Answer:
Point A: (1,3)Point D: (-1,-3)Step-by-step explanation:
The value of y in the (x, y) pair must be 3 times the value of x if the point is to be on the line. That is the case for points A, D.
Andrei wants to fill a glass tank with marbles, and then fill the remaining space with water. WWW represents the volume of water Andrei uses (in liters) if he uses nnn marbles. W=32-0.05nW=32−0.05nW, equals, 32, minus, 0, point, 05, n What is the glass tank's volume?
Before Andrei adds the marbles to the glass tank, the glass tank was empty. This means that the volume of the empty tank is when n = 0 and the volume is 32 liters.
Given that:
[tex]W = 32 - 0.05n[/tex]
A linear function is represented as:
[tex]y = b + mx[/tex]
Where
[tex]b \to[/tex] y intercept
Literally, the y intercept is the initial value of the function.
In this function, the y intercept means the initial volume of the glass tank before filling it with marbles.
Compare [tex]y = b + mx[/tex] and [tex]W = 32 - 0.05n[/tex]
[tex]b = 32[/tex]
This means that the volume of the glass tank is 32 liters.
Read more about linear functions at:
https://brainly.com/question/21107621
Answer:
0.05
Step-by-step explanation:
e of Scores, a publication of the Educational Testing Service, the scores on the verbal portion of the GRE have mean 150 points and standard deviation 8.75 points. Assuming that these scores are (approximately) normally distributed, a. obtain and interpret the quartiles. b. find and interpret the 99th percentile.
Answer:
a) Q1= 144.10
Median = 150
Q3=155.90
b) The 99 percentile would be:[tex]a=150 +2.33*8.75=170.39[/tex]
And represent a value who accumulate 99% of the values below
Step-by-step explanation:
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(150,8.75)[/tex]
Where [tex]\mu=150[/tex] and [tex]\sigma=8.75[/tex]
Part a
Lets begin with the first quartile:
[tex]P(X>a)=0.75[/tex] (a)
[tex]P(X<a)=0.25[/tex] (b)
We can find the quantile in the normal standard distribution and we got z=-0.674.
And we can apply the z score formula and we got:
[tex]z=-0.674<\frac{a-150}{8.75}[/tex]
And if we solve for a we got
[tex]a=150 -0.674*8.75=144.10[/tex]
The median for this case is the mean [tex]Median =150[/tex]
For the third quartile we find the quantile who accumulate 0.75 of the area below and we got z=0.674 and we got:
[tex]a=150 +0.674*8.75=155.90[/tex]
Part b
We can find the quantile in the normal standard distribution who accumulate 0.99 of the area below and we got z=2.33.
And we can apply the z score formula and we got:
[tex]z=2.33<\frac{a-150}{8.75}[/tex]
And if we solve for a we got
[tex]a=150 +2.33*8.75=170.39[/tex]
And represent a value who accumulate 99% of the values below
Section Exercise 11-8 Sales of People magazine are compared over a 5-week period at four Borders outlets in Chicago. Weekly Sales Store 1 Store 2 Store 3 Store 4 102 97 89 100 106 77 91 116 105 82 75 87 115 80 106 102 112 101 94 100 Click here for the Excel Data File Fill in the missing data. (Round your p-value to 4 decimal places, mean values to 1 decimal place, and other answers to 2 decimal places.) Treatment Mean n Std. Dev Store 1 Store 2 Store 3 Store 4 Total One-Factor ANOVA Source SS df MS F p-value Treatment Error Total (a) Based on the given hypotheses choose the correct option. H0: μ1 = μ2 = μ3 = μ4 H1: Not all the means are equal α = 0.05 Reject the null hypothesis if F > 3.24 Reject the null hypothesis if F < 3.24 (b) Determine the value of F. (Round your answer to 2 decimal places.) F-value (c) On the basis of the above-determined values, choose the correct decision from below. Fail to reject the null hypothesis. Reject the null hypothesis. (d) Determine the p-value. (Round your answer to 4 decimal places.) p-value Next Visit question mapQuestion 1 of 2 Total1 of 2 Prev
Answer:
Step-by-step explanation:
Hello!
An ANOVA was conducted to analyze the variable
Y: sales of "People" magazine over a 5-week period
This was studied in 4 Borders outlets in Chicago.
So this test has one factor: "Borders outlets" and four treatments: "Store 1, store 2, store 3 and store 4"
For each store you have the data for the weekly sales over a 5-week period so the sample sizes are:
n₁=n₂=n₃=n₄= 5 weeks
Store 1
∑X₁= 540; ∑X₁²= 58434
X[bar]₁= 108
S₁²= 28.50
S₁= 5.34
Store 2
∑X₂= 437; ∑X₂²= 38663
X[bar]₂= 87.40
S₂²= 117.30
S₂= 10.83
Store 3
∑X₃= 455; ∑X₃²= 41899
X[bar]₃= 91
S₃²= 123.50
S₃= 11.11
Store 4
∑X₄=505; ∑X₄²= 51429
X[bar]₄= 101
S₄²= 106
S₄= 10.30
Totals
N= n₁ + n₂ + n₃ + n₄= 4*5= 20
∑Mean= 108+87.40+91+101= 387.4
∑Variance= 28.50+117.30+123.50+106= 375.30
∑Standard deviation= 5.34+10.83+11.11+10.30= 37.58
Hypothesis test:
H₀: μ₁= μ₂= μ₃= μ₄
H₁: At least one population mean is different.
α: 0.05
The statistic for this test is
[tex]F= \frac{MS_{Treatments}}{MS_{Error}} ~~F_{k-1;N-k}[/tex]
k-1= 3 Df of the treatments, k=4 number of treatments
N-k= 16 Df of errors, N=20 total number of observations in all treatments
[tex]F_{H_0}= \frac{441.78}{93.83}= 4.71[/tex]
The critical region and p-value for this test are one-tailed to the right.
Using the critical value approach:
[tex]F_{k-1;N-k;1-\alpha }= F_{3;16;0.95}= 3.25[/tex]
The decision rule is:
If [tex]F_{H_0}[/tex] ≥ 3.25, reject the null hypothesis.
If [tex]F_{H_0}[/tex] < 3.25, do not reject the null hypothesis.
Using the p-value approach:
Little reminder: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
So under the distribution F₃,₁₆ you have to calculate the probability of the calculated [tex]F_{H_0}[/tex]:
P(F₃,₁₆≥4.71)= 1 - P(F₃,₁₆<4.71)= 1 - 0.9847 = 0.0153
p-value= 0.0153
The decision rule for this approach is
If p-value ≤ α, reject the null hypothesis.
If p-value > α, do not reject the null hypothesis.
The p-value is less than the level of significance, so the decision is to reject the null hypothesis.
I hope this helps!
Choose the function that is a "parent function".
Answer:
f(x)=√x
Step-by-step explanation:
A parent function is something simple with just x
Answer:
f(x)=Square Root of x (choice B or the second one down)
Step-by-step explanation:
All the other choices have either a plus 3 or minus 3. A parent function is not going to have any type of number being added or subtracted to it.
brenna goes on a cave tour with her family.she spots a mysterious crystal that is shaped like a cube the crystal has edge lengths of 5 centimeters what is the volume of the crystal
Answer:
The volume of the crystal is [tex]V=125 \:cm^3[/tex].
Step-by-step explanation:
The volume enclosed by a cube is the number of cubic units that will exactly fill a cube.
To find the volume of a cube recall that a cube has all edges the same length. The volume of a cube is found by multiplying the length of any edge by itself three times. Or as a formula
[tex]V=s^3[/tex]
where:
s is the length of any edge of the cube.
From the information given we know that the crystal has edge lengths of 5 centimeters. Therefore, the volume of the crystal is
[tex]V=5^3=125 \:cm^3[/tex].
Assume that military aircraft use ejection seats designed for men weighing between 133.8 lb and 208.0 lb. If women’s weights are normally distributed with a mean of 172.6 lb and a standard deviation of 42.4 lb, what percentage of women have weights between the ejection seat’s weight limits (that is, 133.8 to 208.0 lb)? Enter your answer as a percent rounded to one decimal place (do not add a "%"); add a trailing zeros as needed. The percentage of women with weights between 133.8 and 208.0 lb is [EjectPct] percent.
Answer:
61.8
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 172.6, \sigma = 42.4[/tex]
What percentage of women have weights between the ejection seat’s weight limits (that is, 133.8 to 208.0 lb)?
We have to find the pvalue of Z when X = 208 subtracted by the pvalue of Z when X = 133.8 for the proportion. Then we multiply by 100 to find the percentage.
X = 208
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{208 - 172.6}{42.4}[/tex]
[tex]Z = 0.835[/tex]
[tex]Z = 0.835[/tex] has a pvalue of 0.798
X = 133.8
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{133.8 - 172.6}{42.4}[/tex]
[tex]Z = -0.915[/tex]
[tex]Z = -0.915[/tex] has a pvalue of 0.180
0.798 - 0.18 = 0.618
0.618*100 = 61.8%
Without the %, the answer is 61.8.
An extremely simple (and surely unreliable) weather prediction model would be one where days are of two types: sunny or rainy. A sunny day is 90% likely to be followed by another sunny day, and a rainy day is 50% likely to be followed by another rainy day. Model this as a Markov chain. If Sunday is sunny, what is the probability that Tuesday (two days later) is also sunny
Answer:
The probability that if Sunday is sunny, then Tuesday is also sunny is 0.86.
Step-by-step explanation:
Let us denote the events as follows:
Event 1: a sunny day
Event 2: a rainy day
From the provided data we know that the transition probability matrix is:
[tex]\left\begin{array}{ccc}1&\ \ \ \ 2\end{array}\right[/tex]
[tex]\text{P}=\left\begin{array}{c}1&2\end{array}\right[/tex] [tex]\left[\begin{array}{cc}0.90&0.10\\0.50&0.50\end{array}\right][/tex]
In this case we need to compute that if Sunday is sunny, what is the probability that Tuesday is also sunny.
This implies that we need to compute the value of P₁₁².
Compute the value of P² as follows:
[tex]P^{2}=P\cdot P[/tex]
[tex]=\left[\begin{array}{cc}0.90&0.10\\0.50&0.50\end{array}\right]\cdot \left[\begin{array}{cc}0.90&0.10\\0.50&0.50\end{array}\right]\\\\=\left[\begin{array}{cc}0.86&0.14\\0.70&0.30\end{array}\right][/tex]
The value of P₁₁² is 0.86.
Thus, the probability that if Sunday is sunny, then Tuesday is also sunny is 0.86.
Please help! Will mark brainliest ! Thank you! Please explain so I can actually understand the question too :)
Answer:
A
Step-by-step explanation:
They are congruent because of the SSS theorem. The chords are congruent because the angles are congruent (angles are congruent because they are vertical angles). Congruent central angles have congruent chords. The other two sides are congruent because they are all radii of the circle and radius are always congruent.
Answer:
A
Step-by-step explanation:
The triangles are isosceles and congruent too. ( both have 2 sides congruent as radius and the angles between them are congruent- SAS)
Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by y 1 3 ex2 /3 , y 0, x 0, and x 3 about the y-axis. Round your answer to three decimal places.
Answer:
Step-by-step explanation:
[tex]y = f(x) =\frac{1}{\sqrt{3 \pi} } e^{-x^{2/3}}[/tex]
y = 0, x = 0 and x = 3
Consider an element of thickness dx at a distance x from the origin. By Cylindirical Shell Method, the volume of the element is given by
[tex]dV=(2\pi rdr)h=(2\pi xdx)f(x) => dV=(2\pi xdx) \frac{1}{\sqrt{3\pi}}e^{-x^{\frac{2}{3}}}[/tex]
[tex]dV=2\sqrt{\frac{\pi}{3}}xe^{-x^{\frac{2}{3}}}dx[/tex]
Integrate the above integral over the limits x=0 to x=3 which implies
[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}xe^{-x^{\frac{2}{3}}}dx[/tex]
Solve by subsititution
[tex]Let,\\ -x^{\frac{2}{3}}=y => \frac{-2}{3}x^{\frac{-1}{3}}dx=dy => x^{\frac{-1}{3}}dx=\frac{-3}{2}dy[/tex]
Also, apply the new limits
[tex]At,\\\\ x=0, y=0 \ and \ At, x=3, y=-\sqrt[3]{9}[/tex]
This implies,
[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}x^{\frac{4}{3}}e^{-x^{\frac{2}{3}}}x^{\frac{-1}{3}}dx=2\sqrt{\frac{\pi}{3}}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}(\frac{-3}{2})dy[/tex]
[tex]V=-\sqrt{3\pi}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]
Let,
[tex]I=\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]
Integrate by parts the above integral
[tex]u=y^2 \ and \ dv=e^ydy => du=2y \ and \ v=e^y[/tex]
Integrate by parts formula
[tex]\int udv=uv-\int vdu => y^2e^y-\int 2ye^ydy[/tex]
Again integrate by parts
[tex]u=y \ and \ dv=e^ydy => du=1 \ and \ v=e^y[/tex]
Integrate by parts formula
[tex]\int udv=uv-\int vdu => y^2e^y-2[ye^y-e^y]=e^y[y^2-2y+2][/tex]
Therefore,
[tex]I=[e^y(y^2-2y+2)]_{0}^{-\sqrt[3]{9}}\\\\=e^{-2.0802}[(2.0802)^2+2(2.0802)+2]-e^{0}[0-0+2]\\\\\frac{(4.3272+4.1604+2)}{8.0061}-2\\\\=\frac{10.4876}{8.0061}-2\\\\=1.3099-2\\\\=-0.6901[/tex]
This implies, the volume is
[tex]V=-\sqrt{3\pi}I\\\\=-\sqrt{3\times 3.142} \times (-0.6901)\\\\=3.0701 \times 0.6901\\\\=2.1186[/tex]
That is, up to three decimal places
[tex]V\approx 2.118[/tex]
.
A students received a score of 50 on his history test. The test had a mean of 69 and a standard deviation of 10. Find the z score and assess whether his score is considered unusual.
1.90; unusual
–1.90; not unusual
–1.90; unusual
1.90; not unusual
Answer:
c) The Z-score = - 1.90 unusual
Step-by-step explanation:
Explanation:-
Let 'X' be the random variable in normal distribution
Given student received a score X = 50
Mean of the Population x⁻ = 69
standard deviation of the Population 'σ' = 10
now
[tex]Z = \frac{x^{-}-mean }{S.D}[/tex]
[tex]Z = \frac{x^{-}-mean }{S.D} = \frac{50 -69}{10} = - 1.90[/tex]
The Z-score = - 1.90
Conclusion:-
The Z-score = - 1.90 unusual
Answer: The Z-score = - 1.90 unusual
Step-by-step explanation:
The bottom of a ladder must be placed 3 ft. from a wall. The ladder is 12 feet long. How far above the ground does the ladder touch the wall? Round your answer to the nearest tenth.
Use the Pythagorean theorem to solve.
Height = sqrt(12^2 -3^2)
Height = sqrt(144-9)
Height = sqrt(135)
Height = 11.6189 = 11.6 feet
Find the length of both of the unknown sides in the triangle shown here.
Give your answer correct to the nearest metre. [5 marks]
Answer:
[tex] (x+11)^2 = (x+3)^2 +16^2[/tex]
And if we solve this equation for x we got:
[tex] x^2 +22x +121 = x^2 +6x +9 +256[/tex]
We can cancel [tex]x^2[/tex] in both sides and we have this:
[tex] 22x -6x= 256+9-121 =144[/tex]
And then we got:
[tex] 16 x= 144[/tex]
[tex] x =\frac{144}{16}= 9[/tex]
And then the length of the sides are 9+11= 20 m for the hypothenuse, 16 for the adjacent side and 9+3 = 12m for the last side.
Lenght of the smaller unknown side: 12m
Lenght of the larger unknown side: 20m
Step-by-step explanation:
For this case we have a right triangle and we can use the Pythagoras Theorem and using the info given by the triangle we can set up the following equation:
[tex] (x+11)^2 = (x+3)^2 +16^2[/tex]
And if we solve this equation for x we got:
[tex] x^2 +22x +121 = x^2 +6x +9 +256[/tex]
We can cancel [tex]x^2[/tex] in both sides and we have this:
[tex] 22x -6x= 256+9-121 =144[/tex]
And then we got:
[tex] 16 x= 144[/tex]
[tex] x =\frac{144}{16}= 9[/tex]
And then the length of the sides are 9+11= 20 m for the hypothenuse, 16 for the adjacent side and 9+3 = 12m for the last side side.
Lenght of the smaller unknown side: 12m
Lenght of the larger unknown side: 20m
The following histogram shows the exam scores for a Prealgebra class. Use this histogram to answer the questions.Prealgebra Exam ScoresScores 70.5, 75.5, 80.5, 85.5, 90.5, 95.5, 100.5Frequency 0, 4, 8, 12, 16, 20, 24Step 1 of 5:Find the number of the class containing the largest number of exam scores (1, 2, 3, 4, 5, or 6).Step 2 of 5:Find the upper class limit of the third class.Step 3 of 5:Find the class width for this histogram.Step 4 of 5:Find the number of students that took this exam.Step 5 of 5:Find the percentage of students that scored higher than 95.595.5. Round your answer to the nearest percent.
Answer:
The number of the class containing the largest score can be found in frequency 24 and the class is 98 - 103
For the third class 78 - 83 ; the upper limit = 83
The class width for this histogram 5
The number of students that took the exam simply refers to the frequency is 84
The percentage of students that scored higher than 95.5 is 53%
Step-by-step explanation:
The objective of this question is to use the following histogram that shows the exam scores for a Pre-algebra class to answer the question given:
NOW;
The table given in the question can be illustrated as follows:
S/N Class Score Frequency
1 68 - 73 70.5 0
2 73 - 78 75.5 4
3 78 - 83 80.5 8
4 83 - 88 85.5 12
5 88 - 93 90.5 16
6 93 - 98 95.5 20
7 98 - 103 100.5 24
TOTAL: 84
a) The number of the class containing the largest score can be found in frequency 24 and the class is 98 - 103
b) For the third class 78 - 83 ; the upper limit = 83 ( since the upper limit is derived by addition of 5 to the last number showing in the highest value specified by the number in the class interval which is 78 ( i.e 78 + 5 = 83))
c) The class width for this histogram 5 ; since it is the difference between the upper and lower boundaries limit of the given class.
So , from above the difference in any of the class will definitely result into 5
d) The number of students that took the exam simply refers to the frequency ; which is (0+4+8+12+16+20+24) = 84
e) Lastly; the percentage of students that scored higher than 95.5 is ;
⇒[tex]\dfrac{20+24}{84} *100[/tex]
= 0.5238095 × 100
= 52.83
To the nearest percentage ;the percentage of students that scored higher than 95.5 is 53%
Answer:
1. 98-103 (6th class)
2. 88
3. 5
4. 84
5. 52%
Step-by-step explanation:
Find attached the frequency table.
The class of exam scores falls between (1, 2, 3, 4, 5, or 6).
The exam score ranged from 68-103
1) The largest number of exam scores = 24
The largest number of exam scores is in the 6th class = 98 -103
Step 2 of 5:
The upper class limit is the higher number in an interval. Third class interval is 83-88
The upper class limit of the third class 88.
Step 3 of 5:
Class width = upper class limit - lower class limit
We can use any of the class interval to find this as the answer will be the same. Using the interval between 73-78
Class width = 78 - 73
Class width for the histogram = 5
Step 4 of 5:
The total of students that took the test = sum of all the frequency
= 0+4+8+12+16+20+24 = 84
The total of students that took the test = 84
Step 5 of 5:Find the percentage of students that scored higher than 95.5
Number of student that scored higher than 95.5 = 20 + 24 = 44
Percentage of students that scored higher than 95.5 = [(Number of student that scored higher than 95.5)/(total number of students that took the test)] × 100
= (44/84) × 100 = 0.5238 × 100 = 52.38%
Percentage of students that scored higher than 95.5 = 52% (nearest percent)
3
Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.
The product of (3 + 2) and a complex number is (17 + 71).
The complex number is
Answer:
5-i
Step-by-step explanation:
Product=multiplication
Let the complex number=x
(3+2i)*x=17+7i
x=17+7i / 3+2i
x=(17-7i)*(3-2i)/(3+2i)*(3+2i)
=51-34i+21i+14i^2 / 9+6i+6i+4i^2
=51+13i+14i^2 / 9+12i+4i^2
= (51+14 - 13i) / 13
= (65 -13i) / 13
= 65 / 13 - 13 i / 13
= 5 - i.
Solve for the value of x
Answer:
x = 8
Step-by-step explanation:
The angle with the expression in it is complementary to the 30° angle, so is 60°. Then we have ...
4 +7x = 60
7x = 56 . . . . . . subtact 4
x = 8 . . . . . . . . .divide by 7
The area of a circle is 153.86 square meters. What is the diameter of the circle? Use 3.14 for π.
Answer:
Option (2). 14 m
Step-by-step explanation:
Formula to get the area of a circle 'A' = [tex]\pi r^{2}[/tex]
where r = radius of the circle
Given in the question,
Area of the circle = 153.86 square meters
By putting the values in the formula,
153.86 = πr²
r = [tex]\sqrt{\frac{153.86}{\pi } }[/tex]
r = [tex]\sqrt{49}[/tex]
r = 7 meters
Diameter of circle = 2 × (radius of the circle)
= 2 × 7
= 14 meters
Therefore, diameter of the circle is 14 meters.
Option (2) is the answer.
Answer:
14m
Step-by-step explanation:
A teacher figures that final grades in the chemistry department are distributed as: A, 25%; B, 25%;C, 40%;D, 5%; F, 5%. At the end of a randomly selected semester, the following number of grades were recorded. Calculate the chi-square test statistic x^2 to determine if the grade distribution for the department is different than expected. Use α = 0.01.
Grade A B C D F
Number 36 42 60 14 8
a. 6.87
b. 0.6375
c. 5.25
d. 4.82
Answer:
[tex]E_{A} =0.25*160=40[/tex]
[tex]E_{B} =0.25*160=40[/tex]
[tex]E_{C} =0.4*160=64[/tex]
[tex]E_{D} =0.05*160=8[/tex]
[tex]E_{F} =0.05*160=8[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(36-40)^2}{40}+\frac{(42-40)^2}{40}+\frac{(60-64)^2}{64}+\frac{(14-8)^2}{8}+\frac{(8-8)^2}{8} =5.25[/tex]
The answer would be:
c. 5.25
Step-by-step explanation:
The observed values are given by:
A: 36
B: 42
C: 60
D: 14
E: 8
Total =160
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is no difference in the proportions for the final grades
H1: There is a difference in the proportions for the final grades
The level of significance assumed for this case is [tex]\alpha=0.01[/tex]
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
Now we just need to calculate the expected values with the following formula [tex]E_i = \% * total[/tex]
And the calculations are given by:
[tex]E_{A} =0.25*160=40[/tex]
[tex]E_{B} =0.25*160=40[/tex]
[tex]E_{C} =0.4*160=64[/tex]
[tex]E_{D} =0.05*160=8[/tex]
[tex]E_{F} =0.05*160=8[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(36-40)^2}{40}+\frac{(42-40)^2}{40}+\frac{(60-64)^2}{64}+\frac{(14-8)^2}{8}+\frac{(8-8)^2}{8} =5.25[/tex]
The answer would be:
c. 5.25
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(categories-1)=(5-1)=4[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{4} >5.25)=0.263[/tex]
The p value is higher than the significance so we have enough evidence to FAIL to reject the null hypothesis
Can anybody please help me with this one??
Answer:
[tex]the \: answer \: is \: d.(x + 4) {}^{2} = 8(y + 4)[/tex]
Which of the following best describes the slope of the line below?
PLSSS HELP
The slope is zero. Slope formula is Y=mx+b and since B is 1.5 and it is a straight line, Y=mx+1.5. What plus 1.5 is 1.5? 0. Hope this helps.
A tank contains 5,000 L of brine with 13 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate.
Required:
a. How much salt is in the tank after t minutes?
b. How much salt is in the tank after 20 minutes?
Answer:
a) [tex]x(t) = 13*e^(^-^\frac{t}{100}^)[/tex]
b) 10.643 kg
Step-by-step explanation:
Solution:-
- We will first denote the amount of salt in the solution as x ( t ) at any time t.
- We are given that the Pure water enters the tank ( contains zero salt ).
- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min
- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.
- The ODE is mathematically expressed as:
[tex]\frac{dx}{dt} =[/tex] ( salt flow in ) - ( salt flow out )
- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0
- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).
- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.
- So any time ( t ) the concentration of salt in the 5,000 L is:
[tex]conc = \frac{x(t)}{1000}\frac{kg}{L}[/tex]
- The amount of salt leaving the tank per unit time can be determined from:
salt flow-out = conc * V( flow-out )
salt flow-out = [tex]\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\[/tex]
salt flow-out = [tex]\frac{x(t)}{100}\frac{kg}{min}[/tex]
- The ODE becomes:
[tex]\frac{dx}{dt} = 0 - \frac{x}{100}[/tex]
- Separate the variables and integrate both sides:
[tex]\int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)[/tex]
- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:
[tex]13 = C*e^0 = C[/tex]
- The solution to the ODE becomes:
[tex]x(t) = 13*e^(^-^\frac{t}{100}^)[/tex]
- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:
[tex]x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg[/tex]
- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg
I need HELP PLEASE HELP ME
Answer:
Graph 2
Step-by-step explanation:
You can see that all the shaded numbers are above negative 25 in that graph. Hope this helped!
If f(x) = 5x + 7 and g(x) = sqrt(x + 6) , which statement is true? A) 9 is NOT in the domain of f g B) 9 IS in the domain of f g
Answer 9 is in the domain of f g
Step-by-step explanation:
graph the equation in a coordinate plane, x+2y=4
Hope this helps!
Stay safe, have a good day :D
0.580 80 repeating as a simplified fraction
Answer:
979
Step-by-step explanation:
Answer:
115/198
Step-by-step explanation:
khan
Which of the following is the solution to |x-1|=8
Answer:
-7,9
Step-by-step explanation:
x-1=-8
x=-7
x-1=8
x=9
The data represents the heights of eruptions by a geyser. Use the heights to construct a stemplot. Identify the two values that are closest to the middle when the data are sorted in order from lowest to highest.
Height of eruption
62 33 50 90
80 50 40 70
50 63 74 53
55 64 60 60
78 70 43 82
Required:
Identify the two values that are closest to the middle when the data are sorted in order from lowest to highest. The values closest to the middle are_________inches and_______inches.
Answer:
[tex] Median = \frac{60+60}{2}=60[/tex]
And we see that the closest values to 60 are 62 and 63 and then the answer would be:
The values closest to the middle are 62 inches and 63 inches.
Step-by-step explanation:
We have the following dataser given:
62 33 50 90 80 50 40 70 50 63 74 53 55 64 60 60 78 70 43 82
We can sort the values from the lowest to the highest and we got::
33 40 43 50 50 50 53 55 60 60 62 63 64 70 70 74 78 80 82 90
Now we see that we have n=20 values and the values closest to the middle and we can use the middle as the median and for this case the median can be calculated from position 10 and 11th and we got:
[tex] Median = \frac{60+60}{2}=60[/tex]
And we see that the closest values to 60 are 62 and 63 and then the answer would be:
The values closest to the middle are 62 inches and 63 inches.
The values closest to these middle elements are 60 and 63 inches
The dataset is given as:
62 33 50 90 80 50 40 70 50 63 74 53 55 64 60 60 78 70 43 82
Next, we sort the data elements in ascending order
33 40 43 50 50 50 53 55 60 60 62 63 64 70 70 74 78 80 82 90
The length of the dataset is 20.
So, the elements at the middle are the 10th and the 11 elements.
From the sorted dataset, these elements are: 60 and 62
Hence, the values closest to these middle elements are 60 and 63
Read more about median at:
https://brainly.com/question/14532771
What is the value of n in the equation: 8n+9= -n+5?
Answer:
n = -1
Step-by-step explanation:
So first subtract 9 to both sides
8n = -n - 9
Now you want the n on one side and the constant on the other
so add the single n to the n side
9n = -9
Divide 9 to both sides to solve for n
n = -1
(DONT REPORT OR ANSWER) pls don’t (what is -2 plus -2)
Answer:
I think it’s -4
Step-by-step explanation:
-2 plus -2 is -4
since 2 plus 2 is 4
Hope this helps ;)
Answer:
It is -4Step-by-step explanation:
It is because 2 plus 2 is 4
so -2 plus -2 is -4
hth!
I need help with this one
Answer:
2 2/3
Step-by-step explanation:
What’s the correct answer for this?
Answer:
The capital B refers to the base of the area
Step-by-step explanation:
Answer:
A
Step-by-step explanation:
The capital B means the area of the base
