SOMEONE HELP ME PLEASE

1.Points A and B are to be mapped onto a number line according to two equations. The solution to the equation is the coordinate

ohaa [14]

.Kim is x years old. Jordan is 7 years older than Kim. Four times Jordan’s age is equal to 200. (a)Write an equation that could be used to solve for Kim’s age, x. (b)How old is Kim? (c)How old is Jordan? Answer:

Kim = x
Jordan = x + 7

4(x+7) = 200
4x + 28 = 200
4x = 200 - 28
4x = 172
x = 172/4
x = 43. Kim's age

Jordan: x + 7 → 43+7 = 50

5 0

3 years ago

Read 2 more answers

(1 point) Suppose we want a 95% confidence interval for the average amount spent on books by freshmen in their first year at col

Ne4ueva [31]

Answer:

a) $n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40$

So the answer for this case would be n=40 rounded up to the nearest integer

b) For this case if we see the formula for the margin of error

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

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

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

$\sigma=16$ represent the population standard deviation

n represent the sample size

Solution to the problem

Part a

The margin of error is given by this formula:

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

And on this case we have that ME =5 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)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40$

So the answer for this case would be n=40 rounded up to the nearest integer

Part b

For this case if we see the formula for the margin of error

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

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

5 0

3 years ago

Statistics 4. Hol (a) The average age of 5 students is 9 years. Out of them, the ages of 4 students are 5, 7, 8, and 15 years. W

olga nikolaevna [1]

Answer:

10 years

Step-by-step explanation:

the average or mean value is the sum of all data points divided by the number of data points.

x = the unknown age of the 5th student.

average = 9 = (5 + 7 + 8 + 15 + x)/5

45 = 5 + 7 + 8 + 15 + x = 35 + x

x = 45 - 35 = 10

the 5th student is 10 years old.

6 0

2 years ago

The sum of three consecutive odd numbers is twenty-three find the odd numbers?

Kazeer [188]

Is it 5+7+11 ???????????????????????????

8 0

2 years ago

A survey of 132 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take

ddd [48]

Answer:

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 66/132 = 0.50

Number of samples n = 132

Confidence level = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

0.50 +/- 1.645√(0.50(1-0.50)/132)

0.50 +/- 1.645√(0.001893939393)

0.50 +/- 0.071589436011

0.50 +/- 0.072

(0.428, 0.572)

The 90% confidence level estimate of the true population proportion of students who responded "yes" is (0.428, 0.572)

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

7 0

3 years ago