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3 years ago

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Jessie spent $10 on 3 hair accessories. Which point represents this relationship? On a coordinate plane, a graph titled Jessie's

Spending has Number of Hair Accessories on the x-axis and Dollars spent on the y-axis. Point A is (3, 10), point B is (3, 20), point C is (5, 3), and point D is (10, 3). A B C D

2 answers:

AlladinOne [14]3 years ago

4 0

Answer:

A. (3, 10)

Step-by-step explanation:

You are told that ...

(x, y) = (number of accessories, dollars)

and you are told that ...

number of accessories = 3

dollars = 10

so the ordered pair is ...

(x, y) = (3, 10) . . . . Point A

Ann [662]3 years ago

4 0

Answer:

its A :)

Step-by-step explanation:

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Alex777 [14]

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3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Contact [7]

Answer:

(A) A 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.

(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

Step-by-step explanation:

We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $390

s = sample standard deviation = $120

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ; </u>

P(-2.101 < $t_1_8$ < 2.101) = 0.95 {As the critical value of t at 18 degrees of

freedom are -2.101 & 2.101 with P = 2.5%}

P(-2.101 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.101) = 0.95

P( $-2.101 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u> 95% confidence interval for</u> $\mu$ = [ $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }$ , $\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }$ ]

= [$332.16, $447.84]

(A) Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would increase because of an increase in the z value.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would decrease because as denominator increases; the whole fraction decreases.

(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion students who purchase their used textbooks = $\frac{210}{500}$ = 0.42

n = sample of students = 500

p = population proportion

<em>Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions</em>

<u>So, 99% confidence interval for the population proportion, p is ; </u>

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

<u> 99% confidence interval for</u> p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ , $0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ ]

= [0.363, 0.477]

Therefore, a 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

8 0

3 years ago

Find the value of each expression for n = 2.

belka [17]

Answer:

1. 22

2. 22

3. 28

Step-by-step explanation:

Substitue the n in each expression with two, then solve the problem for the value

5 0

3 years ago

Which expressions have a value of 1?

Anastaziya [24]

Answer:

b)4^0

d)4^3/4^3

Step-by-step explanation:

a)4^-1

1/4 = .25

b)4^0

Anything to the zero power is 1

1

c)4^1

4

d)4^3/4^3

This is a/a = 1

a

7 0

3 years ago

How do you solve for x? 3 + 2x = 4y

Dimas [21]

Answer:

x= 2y - 3/2

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

8 0

3 years ago

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