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sveticcg [70]
2 years ago
10

Katrina randomly selected 64 students out of her school of 400 students and found that 40 of them want school to start at a late

r time.
With a desired confidence of 99%, which has a z*-score of 2.58, which statements are true? Check all that apply.

E = z*

The sample size is 40.
The sample size is 64.
The point estimate of the population proportion is 0.16.
The point estimate of the population proportion is 0.625.
The margin of error is approximately 12%.
The margin of error is approximately 16%.
Mathematics
2 answers:
SashulF [63]2 years ago
6 0

- The sample size is 64.

- The point estimate of the population proportion is 0.625.

-The margin of error is approximately 16%.

Svetllana [295]2 years ago
4 0

Answer:

Step-by-step explanation:

Given that Katrina randomly selected 64 students out of her school of 400 students and found that 40 of them want school to start at a later time.

Out of the statements given the true statements are :

1) The sample size is 64

2) The point estimate of the population is sample proportion =40/64 =5/8

=0.625

3) Margin of error = std error x 2.58

Std error = sq rt pq/n

p=0.625 and q = 0.375

Hence std error = 0.060

Margin of error = 0.156

Round off to 0.16 = 16%

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4. 19.8
56/4=14
Diagonal of a square is equal to one side multiplied by the square root of two.
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2 years ago
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Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145

5 0
2 years ago
Find the area of the polygon WXYZ with its vertices at W(–3, –2), X(–3, 5), Y(2, 5), and Z(2, –2).
ki77a [65]

Answer:

The area of the polynomial is 35 square unit.

Step-by-step explanation:

It is given that area of the polygon WXYZ with its vertices at W(–3, –2), X(–3, 5), Y(2, 5), and Z(2, –2).

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XY=5

YZ=7

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Using the Pythagoras theorem, the length of diagonal WY and XZ are

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XZ=\sqrt{(XY)^2+(YZ)^2}=\sqrt{5^2+7^2}=\sqrt{74}

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e.

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