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Zielflug [23.3K]
3 years ago
7

What is the value of y in the solution to the following system of equations?

Mathematics
2 answers:
dedylja [7]3 years ago
6 0

Answer:

{x,y} = {33/19,-2/19}

Step-by-step explanation:

mafiozo [28]3 years ago
4 0

Answer: For 5x-3y=-11, the answer is y=11/3 + 5x/3.

For 2x-5y=-14, the answer is y=14/5 + 2x/5

Step-by-step explanation: Move all terms that do not contain y to the right side and then solve.

I hope this helps you out! :)

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An economist uses the price of a gallon of milk as a measure of inflation. She finds that the average price is $3.82 per gallon
salantis [7]

Answer:

(a) The standard error of the mean in this experiment is $0.052.

(b) The probability that the sample mean is between $3.78 and $3.86 is 0.5587.

(c) The probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.

(d) The likelihood that the sample mean is greater than $3.92 is 0.9726.

Step-by-step explanation:

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

n=40\\\mu=\$3.82\\\sigma=\$0.33

As <em>n</em> = 40 > 30, the distribution of sample mean is \bar X\sim N(3.82,\ 0.052^{2}).

(a)

The standard error is the standard deviation of the sampling distribution of sample mean.

Compute the standard deviation of the sampling distribution of sample mean as follows:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

    =\frac{0.33}{\sqrt{40}}\\\\=0.052178\\\\\approx 0.052

Thus, the standard error of the mean in this experiment is $0.052.

(b)

Compute the probability that the sample mean is between $3.78 and $3.86 as follows:

P(3.78

                               =P(-0.77

Thus, the probability that the sample mean is between $3.78 and $3.86 is 0.5587.

(c)

If the difference between the sample mean and the population mean is less than $0.01 then:

\bar X-\mu_{\bar x}

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

Thus, the probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.

(d)

Compute the probability that the sample mean is greater than $3.92 as follows:

P(\bar X>3.92)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{3.92-3.82}{0.052})

                    =P(Z

Thus, the likelihood that the sample mean is greater than $3.92 is 0.9726.

3 0
3 years ago
Which equation is a point slope form equation for line AB ?
zepelin [54]
(-2,3)(4,-1)
slope = (3 + 1)/(-2 - 4) = 4/-6 = -2/3

equation
y + 1 = -2/3(x - 4)
7 0
3 years ago
What is 60 /−3 and how do you solve
SVETLANKA909090 [29]
The answer is -20 and you solve by putting it in the calculator
7 0
2 years ago
Read 2 more answers
True or False? The first distribution shown below has a smaller mean than
lozanna [386]

I'm not really sure because I have not done this in a while but I think True

Step-by-step explanation:

...

5 0
3 years ago
Read 2 more answers
Consider the following hypothesis test:
Gnom [1K]

Answer:

a. z=3.09. Yes, it can be concluded that the population mean is greater than 50.

b. z=1.24. No, it can not be concluded that the population mean is greater than 50.

c. z=2.22. Yes, it can be concluded that the population mean is greater than 50.

Step-by-step explanation:

We have a hypothesis test for the mean, with the hypothesis:

H_0: \mu\leq50\\\\H_a:\mu> 50

The sample size is n=55 and the population standard deviation is 6.

The significance level is 0.05.

We can calculate the standard error as:

\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{55}}=0.809

For a significance level of 0.05, the critical value for z is zc=1.644. If the test statistic is bigger than 1.644, the null hypothesis is rejected.

a. If the sample mean is M=52.5, the test statistic is:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{52.5-50}{0.809}=\dfrac{2.5}{0.809}=3.09

The null hypothesis is rejected, as z>zc and falls in the rejection region.

b. If the sample mean is M=51, the test statistic is:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51-50}{0.809}=\dfrac{1}{0.809}=1.24

The null hypothesis failed to be rejected, as z<zc and falls in the acceptance region.

c. If the sample mean is M=51.8, the test statistic is:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51.8-50}{0.809}=\dfrac{1.8}{0.809}=2.22

The null hypothesis is rejected, as z>zc and falls in the rejection region.

8 0
3 years ago
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