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2 years ago

Plsss helpppcan you pls do this?

Mathematics

1 answer:

Aleks04 [339]2 years ago

7 0

568 and 836 hope that helps I don’t really know

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Wgich value is closest to 85% of 25..is it20,16,18, or 22

olya-2409 [2.1K]

Well 85% of 25 is 21.25. So the closest ones would be 20 or 22. Well let's check out the difference's. 21.25-20= 1.25.....22-21.25=.75. well clearly .75 is smaller than 1.25 so your answer would be 22. 85% of 25 is closest to 22.

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3 years ago

What is 20 - {4 + [4 + (10/2)]}

vampirchik [111]

Answer:

Step-by-step explanation:

Plugged it in the calc!

Hope it helps :)

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3 years ago

In a recent survey, 78% of students at Lincoln Middle School said they plan to attend a camp this summer. Write this percent as

prohojiy [21]

Answer:

Step-by-step explanation:

From the question, we are to represent 78% as a decimal

78% = 78/100

78/100 = 78 * 1/100

78/100 = 78 * 0.01

78/100 = 0.78

<em>hence the percent written as a decimal is 0.78</em>

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3 years ago

Theo is using this spinner to predict the favorite movie genre of randomly chosen students in his school. If Theo spins the spin

marusya05 [52]

Answer:

Option B.

Step-by-step explanation:

In the question the number categories on the spinner are not given.

Consider there are 5 different categories on the spinner.

A = Spinner land on "Action".

So, probability of land on "Action" is

$P(A)=\frac{1}{5}$

Number of spins, n= 60 times

The expected number of land on "Action" is

$\text{Expected value}=n\times P(A)$

$\text{Expected value}=60\times \frac{1}{5}$

$\text{Expected value}=12$

Therefore, the correct option is B.

3 0

3 years ago

Read 2 more answers

What sample size is needed to give a margin of error within plus-or-minus 5% in estimating a population proportion with 90% conf

VARVARA [1.3K]

Answer: 271.

Step-by-step explanation:

When the prior population proportion of success is not available, then the formula to find the sample size is given by :-

$n=00.25(\dfrac{z_{\alpha/2}}{E})^2$

Given : Significance level : $\alpha: 1-0.90=0.1$

By using the standard normal distribution table ,

Critical value : $z_{\alpha/2}=1.645$

Margin of error : $E=0.05$

Then , the required minimum sample size will be :-

$n=0.25(\dfrac{(1.645)}{0.05})^2=270.6025\approx271$

Hence, the required minimum sample size is 271.

4 0

3 years ago

Plsss helpppcan you pls do this?​

Plsss helpppcan you pls do this?