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1 year ago

Of india’s more than 1.2 billion population, about _____ percent are muslims.

Mathematics

1 answer:

beks73 [17]1 year ago

8 0

Fifteen percent of Muslims live in India

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(3-p)^2 How to solve this steps

Zanzabum

Answer:

Look at the attachment

7 0

3 years ago

If the Leaning Tower of Pisa (which is roughly cylindrical) has a height of 55 meters, radius of 8 meters, what is the volume?

valentinak56 [21]

Answer:

Option C. $V=3,520\pi\ m^3$

Step-by-step explanation:

we know that

The volume of a cylinder is equal to

$V=\pi r^{2} h$

we have

$r=8\ m$

$h=55/ m$

substitute

$V=\pi (8)^{2} (55)$

$V=3,520\pi\ m^3$

8 0

3 years ago

Read 2 more answers

spayn [35]

Answer:

1) y= -.5

2)y= 7

3)x= -2

4 0

3 years ago

Simplify the expression 2(3+d-1)=

Alex73 [517]

Answer:

4+2d

Step-by-step explanation:

2(3+d-1)

6+2d-2=6+(-2)+2d

=4+2d

3 0

2 years ago

A sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's p

Grace [21]

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, $H_0$ : p $\leq$ 0.29 {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, $H_1$ : p > 0.29 {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

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

where, $\hat p$ = proportion of chips that do not fail in the first 1000 hours of their use = 32%

n = sample of chips = 1500

So, test statistics = $\frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }$

= 2.491

Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

7 0

3 years ago