(7a-1)^2 please help me quickly

qaws [65]

When I do what the problem statement says, I get 47° for the left angle and 58° for the right one. They are not congruent.

4 0

3 years ago

The following information is obtained from two independent samples selected from two normally distributed populations.

AysviL [449]

A. The point estimate of μ1 − μ2 is calculated using the value of x1 - x2, therefore:

μ1 − μ2 = x1 – x2 = 7.82 – 5.99

μ1 − μ2 = 1.83

B. The formula for confidence interval is given as:

Confidence interval = (x1 –x2) ± z σ

where z is a value taken from the standard distribution tables at 99% confidence interval, z = 2.58

and σ is calculated using the formula:

σ = sqrt [(σ1^2 / n1) + (σ2^2 / n2)]

σ = sqrt [(2.35^2 / 18) + (3.17^2 / 15)]

σ = 0.988297

Going back to the confidence interval:

Confidence interval = 1.83 ± (2.58) (0.988297)

Confidence interval = 1.83 ± 2.55

Confidence interval = -0.72, 4.38

3 0

3 years ago

The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in

Stolb23 [73]

Answer:

$ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75$

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

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".

$\bar X=34.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 represent the sample size

Solution to the problem

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=300-1=299$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that $t_{\alpha/2}=1.653$