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3 years ago

WHOEVER ANSWERS CORRECT WILL GET BRAINLIST

Mathematics

1 answer:

Luba_88 [7]3 years ago

3 0

Answer: 119%

Step-by-step explanation: Remember that a percent is a ratio that compares a number to 100.

Here, since our first grid has 100 out of 100

boxes shaded, it represents 100%.

Notice however that our second grid has 19 out of 100

boxes shaded so it represents 19%.

Finally, adding our percents together, we have 100% + 19% or 119%.

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A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimat

Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A. Yes, using the additional survey information from part (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

For this problem, we consider that we want it to be within 4%.

Item a:

The sample size is n for which M = 0.04.

There is no estimate, hence $\pi = 0.5$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2$

$n = 600.25$

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is $\pi = 0.96$ , hence:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}$

$\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2$

$n = 92.2$

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to $\pi = 0.5$ , the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151

8 0

2 years ago

. can someone help pls

disa [49]

Answer:

The answer is 3,906.25

Step-by-step explanation:

All you have to do is divide 15,625 by 4! :)

6 0

2 years ago

Two friends share 1/3 of a pitcher of

Oxana [17]

Answer:

⅙ pitcher

Step-by-step explanation:

½ of ⅓ pitcher = ½ × ⅓ pitcher = ⅙ pitcher

7 0

2 years ago

How do I find the answer for 1/2 of 3/5

Tems11 [23]

The answer is 0.3. To find this answer, 1/2 / 3/5 = 56 ≅ 0.8333333

7 0

3 years ago

Find the probability of picking 4 consonant and 1 vowels when 5 letters are picked (without replacement) from a set of alphabet

Ganezh [65]

The required probability is 0.45

We have total 26 letters in english, of which 21 are consonants and 5 are vowels.

According to the question, we have to select 5 of them of which 4 will be consonants and 1 will be vowel.

Suppose we want to have the vowel in the first selection, so the probability of picking a vowel is equal to the quotient between the number of vowels and the number of total number of letters.

So the probability is 5/26

Now, a letter has been selected, so in the set, we have 25 letters left.

In the next 4 selections, we must select consonants.

In the second selection the probability is 21/25

In the third selection the probability is 20/24

In the fourth selection the probability is 19/23

In the last selection the probability is 18/22

So, the probability is (5/26)×(21/25)×(20/24)×(19/23)×(18/22)

Now, remember that we take that the vowel must be in the first place, but it can be in any five places, so if we add the permutations of the vowel letter we have,

P = 5×(5/26)×(21/25)×(20/24)×(19/23)×(18/22) = 0.45

Learn more about probability here :

brainly.com/question/251701

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1 year ago