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3 years ago

Alice rolls a fair dice and flips a fair coin. What is the probability of obtaining a factor of 12 and a tail?

Mathematics

1 answer:

Semenov [28]3 years ago

4 0

Answer:

5/12

Step-by-step explanation:

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8 5/6 + 2 3/4 +__=14 this is fractions so the answer should be a fraction please help

Mariulka [41]

Answer:

2 5/12

Step-by-step explanation:

8 5/6 + 2 3/4 = 8 10/12 + 2 9/12 = 10 19/12 = 10 + 1 7/12 = 11 7/12

13 12/12 - 11 7/12 = 2 5/12

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Step-by-step explanation:

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2 years ago

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2 years ago

A store offered a 20% price discount on all items on Friday. By what fraction of the discounted price of an item did the price i

grin007 [14]

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What were two of the important points found in the Monroe Doctrine?

All European nations had to give up their colonies and allow for individual autonomy.

The United States would not allow additional colonization of Latin American nations.

America would not interfere with already established Latin American colonies.

All trade agreements with Latin American nations had to go through the United States.

Step-by-step explanation:

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2 years ago

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%.

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3 years ago