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3 years ago

The mass

Mathematics

2 answers:

saul85 [17]3 years ago

5 0

Do you have a picture for this question??

7nadin3 [17]3 years ago

3 0

Step-by-step explanation:

Of?

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Which expression can be used to find the area of triangle RTS?

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answer is 13 but I can't understand the question

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3 years ago

Shira has 20 books on her bookshelf. Of those 15 are about horses, the rest are nature books. What fraction of Shiras books are

givi [52]

Answer:

1/4

Step-by-step explanation:

15 out of the 20 books are horses. 5 of those are nature. 5/20 is the fraction, but can be simplified to 1/4

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3 years ago

The coordinates of a rectangle are (4,0), (8,0), (4, 10) and (8, 10). Which transformation results in a reduction in the

Cerrena [4.2K]

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It all depends on the reduction, like by 2 it would shrink half the size or 3it would get smaller but not bt much. So all deoends

Step-by-step explanation:

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3 years ago

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Can someone please explain to me how to solve this problem? I have the answer already all I need is to know how to solve the pro

omeli [17]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

Suppose that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage

Zigmanuir [339]

Answer:

We conclude that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

We are given that in a random selection of 100 colored candies, 28% of them are blue. The candy company claims that the percentage of blue candies is equal to 29%.

Let p = population percentage of blue candies

So, Null Hypothesis, $H_0$ : p = 29% {means that the percentage of blue candies is equal to 29%}

Alternate Hypothesis, $H_A$ : p $\neq$ 29% {means that the percentage of blue candies is different from 29%}

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

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

where, $\hat p$ = sample proportion of blue coloured candies = 28%

n = sample of colored candies = 100

So, the test statistics = $\frac{0.28-0.29}{\sqrt{\frac{0.29(1-0.29)}{100} } }$

= -0.22

The value of the z-test statistics is -0.22.

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

P-value = P(Z < -0.22) = 1 - P(Z $\leq$ 0.22)

= 1 - 0.5871 = 0.4129

Now, at a 0.10 level of significance, the z table gives a critical value of -1.645 and 1.645 for the two-tailed test.

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

Therefore, we conclude that the percentage of blue candies is equal to 29%.

4 0

3 years ago