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2 years ago

12

Kristoff ran more than 16 miles. Kristoff wants to write an inequality for the number of miles he ran. What constant should he u

se in the inequality?

2 answers:

Levart [38]2 years ago

7 0

Hope this helps you ;)

Elden [556K]2 years ago

3 0

its 16, btw i dont really know much about this but im 100 percent is 16

Hope it helps

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A rectangle is transformed according to the rule R0, 90º. The image of the rectangle has vertices located at R'(–4, 4), S'(–4, 1

mars1129 [50]

<span>(4, 3) Is your answer. </span>

6 0

1 year ago

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Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

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 of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

1 year ago

A survey of the men in a certain town showed that 35% of the men have blond hair and 14% of them have blond hair and blue eyes.

natita [175]

If you are given he has blond hair, so the total probability will be 35%. And among them, he needs to have blue eyes, its probability is 14% among the 35%. So the final probability is 14%/35%=40%.

7 0

2 years ago

A quarter is approximately 0.07 inches thick. A dollar bill is approximately 4.3 × 10-3 inches thick.

nlexa [21]

Answer:

m = 6.57 and n = -2

Step-by-step explanation:

To answer this question you will covert the thickness of a dollar to standard notation and then compare the two values.

0.07 =compared to 0.0043 = 0.0657

0.0657 = 6.57 x 10^-2

A quarter is thicker than a dollar by 6.57 x 10^-2.

m = 6.57 and n = -2

8 0

1 year ago

Determine the number of atoms in<br> 98.3 g mercury, Hg

Arisa [49]

The answer is 2.95 × 10²³ atoms

Atomic mass is 200.59 g.
So, 1 mole has 200.59 g. Let's calculate how many moles have 98.3 g:
1M : 200.59g = x : 98.3g
x = 98.3 g * 1 M : 200.59 g = 0.49 M

To calculate this, we will use Avogadro's number which is the number of units (atoms, molecules) in 1 mole of substance:

6.023 × 10²³ atoms per 1 mole
<span>How many atoms are in 0.49 mole:
</span>6.023 × 10²³ atoms : 1M = x : 0.49M
x = 6.023 × 10²³ atoms : 1M * 0.49M = 2.95 × 10²³ atoms

8 0

2 years ago

Read 2 more answers