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3 years ago

22,647 inches into miles Please explain how you found the answer thank you

Mathematics

1 answer:

lys-0071 [83]3 years ago

7 0

Answer:

0.35743371 Miles

Step-by-step explanation:

Used a conversion calculator.

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Find the pressure exerted by a force of 240 N on an area of 30 cm2. Find the answer in SI unit

NeTakaya

Answer:

Pressure is 80,000 pascals.

Step-by-step explanation:

Area:

$= { \bf{ 30 \times {10}^{ - 4} \: {m}^{2} }}$

Pressure:

${ \bf{pressure = \frac{force}{area} }} \\ \\ { \tt{ = \frac{240}{30 \times {10}^{ - 4} } }} \\ = { \tt{80000 \: pascals}}$

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3 years ago

Please help a girl out:((((

oksian1 [2.3K]

The answerrrrr is 117/5.

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2 years ago

PLEASEEEE.. MY TEACHER IS ASKING FOR THE ANSWER

VARVARA [1.3K]

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3 years ago

Stephen's lunch bill is currently at $8.33. Stephen orders a fruit salad for take-out, and wants to leave $2.25 as a tip for his

Salsk061 [2.6K]

Answer:

$1.53

Step-by-step explanation:

The computation of the change in the bill is shown below:

Total money he has to pay is

= $8.33 + $2.25

= $13.47

And, he have

= $10 + $5

= $15

So, the change would be

= $15 - $13.47

= $1.53

6 0

3 years ago

A manager of a grocery store wants to determine if consumers are spending more than the national average. The national average i

strojnjashka [21]

The valid conclusions for the manager based on the considered test is given by: Option

<h3>When do we perform one sample z-test?</h3>

One sample z-test is performed if the sample size is large enough (n > 30) and we want to know if the sample comes from the specific population.

For this case, we're specified that:

Population mean = $\mu$ = $150
Population standard deviation = $\sigma$ = $30.20
Sample mean = $\overline{x}$ = $160
Sample size = n = 40 > 30
Level of significance = $\alpha$ = 2.5% = 0.025
We want to determine if the average customer spends more in his store than the national average.

Forming hypotheses:

Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get: $H_0: \mu_0 \leq \mu = 150$
Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically $H_1: \mu_0 > \mu = 150$

where $\mu_0$ is the hypothesized population mean of the money his customer spends in his store.

The z-test statistic we get is:

$z = \dfrac{\overline{x} - \mu_0}{\sigma/\sqrt{n}} = \dfrac{160 - 150}{30.20/\sqrt{40}} \approx 2.094$

The test is single tailed, (right tailed).

The critical value of z at level of significance 0.025 is 1.96

Since we've got 2.904 > 1.96, so we reject the null hypothesis.

(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).

Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.

Learn more about one-sample z-test here:

brainly.com/question/21477856

3 0

2 years ago

Read 2 more answers