions and rational functions is that polynomial functions have a defined end behavior, while rational functions do not. The end behavior of a polynomial function depends on the degree and leading coefficient of the function. Rational functions, however, can approach vertical asymptotes as x approaches certain values, making the end behavior undefined.
Another difference is that the domain of a polynomial function is all real numbers, while the domain of a rational function excludes any value of x that makes the denominator zero. This means that the domain of a rational function can have "holes" in the graph where the function is undefined.
In summary, polynomial functions and rational functions are both important types of functions in mathematics, but they have fundamental differences in their graphs. Polynomial functions have a smooth curve and defined end behavior, while rational functions can have breaks and holes in the graph and undefined end behavior.
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what is the probability of winning a state lottery game where the winning number is made up of four digits from 0 to 9 chosen at random?
The probability of winning this lottery game is 1/10,000 or 0.0001 (0.01% chance). The probability of winning a state lottery game where the winning number is made up of four digits from 0 to 9 chosen at random can be calculated as follows.
First, we need to determine the total number of possible outcomes. There are 10 digits (0 to 9) and we are choosing four of them, so the total number of possible outcomes is 10 x 10 x 10 x 10 = 10,000.
Next, we need to determine the number of favorable outcomes, which is the number of ways to choose four digits from 0 to 9. This is a combination problem, and we can use the formula nCr = n! / r!(n-r)! where n is the total number of options and r is the number of choices. So in this case, n = 10 and r = 4, giving us 10C4 = 10! / 4!(10-4)! = 210 favorable outcomes.
Finally, we can calculate the probability of winning by dividing the number of favorable outcomes by the total number of outcomes:
Probability of winning = favorable outcomes / total outcomes
Probability of winning = 210 / 10,000
Probability of winning = 0.021 or 2.1%
So the probability of winning a state lottery game where the winning number is made up of four digits from 0 to 9 chosen at random is 0.021 or 2.1%.
Hi! The probability of winning a state lottery game with a four-digit winning number, where each digit ranges from 0 to 9, can be calculated as follows:
There are 10 choices (0 to 9) for each of the four digits. Thus, the total number of possible combinations is 10 x 10 x 10 x 10 = 10,000. Since there is only one winning number, the probability of selecting that number at random is 1 out of the total possible combinations.
So, the probability of winning this lottery game is 1/10,000 or 0.0001 (0.01% chance).
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We suspect that automobile insurance premiums (in dollars) may be steadily decreasing
with the driver's driving experience (in years), so we choose a random sample of drivers
who have similar automobile insurance coverage and collect data about their ages and
insurance premiums.
A. matched pairs t-test
B. two-sample t-test
C. ANOVA
D. chi-squared test for independence
E. inference for regression
A statistical technique called the chi-square test is used to compare actual outcomes with predictions. The goal of this test is to establish whether a discrepancy between observed and expected data is the result of chance or a correlation between the variables you are researching.
T tests and chi-square tests can both evaluate differences between two groups. However, a t test is utilised when there are two groups in a categorical variable and a dependent quantitative variable. In cases where there are two categorical variables, a chi-square test of independence is applied.A statistical technique called the chi-square test is used to compare actual outcomes to predictions.
Typically, it involves a contrast between two sets of statistical data. Karl Pearson developed this test in 1900 for the analysis and distribution of categorical data. As a result, Pearson's chi-squared test was cited.
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which of the following is true of relationships between variables?which of the following is true of relationships between variables?a negative relationship exists between two variables if low levels of one variable are associated with low levels of another.in a linear relationship between two variables, the strength and the direction of the relationship change over the range of both variables.a linear relationship is much simpler to work with than a curvilinear relationship.relationships between variables lack direction.the larger the size of the correlation coefficient between two variables, the weaker the association between them.
The statement that is true of relationships between variables is "Marketers are often interested in describing the relationship between variables they think influence purchases of their products." (option b).
In many fields, researchers and professionals seek to understand the relationships between different variables. A variable is any characteristic or feature that can vary and can be measured or observed. Understanding the relationship between variables can help in predicting, explaining, and controlling different phenomena. In this context, it's important to distinguish between different types of relationships and to use appropriate statistical methods to describe and test these relationships.
This statement is true. Marketers often want to understand the relationship between different variables and how they influence consumer behavior. For example, they might want to know how price, quality, brand reputation, and advertising affect the likelihood of a consumer purchasing their product. By understanding these relationships, marketers can develop more effective marketing strategies.
Hence the correct option is (b).
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Complete Question:
Which of the following is true of relationships between variables?
a) A curvilinear relationship is much simpler to work with than a linear relationship.
b) Marketers are often interested in describing the relationship between variables they think influence purchases of their products.
c) A negative relationship exists between two variables if low levels of one variable are associated with low levels of another.
d) The strength of association is determined by the size of the correlation coefficient, with smaller coefficients indicating a stronger association.
e) The null hypothesis for the Pearson correlation coefficient states that there is a strong association between two variables.
In robust optimization, a constraint that cannot be violated is known as a
a. optional constraint
b. soft constraint
c. hard constraint
In robust optimization, a constraint that cannot be violated is known as a hard constraint.
Hard constraints must be satisfied by any feasible solution to the optimization problem, while soft constraints are allowed to be violated, but at a cost. Optional constraints are constraints that can be included or excluded from the problem formulation depending on the specific needs of the application.
On the other hand, optional constraints or soft constraints are those that can be violated to some extent without significantly affecting the overall objective of the optimization problem. Soft constraints are used to express preferences or goals that are desirable but not strictly necessary for the problem to be solved.
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An airliner carries 50 passengers and has doors with a height of 70 in. Heights of men are normally distributed with a mean of 69. 0 in and a standard deviation of 2. 8 in. Complete parts (a) through (d). A. If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending. The probability is 0. 6406. (Round to four decimal places as needed. ) b. If half of the 50 passengers are men, find the probability that the mean height of the 25 men is less than 70 in. The probability is 0. 9633. (Round to four decimal places as needed. )
The probability is 0.6480.
The probability is 0.9629.
How to solve for the probability1. This can be computed using the standard normal distribution as follows:
z = (70 - 69.0) / 2.8 = 0.357
Using a standard normal table or calculator, we find that P(Z ≤ 0.357) ≈ 0.6480. Therefore, the probability that a male passenger can fit through the doorway without bending is approximately 0.6480.
2. = 2.8/√25 = 0.56 inches.
We want to find P(x < 70), which is the probability that the mean height of the 25 men is less than 70 inches. This can be standardized using the standard normal distribution as follows:
z = (70 - 69.0) / 0.56 = 1.79
Using a standard normal table or calculator, we find that P(Z < 1.79) ≈ 0.9629. Therefore, the probability that the mean height of the 25 men is less than 70 inches is approximately 0.9629.
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Suppose that a conservative 95% confidence interval for the proportion of first-year students at a school who played in intramural sports is 68% plus or minus 4%. Find the sample size, n, that was used to obtain this confidence interval.
The sample size used to obtain the 95% confidence interval for the proportion of first-year students at a school who played in intramural sports is approximately 360.
To find the sample size, n, we need to use the formula:
n = (z^2 * p * q) / E^2
Where:
z = the z-score corresponding to the desired confidence level (in this case, it is 1.96 for a 95% confidence level)
p = the proportion of first-year students who played in intramural sports (given as 0.68)
q = the complement of p (q = 1 - p)
E = the margin of error (given as 0.04)
Substituting the given values into the formula, we get:
n = (1.96^2 * 0.68 * 0.32) / 0.04^2
n = 360.15
Therefore, the sample size used to obtain the 95% confidence interval for the proportion of first-year students at a school who played in intramural sports is approximately 360.
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for a single randomly selected movie, find the probability that this movie's production cost is between 64 and 70 million dollars.
Probability of selecting a movie with a production cost between 64 and 70 million dollars is 0.1915 or 19.15%.
To find the probability that a single randomly selected movie's production cost is between 64 and 70 million dollars, we need to know the distribution of production costs for movies. Let's assume that the distribution is approximately normal.
We also need to know the mean and standard deviation of production costs. Let's assume that the mean production cost is 60 million dollars and the standard deviation is 10 million dollars.
Using these parameters, we can standardize the range of production costs we're interested in by subtracting the mean and dividing by the standard deviation:
z1 = (64 - 60) / 10 = 0.4
z2 = (70 - 60) / 10 = 1
We can then use a standard normal distribution table or calculator to find the area under the curve between these two standardized values:
P(0.4 ≤ Z ≤ 1) ≈ 0.1915
This means that the probability of selecting a movie with a production cost between 64 and 70 million dollars is approximately 0.1915 or 19.15%.
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Which r-value represents the most moderate correlation?.
The r-value that represents the most moderate correlation would be around 0.5. This value indicates a moderate positive correlation, meaning that there is a moderate relationship between two variables that are moving in the same direction.
An r-value, or correlation coefficient, represents the strength and direction of a linear relationship between two variables. The r-value ranges from -1 to 1, where:
-1 indicates a strong negative correlation,
0 indicates no correlation, and
1 indicates a strong positive correlation.
A moderate correlation falls in the middle of this range. For example, an r-value of approximately 0.5 (positive moderate correlation) or -0.5 (negative moderate correlation) would represent a moderate correlation between the two variables.
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