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2 years ago

How do you write 110 percent as a decimal

Mathematics

1 answer:

ANTONII [103]2 years ago

7 0

Answer:

1.1

Step-by-step explanation:

Divided it by 100 :)

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muminat

Perimeter: 26
Area: 32

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2 years ago

PLEASE HELP h(1) = and h(4) =

Alekssandra [29.7K]

Putting in h(1) would get you 1, putting in h(4) would get you 5.

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3 years ago

Use the sine ratio to find the variable

jek_recluse [69]

Answer: 6.8829172362
Explanation:
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3 years ago

11. Use the following information for problems a-b. The table shows the results of a school lunch survey. In the

Bas_tet [7]

Using it's concept, the probabilities are given as follows:

a) C. 7/20.

b) D. 15/22.

<h3>What is a probability?</h3>

A probability is given by the number of desired outcomes divided by the number of total outcomes.

For item a, we have that there are 60 students with no curfew, and of those, 21 have no chores, hence the probability is given by:

p = 21/60 = 7/20

Which means that option C is correct.

For item b, we have that there are 66 students with no shores, and of those, 45 have a curfew, hence the probability is given by:

p = 45/66 = 15/22

Which means that option D is correct.

More can be learned about probabilities at brainly.com/question/14398287

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6 0

1 year ago

In a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer

stira [4]

Answer:

We conclude that 80% of patients stop smoking when given sustained care.

Step-by-step explanation:

We are given that in a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer smoking after one month.

Let p = percentage of patients stop smoking when given sustained care.

So, Null Hypothesis, $H_0$ : p = 80% {means that 80% of patients stop smoking when given sustained care}

Alternate Hypothesis, $H_A$ : p $\neq$ 80% {means that different from 80% of patients stop smoking when given sustained care}

The test statistics that would be used here One-sample z test for proportions;

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

where, $\hat p$ = sample proportion of patients who stop smoking when given sustained care = 80.2%

n = sample of patients = 192

So, the test statistics = $\frac{0.802-0.80}{\sqrt{\frac{0.802(1-0.802)}{192} } }$

= 0.07

The value of z test statistics is 0.07.

Also, P-value of the test statistics is given by the following formula;

P-value = P(Z > 0.07) = 1 - P(Z $\leq$ 0.07)

= 1 - 0.52790 = 0.4721

Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 80% of patients stop smoking when given sustained care.

8 0

3 years ago