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2 years ago

Sam typed 570 words in 15 minutes. if he continues to type at this rate, how many words will he have typed in 2 hours?

2 answers:

5 0

What we can do here is take the number 570 and divide it by 15 to get a unit value. 570/15 =38. Sam can type 38 words per minute. So there are 120 minutes in 2 hours, so multiply the amount per minute (38) by 120. 38*120 is 4560. Answer 2.

alisha [4.7K]2 years ago

4 0

Hello! Well first what we can do is take the amount minutes and make it into two hours. Two hours equate to 120 minutes, dividing that by 15 minutes will give us 8. Now we take 8 and multiply that by the amount of words he's typed in those 15 minutes, which means that he will type 4,560 words in two hours.

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From 1000, create a pattern by subtracting 8 from each number and stop at 5 numbers

shusha [124]

Well subtract 8 from 1000 and then do that five times. so 992,984,976,968, and 960

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2 years ago

Enjoy the points :D eeeeeeeeeeeeee

balandron [24]

The answer is 4/8 because half of 16 is 8 and half of 8 is 4, therefore they are equivalent

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2 years ago

Read 2 more answers

A researcher finds that of 1000 people who said that they attend a religious service at least once a week, A stopped to help a p

Ne4ueva [31]

Answer:

There is enough evidence to support the claim that the proportions are not equal. (P-value: 0.048).

Step-by-step explanation:

The question is incomplete:

"A researcher finds that of 1000 people who said that they attend a religious service at least once a week, 31 stopped to help a person with car trouble. Of 1200 people interviewed who had not attended a religious service at least once a month, 22 stopped to help a person with car trouble. At the 0.05 significance level, test the claim that the two proportions are different."

This is a hypothesis test for the difference between proportions.

The claim is that the proportions are not equal.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=1000 has a proportion of p1=0.031.

$p_1=X_1/n_1=31/1000=0.031$

The sample 2, of size n2=1200 has a proportion of p2=0.018.

$p_2=X_2/n_2=22/1200=0.018$

The difference between proportions is (p1-p2)=0.013.

$p_d=p_1-p_2=0.031-0.018=0.013$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{31+22}{1000+1200}=\dfrac{53}{2200}=0.024$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.024*0.976}{1000}+\dfrac{0.024*0.976}{1200}}\\\\\\s_{p1-p2}=\sqrt{0+0}=\sqrt{0}=0.007$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.013-0}{0.007}=\dfrac{0.013}{0.007}=1.98$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(z>1.98)=0.048$

As the P-value (0.048) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportions are not equal.

6 0

2 years ago

At the Science Museum, there is an exhibition with 150 displays about energy.

tatuchka [14]

1/5

This is a matter of elimination. So let's figure out how many of each display there is. 30% of 150 = 0.30 * 150 = 45 displays about electricity. 2/5 * 150 = 0.40 * 150 = 60 displays about renewable energy 1/10 * 150 = 150/10 = 15 displays about gas. The rest of the displays are about nuclear power. So 150 - 45 - 60 - 15 = 30 displays about nuclear power. Since we want a fraction, we get 30/150 But 30/150 isn't in it's simplest form. The largest number that will evenly divide both 30 and 150 is 30. So divide both the numerator and denominator by 30 (30/30)/(150/30) = 1/5 So 1/5 of the displays are about nuclear power.

8 0

2 years ago

Which expression represents -8x2 + 54x + 140 in factored form?

ra1l [238]

Answer:

-2(x+2)(4x-35)

Step-by-step explanation:

Factor out the 2 and by grouping.

3 0

2 years ago