Noah is writing an exam for his 8th grade students. The exam is worth 100 points and Noah wants 35 questions on the exam. He plans to mix short answer questions, worth 3 points, with multiple choice questions worth 2 points. Create a system of equations to tell us how many of each type of question Noah can have on the test. Let x= the number of short answer questions and y= the number of multiple choice questions

Answer:

On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real proportion at least at 95% of confidence.

Step-by-step explanation:

1) Notation and definitions

[tex]X=34[/tex] number of people living at USA with a particular disease.

[tex]n=527[/tex] random sample taken

[tex]\hat p=\frac{34}{527}=0.0645[/tex] estimated proportion of people living at USA with a particular disease

[tex]p[/tex] true population proportion of people living at USA with a particular disease.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.0645 - 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0435[/tex]

[tex]0.0645 + 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0855[/tex]

The 95% confidence interval would be given by (0.0435;0.0855)

On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real mean at least at 95% of confidence.

Answer:

The projected enrollment is [tex]\lim_{t \to \infty} E(t)=10,000[/tex]

Step-by-step explanation:

Consider the provided projected rate.

[tex]E'(t) = 12000(t + 9)^{\frac{-3}{2}}[/tex]

Integrate the above function.

[tex]E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt[/tex]

[tex]E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c[/tex]

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

[tex]2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c[/tex]

[tex]2000=-\frac{24000}{3}+c[/tex]

[tex]2000=-8000+c[/tex]

[tex]c=10,000[/tex]

Therefore, [tex]E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000[/tex]

Now we need to find [tex]\lim_{t \to \infty} E(t)[/tex]

[tex]\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000[/tex]

[tex]\lim_{t \to \infty} E(t)=10,000[/tex]

Hence, the projected enrollment is [tex]\lim_{t \to \infty} E(t)=10,000[/tex]

Answer:

[tex]X_1 + X_2 \sim (\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2 )[/tex]

Step-by-step explanation:

We are given the following in the question:

[tex]X_1[/tex] is a random normal variable with mean and variance

[tex]\mu_1\\\sigma_1^2[/tex]

[tex]X_1 \sim N(\mu_1,\sigma_1^2)[/tex]

[tex]X_2[/tex] is a random normal variable with mean and variance

[tex]\mu_2\\\sigma_2^2[/tex]

[tex]X_2 \sim N(\mu_2,\sigma_2^2)[/tex]

[tex]X_1, X_2[/tex] are independent events.

Let

[tex]Z =X_1 + X_2[/tex]

Then, Z will have a normal distribution with mean equal to the sum of the two means and its variance equal the sum of the two variances.

Thus, we can write:

[tex]\mu = \mu_1 + \mu_2\\\sigma^2 = \sigma_1^2 + \sigma_2^2\\Z \sim (\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2 )[/tex]

Final answer:

The sum of two independent normal random variables X1 and X2 is also a normal random variable with a mean of μ1 + μ2 and variance of σ²₁ + σ²₂, denoted by N(μ1 + μ2, σ²₁ + σ²₂).

Explanation:

The distribution of the sum of two independent normal random variables, X1 + X2, is itself a normal random variable. Given that X1 is a normal random variable with mean μ1 and variance σ²₁, and X2 is a normal random variable with mean μ2 and variance σ²₂, the sum X1 + X2 will have a mean of μ1 + μ2 and variance of σ²₁ + σ²₂ because of the properties of the normal distribution and the independence of X1 and X2.

The resulting distribution can be denoted as N(μ1 + μ2, σ²₁ + σ²₂). This conclusion comes from the central limit theorem, which states that the sum of independent random variables tends towards a normal distribution as the number of variables increases, and the means and variances add up.

P-value for the airline's claim that the no-show rate for passengers booked on its flights is less than 6% is 0.2061. Hence, option (B) is correct.

Given that, an airline claims that the no-show rate for passengers booked on its flights is less than 6%.

From this claim, it is clear that:

Null hypothesis :  Proportion of no-show rate for passengers booked on its flights, P = 6%. i.e. H₀: P = 0.06

Alternative hypothesis: Proportion of no-show rate for passengers booked on its flights P < 6%, i.e. P < 0.06.

Out of 380 randomly selected reservations, 19 were no-shows

Sample proportion, [tex]\hat{p} = \frac{19}{380}[/tex]  = 0.05

Then, the standard error for the sample size of 380 is:

[tex]\text{S.E.} = \sqrt{{\frac{p(1-p)}{n}[/tex]

[tex]\text{S.E.} = \sqrt{{\frac{0.06(1-0.06)}{380}[/tex]

[tex]\text{S.E.} = 0.012[/tex]

Now calculating the test statistic

[tex]z = \frac{( \hat{p} - P)}{S.E}[/tex]

[tex]z = \frac{(0.05 - 0.06)}{0.012 }[/tex]

z = -0.833

p value for z = -0.083 is 0.2061 (From the normal table).

Hence, the p-value for a test of the airline's claim is 0.2061. Option (B) is correct.

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Final answer:

To find the P-value for the hypothesis test, calculate the sample proportion and standard error, then use the standard normal distribution to calculate the z-score and find the corresponding P-value.

Explanation:

To find the P-value for the hypothesis test, we need to use the given data and perform calculations.

First, we need to calculate the sample proportion, which is the number of no-shows divided by the total number of reservations: 19/380 = 0.05.

Next, we calculate the standard error of the sample proportion using the formula: √((p' * (1 - p')) / n), where p' is the sample proportion and n is the sample size. In this case, the standard error is √((0.05 * (1 - 0.05)) / 380) = 0.014.

Finally, we use the standard normal distribution to calculate the z-score and find the corresponding P-value. In this case, the observed proportion is less than the claimed proportion, so we use a one-tailed test and calculate the z-score as (observed proportion - claimed proportion) / standard error = (0.05 - 0.06) / 0.014 = -0.714. Looking up the P-value for a z-score of -0.714, we find that it is approximately 0.4714.

Therefore, the P-value for the test of the airline's claim is approximately 0.4714.

Answer:

Answer is 13.3 ft

Step-by-step explanation:

i explained in the image below

Answer:

90% confidence interval: (2186.53;2881.47)

95% confidence interval: (2118.73;2949.27)

99% confidence interval: (1987.37;3080.63)

For the last part is not the best way say : "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."

The best interpretation is this one: "We are 95% confident that the actual mean for the rents of unfurnished one-bedroom apartments in the Boston area is between (2118.73;2949.27)"

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=2534[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=670[/tex] represent the population standard deviation

n=10 represent the sample size  

90% confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

Now we have everything in order to replace into formula (1):

[tex]2534-1.64\frac{670}{\sqrt{10}}=2186.53[/tex]    

[tex]2534+1.64\frac{670}{\sqrt{10}}=2881.47[/tex]

So on this case the 90% confidence interval would be given by (2186.53;2881.47)

95% confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]

Now we have everything in order to replace into formula (1):

[tex]2534-1.96\frac{670}{\sqrt{10}}=2118.73[/tex]    

[tex]2534+1.96\frac{670}{\sqrt{10}}=2949.27[/tex]

So on this case the 95% confidence interval would be given by (2118.73;2949.27)

99% confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]

Now we have everything in order to replace into formula (1):

[tex]2534-2.58\frac{670}{\sqrt{10}}=1987.37[/tex]    

[tex]2534+2.58\frac{670}{\sqrt{10}}=3080.63[/tex]

So on this case the 99% confidence interval would be given by (1987.37;3080.63)

For the last part is not the best way say : "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."

The best interpretation is this one: "We are 95% confident that the actual mean for the rents of unfurnished one-bedroom apartments in the Boston area is between (2118.73;2949.27)"

Answer:

H₀: σ² ≥ 17956

H₁: σ² < 17956

Step-by-step explanation:

Hello!

You are asked to test if the population standard deviation of a certain population. Now keep in mind that wherever you want to make a hypothesis test for a population parameter, you have to have a known distribution that includes this parameter. Since the population standard deviation is no parameter of any distribution, what you have to do is test the population variance. Any decision you make about the population variance can be extrapolated to the population standard deviation.

To make a hypothesis test for the population variance, you need a variable with normal distribution and the statistic to use is a Chi-square statistic.

The hypothesis is that the population standard deviation is less than 134, symbolically: σ < 134

Translated in terms of the population variance: σ² < 17956

H₀: σ² ≥ 17956

H₁: σ² < 17956

α: 0.05

χ²=  (n-1)S²  ~χ²[tex]_{n-1}[/tex]

σ²

χ²=  (27-1)(126)²  = 22.988

17956

The test is one tailed (left)

χ²[tex]_{n-1;α}[/tex] =χ²[tex]_{26;0.05}[/tex] = 15.379

If the calculated Chi-square value is ≤ than the critical value, you reject the null hypothesis.

If the calculated Chi-square value is > than the critical value, you don't reject the null hypothesis.

Since the value is greater than the critical value, you do not reject the null hypothesis. So at a 5% level, there is not enough evidence to reject the null hypothesis, this means the population variance is at least 17956. On the same level, you can conclude that the population standard deviation is at least 134.

I've made the test so that you have an example of how to do it.

I hope it helps!

Answer:

5/8,-5√2,{(2x+3)(x-5)}

Step-by-step explanation:

1) 5x-10/8x-16

=5(x-2)/8(x-2)

=5/8

2) √32-3√18

=4√2-3√18

=4√2-9√2

=(4-9)√2

= -5√2

4) 2x^2-7x-15

=2x^2+3x-10x-15

=(2x+3),(x-5)

Answer:

10 ft x 10 ft

Area = 100 ft^2

Step-by-step explanation:

Let 'S' be the length of the southern boundary fence and 'W' the length of the eastern and western sides of the fence.

The total area is given by:

[tex]A=S*W[/tex]

The cost function is given by:

[tex]\$ 120 = \$3*2W+\$6*S\\20 = W+S\\W = 20-S[/tex]

Replacing that relationship into the Area function and finding its derivate, we can find the value of 'S' for which the area is maximized when the derivate equals zero:

[tex]A=S*(20-S)\\A=20S-S^2\\\frac{dA}{dS} = \frac{d(20S-S^2)}{dS}\\0= 20-2S\\S=10[/tex]

If S=10 then W =20 -10 = 10

Therefore, the largest area enclosed by the fence is:

[tex]A=S*W\\A=10*10 = 100\ ft^2[/tex]

Answer:

                          Hamburger       Chicken

Adults                    65                        60         125

children                 55                        20          75

                             120                        80         200

a)What is the probability that a randomly selected individual is an adult?

Total no. of adults = 125

Total no. of people  200

The probability that a randomly selected individual is an adult = [tex]\frac{125}{200}=0.625[/tex]

b) What is the probability that a randomly selected individual is a child and prefers chicken?

No. of child prefers chicken = 20

The probability that a randomly selected individual is a child and prefers chicken= [tex]\frac{20}{200}=0.1[/tex]

c)Given the person is a child, what is the probability that this child prefers a hamburger?

No. of children prefer hamburger = 55

No. of child = 75

The probability that this child prefers a hamburger= [tex]\frac{35}{75}=0.46[/tex]

d) Assume we know that a person has ordered chicken, what is the probability that this individual is an adult?

No. of adults prefer chicken = 60

No. of total people like chicken = 80

A person has ordered chicken, the probability that this individual is an adult= [tex]\frac{60}{80}=0.75[/tex]

Final answer:

The probability that a randomly selected individual is an adult is 0.625. The probability that a randomly selected individual is a child and prefers chicken is 0.10. Given a child, the probability of preferring a hamburger is approximately 0.733, and given that a person ordered chicken, the probability that they are an adult is 0.75.

Explanation:

Let's address the questions based on the contingency table provided:

Probability Calculations

a) The probability that a randomly selected individual is an adult can be calculated as follows:

The number of adults: 125

The total number of respondents: 200

Probability(Adult) = Number of Adults / Total Number of Respondents = 125 / 200 = 0.625

b) The probability that a randomly selected individual is a child and prefers chicken:

The number of children who prefer chicken: 20

The total number of respondents: 200

Probability(Child and Chicken) = Number of Children who prefer Chicken / Total Number of Respondents = 20 / 200 = 0.10

c) If the person is a child, the probability that this child prefers a hamburger:

The number of children who prefer hamburger: 55

The total number of children: 75

Probability(Hamburger | Child) = Number of Children who prefer Hamburger / Total Number of Children = 55 / 75 ≈ 0.733

d) Given that a person has ordered chicken, the probability that this individual is an adult:

The number of adults who prefer chicken: 60

The total number of chicken preferences: 80

Probability(Adult | Chicken) = Number of Adults who prefer Chicken / Total Number of Chicken Preferences = 60 / 80 = 0.75

Answer:

Option D is right

Step-by-step explanation:

given that a two sample test for mean of independent samples to be done.

We create hypotheses as:

[tex]H_0: \ bar x = \bar y[/tex] vs alternate suitably right or left or two tailed according to the needs.

The conditions needed for conducting this test would be

I. Both populations are approximately normally distributed

II. Both sample sizes greater than 30

III. Population of differences is approximately normally distributed

i.e. either i, ii or III

Option D is right.

Parameterize the line segments by

[tex]\vec r(t)=(1-t)\langle1,1,1\rangle+t\langle2,2,2\rangle=\langle1+t,1+t,1+t\rangle[/tex]

and

[tex]\vec s(t)=(1-t)\langle2,2,2\rangle+t\langle3,6,8\rangle=\langle2+t,2+4t,2+6t\rangle[/tex]

both with [tex]0\le t\le1[/tex]. Then

[tex]\vec r'(t)=\langle1,1,1\rangle[/tex]

[tex]\vec s'(t)=\langle1,4,6\rangle[/tex]

so that the work done by [tex]\vec F(x,y,z)=\langle\sin(x+y),xy,x^2z\rangle[/tex] over the respective line segments is given by

[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r=\int_0^1\langle\sin(2+2t),(1+t)^2,(1+t)^3\rangle\cdot\langle1,1,1\rangle\,\mathrm dt[/tex]

[tex]=\displaystyle\int_0^1\sin(2+2t)+(1+t)^2+(1+t)^3\,\mathrm dt=\boxed{\frac{73+6\cos2-6\cos4}{12}}[/tex]

and

[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec s=\int_0^1\langle\sin(4+5t),(2+t)(2+4t),(2+t)^2(2+6t)\rangle\cdot\langle1,4,6\rangle\,\mathrm dt[/tex]

[tex]=\displaystyle\int_0^1\sin(4+5t)+4(2+t)(2+4t)+6(2+t)^2(2+6t)\,\mathrm dt=\boxed{\frac{3695+3\cos4-3\cos9}{15}}[/tex]

(both measured in Newton-meters)

To compute the work done by the force in this scenario, we need to apply the principle of line integrals of force with respect to displacement, calculating work for each segment of the trajectory separately and then adding these results.

Computing work done by the given force while moving an object generally involves utilizing the line integral of the force with respect to displacement. The principle is simplistically shown in the equation W = F⋅d = Fd cos θ, where W represents work, F denotes force, and d symbolizes displacement, with '⋅' denoting the dot product, and cos θ representing the cosine of the angle between the force and displacement vectors. F can break down into its components, such as Fx, Fy, Fz, and similarly, d can be broken down to dx, dy, dz. Using these, we can express work for three dimensions as dW = Fxdx + Fydy + Fzdz which extends the notion of work done to three-dimensional space. This concept is categorically illustrated while calculating the infinitesimal work done by a variable force.

The trajectory (path) for this problem can be divided into two line segments, one from (1, 1, 1) to (2, 2, 2) and the other from (2, 2, 2) to (3, 6, 8). The work for each segment can be calculated independently, based on the variable force function and the displacement during each segment, after which the two results can be added up to determine the total work done.

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Answer:

53¢

Step-by-step explanation:

First, I'll put these in order.

20¢;30¢; 30¢;75¢;40¢;40¢;40¢;40¢;50¢;55¢55¢65¢;65¢; $1.50;  

Then, I'll combine like terms.

30+30=60

40+40+40+40(or 40 x 4)=160

55+55=110

65+65=130

60+160+110+130+20+75+50+$1.50=$7.55/14=53¢

PLZ correct me if i'm wrong :-D

Answer: c) [tex]H_0:\mu= 1.5\\\\H_a:\mu \neq1.5[/tex]

Step-by-step explanation:

Given: Nestor Milk Powder is sold in packets with an advertised mean weight of 1.5 kgs.  

i.e. [tex]\mu= 1.5[/tex]  

A consumer group wishes to check the accuracy of the advertised mean and takes a sample of 52 packets finding an average weight of 1.49 kgs.

So 1.49 is sample mean but hypothesis is the statement about the parameter which is [tex]\mu[/tex].

i.e. he wanted to check whether [tex]\mu= 1.5[/tex] or [tex]\mu \neq1.5[/tex]

Since null hypothesis[tex](H_0)[/tex] contains equality and alternative hypothesis[tex](H_a)[/tex] is against it.

Therefore, the set of hypotheses that should be used to test the accuracy of advertised weight would be :

[tex]H_0:\mu= 1.5\\\\H_a:\mu \neq1.5[/tex]

Hence, the correct answer is option c) oH: μ= 1.5; 1H: μ≠1.5

Answer:

21/64

Step-by-step explanation:

First, we need to note that the function f(x) = x² is increasing on (0, +∞), and it is decreasing on (-∞,0)

The first interval generated by the partition is [-1, -1/2], since f is decreasing for negative values, we have that f takes its minimum values at the right extreme of the interval, hence -1/2.

The second interval is [-1/2, 1/2]. Here f takes its minimum value at 0, because f(0) = 0, and f is positive otherwise.

Since f is increasing for positive values of x, then, on the remaining 2 intervals, f takes its minimum value at their respective left extremes, in other words, 1/2 and 3/4 respectively.

We obtain the lower Riemman sum by multiplying this values evaluated in f by the lenght of their respective intervals and summing the results, thus

LP(f) = f(-1/2) * ((-1/2) - (-1)) + f(0) * (1/2 - (-1/2)) + f(1/2)* (3/4 - 1/2) + f(3/4) * (1- 3/4)

= 1/4 * 1/2 + 0 * 1 + 1/4 * 1/4 + 9/16 * 1/4 = 1/8 + 0 + 1/16 + 9/64 = 21/64

As a result, the lower Riemann sum on the partition P is 21/64

Final answer:

To compute the lower Riemann sum for the given function f(x) = x² over the interval x ∈ [-1,1] with respect to the partition P = [-1, -1/2, 1/2, 3/4 ,1], we need to find the height of each subinterval and multiply it by the width of the subinterval. The lower Riemann sum is 5/8.

Explanation:

To compute the lower Riemann sum for the given function f(x) = x² over the interval x ∈ [-1,1] with respect to the partition P = [-1, -1/2, 1/2, 3/4 ,1], we need to find the height of each subinterval and multiply it by the width of the subinterval.

First, let's calculate the width of each subinterval:

width of subinterval 1: (-1) - (-1) = 0

width of subinterval 2: (-1/2) - (-1) = 1/2

width of subinterval 3: (1/2) - (-1/2) = 1

width of subinterval 4: (3/4) - (1/2) = 1/4

width of subinterval 5: (1) - (3/4) = 1/4

Next, let's calculate the height of each subinterval by substituting the left endpoint of each subinterval into the function:f((-1/2))^2 = 1/4, f(1/2)² = 1/4, f(3/4)² = 9/16, f(1)² = 1

Finally, we compute the lower Riemann sum by multiplying the height by the width for each subinterval and summing them up:

(0 * 0) + (1/2 * 1/4) + (1 * 9/16) + (1/4 * 1) = 5/8.

Answer:

We conclude that the plant should shut down.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 60

Sample mean, [tex]\bar{x}[/tex] = 61.498

Sample size, n = 100

Alpha, α = 0.05

Population standard deviation, σ = 6

a) First, we design the null and the alternate hypothesis  such that the power plant will be shut down when the null hypothesis is rejected.

[tex]H_{0}: \mu \leq 60\\H_A: \mu > 60[/tex]

We use One-tailed(right) z test to perform this hypothesis.

b) Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{61.498 - 60}{\frac{6}{\sqrt{100}} } = 2.49[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]

Since,  

[tex]z_{stat} > z_{critical}[/tex]

We reject the null hypothesis and accept the alternate hypothesis. Thus, the temperature of waste water discharged is greater than 60°F. We conclude that the power plant will shut down.

Calculating the p-value from the z-table:

P-value = 0.0063

Since,

P-value < Significance level

We reject the null hypothesis and accept the alternate hypothesis. Thus, the temperature of waste water discharged is greater than 60°F. We conclude that the power plant will shut down.

Answer:

a) [tex]P(X\leq 12)=0.8586[/tex]

b) [tex]P(X\geq 5)=0.9802[/tex]

c) [tex]P(5\leq X\leq 12)=0.8389[/tex]

Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=20, p=0.48)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

We need to check the conditions in order to use the normal approximation.

[tex]np=20*0.48=9.6 \approx 10 \geq 10[/tex]

[tex]n(1-p)=20*(1-0.48)=10.4 \geq 10[/tex]

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

[tex]E(X)=np=20*0.48=9.6[/tex]

[tex]\sigma=\sqrt{np(1-p)}=\sqrt{20*0.48(1-0.48)}=2.234[/tex]

Part a

We want this probability:

[tex]P(X\leq 12)[/tex]

We can use the z score given by this formula [tex]Z=\frac{x-\mu}{\sigma}[/tex].

[tex]P(X\leq 12)=P(\frac{X-\mu}{\sigma}\leq \frac{12-9.6}{2.234})=P(Z\leq 1.074)=0.8586[/tex]

Part b

We want this probability:

[tex]P(X\geq 5)[/tex]

We can use again the z score formula and we have:

[tex]P(X\geq 5)=1-P(X<5)=1-P(\frac{X-\mu}{\sigma}< \frac{5-9.6}{2.234})=1-P(Z<- 2.059)=0.9802[/tex]

Part c

We want this probability:

[tex]P(5\leq X\leq 12)=P(\frac{5-9.6}{2.234}\leq \frac{X-\mu}{\sigma}\leq \frac{12-9.6}{2.234})=P(-2.059\leq Z \leq 1.074)[/tex]

[tex]=P(Z<1.074)-P(Z<-2.059)=0.8586-0.0197=0.8389[/tex]

Answer:

The function y must be decreasing (or equal to 0) on any interval on which it is defined.

Step-by-step explanation:

The derivative of a function gives us the rate at which that function is changing. In this case, -y^2, yields a negative value for every possible value of y, thus, the rate of change is always negative and the function y is decreasing (or equal to 0) on any interval on which it is defined.

The differential equation y' = -[tex]y^2[/tex] implies that y is either decreasing or constant wherever it is defined, because the derivative y' is non-positive.

By examining the differential equation y' = -[tex]y^2[/tex], we can infer some characteristics about the solutions without solving it. If y is a solution to this equation, then y' represents the derivative of y with respect to x. This derivative tells us about the rate of change of the function y.

Since the right side of the equation is -[tex]y^2[/tex], and a square of a real number is always non-negative, multiplying by -1 makes it non-positive. This implies that the derivative y' is either less than or equal to zero. Therefore, wherever the function y is defined, it must be either decreasing or constant (equal to zero). If y is positive, y will decrease because of the negative sign in front of the square. If y is negative, squaring it results in a positive number, but the negative sign still ensures that the rate of change is non-positive.

Conclusion: the function y is decreasing or remains constant on any interval it is defined; it cannot be increasing.

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Answer: B. 461 acres

Step-by-step explanation:

Given : In an agricultural study, the average amount of corn yield is normally distributed with a mean of 185.2 bushels of corn per acre, with a standard deviation of 23.5 bushels of corn.

i.e. [tex]\mu=185.2\ \ , \ \sigma=23.5[/tex]

Let x denotes the amount of corn yield.

Now, the probability that the amount of corn yield is more than 190 bushels of corn per acre.

[tex]P(x>190)=P(\dfrac{x-\mu}{\sigma}>\dfrac{190-185.2}{23.5})[/tex]

[Formula : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]]

[tex]=P(z>0.2043)=1-P(z<0.2043)[/tex]  [∵ P(Z>z)=1-P(Z<z)]

[tex]1-0.5809405[/tex]   [using z-value calculator or table]

[tex]=0.4190595[/tex]

Now, If a study included 1100 acres then the expected number to yield more than 190 bushels of corn per acre :-

[tex]0.4190595\times1100=460.96545\approx461\text{ acres}[/tex]

hence, the correct answer is B. 461 acres .

Answer:

B.  0.10 - 0.20 = 0.10.

Step-by-step explanation:

100/1000 = 0.10 of the population were born , and

200/1000 = 0.20 of the population died so it is:

0.10 - 0.20 = 0.10.

Answer: B

Step-by-step explanation:

The total population was initially 1000 individuals. In this population, a total of of 100 new individuals were born over the course of one year. The proportion of new individuals that make up the population is 100/1000 = 0.1

Since they were added, then it is positive and it is a gain for the population.

A total of 200 individuals also died over the course of one year. The proportion that died = 200/1000 = 0.2. This would be negative because it is a loss for the population.

The population growth rate will be gain - loss. This becomes

0.1 - 0.2 = -0.1

Answer:

A. Increase by 2

Step-by-step explanation:

Given that a  fitted multiple regression equation is

[tex]Y = 28 + 5X_1 -4X_2 + 7X_3 + 2X_4[/tex]

This is a multiple regression line with dependent variable y and independent variables x1, x2, x3 and x4

The coefficients of independent variables represent the slope.

In other words the coefficients represent the rate of change of y when xi is changed by 1 unit.

Given that x3 and x4 remain unchanged and x1 increases by 2 and x2 by 2 units

Since slope of x1 is 5, we find for one unit change in x1 we can have 5 units change in y

i.e. for 2 units change in x1, we expect 10 units change in Y

Similarly for 2 units change in x2, we expect -2(4) units change in Y

Put together we have

[tex]10-8 =2[/tex] change in y

Since positive 2, there is an increase by 2

A. Increase by 2

Answer:

a)  [tex]\bar X=9.07[/tex]

b) The 95% confidence interval is given by (8.197;9.943)  

c) [tex]m=2.14 \frac{1.580}{\sqrt{15}}=0.873[/tex]

d)  3 possible ways

1) Increasing the sample size n.  

2) Reducing the variability. If we have more data probably we will have less variation.

3) Lower the confidence level. Because if we have lower confidence then the quantile from the t distribution would belower and tthe margin of error too.

Step-by-step explanation:

Notation and definitions  

n=15 represent the sample size  

[tex]\bar X= 9.07[/tex] represent the sample mean  

[tex]s=1.580[/tex] represent the sample standard deviation  

m represent the margin of error  

Confidence =95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a: Find the point estimate of the population mean.

The point of estimate for the population mean [tex]\mu[/tex] is given by:

[tex]\bar X =\frac{\sum_{i=1}^{n} x_i}{n}[/tex]

The mean obteained after add all the data and divide by 15 is [tex]\bar X=9.07[/tex]

Calculate the critical value tc  

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. The degrees of freedom are given by:  

[tex]df=n-1=15-1=14[/tex]  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.025,14)" for [tex]t_{\alpha/2}=-2.14[/tex]  

"=T.INV(1-0.025,14)" for [tex]t_{1-\alpha/2}=2.14[/tex]  

The critical value [tex]tc=\pm 2.14[/tex]  

Part c: Calculate the margin of error (m)  

First we need to calculate the standard deviation given by this formula:

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex]

s=1.580

The margin of error for the sample mean is given by this formula:  

[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]  

[tex]m=2.14 \frac{1.580}{\sqrt{15}}=0.873[/tex]  

Part b: Calculate the confidence interval  

The interval for the mean is given by this formula:  

[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]  

And calculating the limits we got:  

[tex]9.07 - 2.14 \frac{1.580}{\sqrt{15}}=8.197[/tex]  

[tex]9.07 + 2.14 \frac{1.580}{\sqrt{15}}=9.943[/tex]  

The 95% confidence interval is given by (8.197;9.943)  

Part d: How can we reduce the margin of error?

We can reduce the margin of error on the following ways:

1) Increasing the sample size n.  

2) Reducing the variability. If we have more data probably we will have less variation.

3) Lower the confidence level. Because if we have lower confidence then the quantile from the t distribution would belower and tthe margin of error too.

The publisher's point estimate for average reading time is 9.2 minutes. The 95% confidence interval for this estimate is between 8.18 and 10.22 minutes. The margin of error is approximately 1.04 minutes, which can be reduced by increasing the sample size.

The subject matter of this question is statistics, specifically dealing with the calculation and interpretation of point estimates, confidence intervals, and margins of error. Here are the steps to solve your question:

Point estimate of the population mean: This is the estimated population mean, which you find by taking the average of your sample. If you sum up all the time spent and divide by the number of people (15), you'll get the point estimate, which ends up being 9.2 minutes.

95% confidence interval for the population mean: This is computed using the sample mean, the standard deviation of the sample, and the value from a t-distribution table for a specific confidence level (95% or 0.05 significance level in this case). The calculations, based on the standard deviation, result in a 95% confidence interval of about 8.18 to 10.22.

Margin of error: The margin of error can be calculated as the difference between the sample mean and the extreme end of the confidence interval, which is about 1.04 in this case.

Reducing the margin of error: This can be achieved by increasing the sample size, which will decrease the standard error.

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Answer:

the probability is 38,46%

Step-by-step explanation:

If all decks are put together and shuffled , then card is picked at random regardless of the number, then the probability that the card is red is

probability = number of red cards / total number of cards = 5/(5+2+6) = 5/13=0.3846= 38,46%

The ideal weight of a 6-foot male (according to the function) is approximately 75.4 kilograms. By manipulating the weight function, we can express height as a function of weight. To verify this function, we substitute it back into the original equation, ensuring our original input value is retrieved.

The question is asking to find the ideal weight of a 6-foot male using the function W(h) = 49 + 2.2(h- 60). The height in inches for a 6-foot male is 72 inches (as 1 foot equals 12 inches). Substituting into the formula we obtain:

W(72) = 49 + 2.2(72 - 60) = 49 + 2.2*12.

Completing the calculation, the ideal weight is about 75.4 kg (rounded to the nearest tenth).

To express height h as a function of weight W, we need to rearrange the function W(h). Subtracting 49 from both sides yields 2.2(h - 60) = W - 49. Then divide both sides by 2.2 to isolate h, resulting in h = (W - 49) / 2.2 + 60.

Verification that W(h(W)) = W and h(W(h)) = h will require substituting the functions back into each other, and determining that the original input is returned.

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Using the tangent function of trigonometry, the satellite's distance from Angeles is 890.53 miles.

Distance between Phoenix and Los Angeles = 340 miles

1 mile = 5,280 feet

340 miles = 1,795,200 feet (340 miles * 5280 feet/mile)

Angle of elevation at Phoenix = 60°

Angle of elevation at Los Angeles = 75°

Tangent Equation:

tan(75°) = x/d

Solving the equation for x:

x = d * tan(75°)

Substituting the distance value into this equation for x:

x = 1,795,200 * tan(75°) ≈ 4,702,000 feet ≈ 890.53 miles (4,702,000/5,280)

Thus, we can conclude that the distance of the satellite at Los Angeles is approximately, 4,702,000 feet or 890.53 miles.

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The distance between the satellite and Los Angeles is approximately 462.215 miles.

To find the distance between the satellite and Los Angeles, we can use trigonometry. Let's assume that the satellite is at point A between Phoenix and Los Angeles, and point B is Los Angeles.

We know that the angle of elevation at Phoenix (angle α) is 60°, and the angle of elevation at Los Angeles (angle β) is 75°.

We can use the tangent function to find the distance from Los Angeles to the satellite:

tan(β) = opposite/adjacent

We know that the opposite side is the distance between Phoenix and Los Angeles, which is 340 miles. So we can write:

tan(75°) = 340/AB

Rearranging the equation, we get:

AB = 340/tan(75°)

Using a calculator, we find that AB ≈ 462.215 miles.

Therefore, the satellite is approximately 462.215 miles away from Los Angeles.

[tex]\dfrac{\mathrm dx}{\mathrm dt}=-5y[/tex]

[tex]\dfrac{\mathrm dy}{\mathrm dt}=-5x\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}=-5\dfrac{\mathrm dx}{\mathrm dt}[/tex]

[tex]\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}-25y=0[/tex]

This ODE is linear in [tex]y(t)[/tex] with the characteristic equation and roots

[tex]r^2-25=0\implies r=\pm5[/tex]

so that

[tex]y(t)=C_1e^{5t}+C_2e^{-5t}[/tex]

Then

[tex]\dfrac{\mathrm dx}{\mathrm dt}=-5C_1e^{5t}-5C_2e^{-5t}[/tex]

[tex]\implies x(t)=-C_1e^{5t}+C_2e^{-5t}[/tex]

Given that [tex]x(0)=1[/tex] and [tex]y(0)=3[/tex], we find

[tex]\begin{cases}1=-C_1+C_2\\3=C_1+C_2\end{cases}\implies C_1=1,C_2=2[/tex]

and the particular solution to this system is

[tex]\begin{cases}x(t)=-e^{5t}+2e^{-5t}\\y(t)=e^{5t}+2e^{-5t}\end{cases}[/tex]

The value of x and y would be [tex]x(t) =-e^{5t}+ 2e^{-5t}\\\\[/tex] and [tex]y(t) = e^{5t}+ 2e^{-5t}[/tex].

An equation containing derivatives of a variable with respect to some other variable quantity is called differential equations.

The derivatives might be of any order, some terms might contain product of derivatives and the variable itself, or with derivatives themselves. They can also be for multiple variables.

we have the following differential equations

[tex]\dfrac{dx}{dt} =-5y\\\\\dfrac{dy}{dt}=-5x[/tex]

by differentiating the second equation we have

[tex]\dfrac{d^2y}{dt^2}=-5\dfrac{dx}{dt}\\\\\dfrac{d^2y}{dt^2}-25y = 0[/tex]

[tex]r^2 - 25 = 0\\\\r = \pm5[/tex]

[tex]y(t) = C_1e^{5t}+ C_2e^{-5t}[/tex]

and by using the characteristic polynomial

[tex]x(t) =-e^{5t}+ 2e^{-5t}\\\\y(t) = e^{5t}+ 2e^{-5t}[/tex]

Learn more about differential equations;

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Answer:

[tex]P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}[/tex]

Step-by-step explanation:

This is a typical example where the Poisson distribution is a good choice to model the situation.

In this case we have an interval of time of 50 milliseconds as average time for the server to address one request and 50 requests per second.  

By cross-multiplying we determine the expected value of requests every 50 milliseconds.  

We know 1 second = 1,000 milliseconds

50 requests __________ 1000 milliseconds

 x requests __________ 50 milliseconds

50/x = 1000/50 ===> x = 2.5  

and the expected value is 2.5 requests per interval of 50 milliseconds.

According to the Poisson distribution, the probability of k events in 50 milliseconds equals

[tex]\bf P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}[/tex]

Answer: this corresponds to a chi-square test for Goodness of Fit.

Step-by-step explanation:

Goodness of fit Chi-squared test gives confirmation on whether a separately observed frequency distribution differs from a theoretical or general distribution.

In this case, the researcher studies a fraction of income earners - low income earners. He is interested in knowing whether the distribution of pay of low income earners is different from the distribution of pay of ALL income earners.

This survey hence corresponds to a chi-square test for goodness of fit.

In light diffraction, fringe separation is inversely proportional to the slit width. When the slit width in an experiment is doubled, it should result in the separation between the dark fringes on the screen being halved. Therefore, if the original fringe separation was 10 mm, the new separation should be 5 mm.

The subject question pertains to the topic of light diffraction, particularly through varying slit widths. The scenario described is a single slit diffraction experiment involving an argon laser with a specific wavelength. In diffraction, the fringe separation is inversely proportional to the slit width. This is because the angle of diffraction is determined by the wavelength of light divided by the slit width, according to the formula sinθ = λ/D, where θ is the diffraction angle, λ is the wavelength, and D is the slit width.

Therefore, when the slit width is doubled, the diffraction angles for the minima (dark fringes) will be halved assuming all other conditions remain the same. Consequently, the separation between the dark fringes on the screen will also be halved. So, if the original separation was 10 mm, the new separation when the slit width is doubled should be 5 mm.

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Answer:

b. (1.04, 1.68)

Step-by-step explanation:

Hello!

With the given data I've estimated the regression line

Y: Sales

X: Year

Yi= -2691.32 + 1.36Xi

Where

a= -2691.32

b= 1.36

The 95% CI calculated using the statistic b±[tex]t_{n-1;1-\alpha/2}[/tex]*(Sb/√n) is [1.04;1.68]

I hope it helps.

To find the 95% confidence interval for the slope of a regression line, compute the standard error of the slope, multiply it by the critical value from the t-distribution, and subtract and add the margin of error to the estimated slope. In this case, the 95% confidence interval for the slope is (0.364, 1.676).

To find the 95% confidence interval for the slope of the regression line, you need to compute the standard error of the slope. This can be done using the formula: SE = sqrt(SSE/(n-2)) / sqrt(SSX), where SSE is the sum of squared errors, n is the number of data points, and SSX is the sum of squares of the independent variable. Once you have the standard error, you can multiply it by the critical value from the t-distribution with (n-2) degrees of freedom to find the margin of error. Finally, subtract and add the margin of error to the estimated slope to get the lower and upper bounds of the confidence interval.

In this case, the slope of the regression line was estimated to be 1.02. You determined that the standard error of the slope is 0.255. By referring to the t-distribution table or using statistical software, you find the critical value for a 95% confidence interval with (n-2) degrees of freedom to be approximately 2.571. Therefore, the margin of error is 2.571 * 0.255 = 0.656. Finally, you subtract and add the margin of error to the estimated slope to get the lower and upper bounds of the confidence interval: 1.02 - 0.656 = 0.364 and 1.02 + 0.656 = 1.676. So the 95% confidence interval for the slope is (0.364, 1.676).

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