A girl weighs 5/8 as much as her brother does. If the girl weighs 110 lb, how much does her brother weigh?

A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. The number of subjects needed to estima

Ludmilka [50]

Answer:

$n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65$

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

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".

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

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

ME represent the margin of error

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

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

2) Solution to the problem

Since the new 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: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

Assuming that the deviation is known we can express the margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

Replacing into formula (b) we got:

$n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65$

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

8 0

3 years ago

Susan made 75 pieces of peanut brittle to share equally with her friends. She put 4 pieces of peanut brittle in each bag. She ga

BabaBlast [244]

Answer:

72/4=18 with remainder of 3 peices given to her mom

Step-by-step explanation:

6 0

3 years ago

Find the slope of the line: (4, 4), (5, 3)

fredd [130]

Answer:

-1

Step-by-step explanation:

we would use the slope formula which is y2-y2/x2-x1

so 3-4/5-4= -1

8 0

3 years ago

PLS ANSWER TODAY TYSM :p<br><br> -5y+1-3y+2=-9

OLga [1]

We just combine the terms, like this:

-8y = -12

The answer is: y = 1.5 or 3/2.

Hope this helped! c:

7 0

4 years ago

Greg has an MP3 player called the Jumble. The Jumble randomly selects a song for the user to listen to. Greg's Jumble has 6 clas

zhannawk [14.2K]

An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or …show more content…
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side.
The Moon takes 27.3 days (sidereal month) to complete its actual orbit around the Earth. Like the Sun, the Moon rises in the east and sets in the west each day.

7 0

3 years ago