Which of the following illustrates the truth value of the glven statements?

The University College is interested in the average number of hours per week that freshmen spend going to parties. They took a r

tekilochka [14]

Answer:

And the confidence interval is given by:

$4 \leq \mu \leq 8$

We can estimate the margin of error like this:

$ME = \frac{8-4}{2}= 2$

And the margin of error is given by:

$ME=z_{\alpha/2} SE$

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

And solving for the standard error we got:

$SE= \frac{2}{1.96}= 1.02$

Step-by-step explanation:

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

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

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

And the confidence interval is given by:

$4 \leq \mu \leq 8$

We can estimate the margin of error like this:

$ME = \frac{8-4}{2}= 2$

And the margin of error is given by:

$ME=z_{\alpha/2} SE$

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

And solving for the standard error we got:

$SE= \frac{2}{1.96}= 1.02$

5 0

3 years ago

Hey! i’ll give brainliest please help

netineya [11]

Answer:

The last one looks correct

Step-by-step explanation:

Because if they have stronger winds then that would improve their flight making them get use to it, and will for sure get stronger over time

8 0

2 years ago

Brad has a garden with dimensions 12 feet long x 7 feet wide. He wants to enlarge the garden to an area of 119 square feet. How

dimulka [17.4K]

Answer:

the length will need to be increased by 5 feet

8 0

2 years ago

The farmer graphs a step function to model the situation. Which of the following describes the farmer’s graph for the step that

marshall27 [118]

Missing data:
Almond Prices
Number of Pounds

Cost (per pound)
up to 10
$20

more than 10, and up to 20
$15

more than 20, and up to 30
$10

more than 30
$5

A) a horizontal line with an open circle at (20, 10) and a closed circle at (30, 10); $207.50
B) a horizontal line with a closed circle at (20, 10) and an open circle at (30, 10); $207.50
C) a horizontal line with an open circle at (21, 10) and a closed circle at (30, 10); $311.25
D) a horizontal line with a closed circle at (21, 10) and an open circle at (30, 10); $311.25

20.75 pounds is in the more than 20, and up to 30 ; $10 range.
y = 10(20.75)
y = 207.50

open circle means less than or greater than. close circle means "less than or equal to" or "greater than or equal to"

My answer : A) a horizontal line with an open circle at (20, 10) and a closed circle at (30, 10); $207.50

4 0

3 years ago

Find the solution of the differential equation that satisfies the given initial condition. (du)/(dt) = (2t + sec^2 t)/(2u), u(0)

Delicious77 [7]

Ah okay so in differential equations you usually want the top variable isolated. To do this, multiply by dt and 2u and you get
$2udu = 2t + { \sec(t) }^{2} dt$
Now just integrate both sides. The integral of 2u with respect to u is u². The integral of (2t + sec²(t) with respect to t is t² + ∫sec²(t)dt. The last part is just tan(x) because d/dt(tan(t)) is sec²(t) so just integrating gets us back. Now we have
${u}^{2} + c = {t}^{2} + \tan(t) + k$
Where c and k are arbitrary constants. Subtracting c from k and you get
${u}^{2} = {t}^{2} + \tan(t) + b$
Where b is another constant. To find b, just plug in u(0) = -1 where u is -1 and t is 0. This becomes
$1 = \tan(0) + b$
tan(0) is 0 so b = 1. Take the plus or minus square root on both sides and you finally get
$u = \: { {t}^{2} } + \tan(t) + 1$
But Brainly didn't let me do but juat remember there is a plus or minus square root on the left.

3 0

2 years ago