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ddd [48]
3 years ago
7

Suppose a couple decided to keep having children until they had a girl. Which

Mathematics
1 answer:
Serhud [2]3 years ago
8 0

Answer:

The couple having 3 children

Step-by-step explanation:

For each children, there is a 50% probability that it is a girl and a 50% probability that it is a boy.

Suppose a couple decided to keep having children until they have a girl.

The probability that they have 3 children, that is 2 boys before a girl is:

The desired outcome is two boys then a girl. Each of these outcomes has a 50% probability. So

The probability that they have 4 children, that is 3 boys before a girl is:

The desired outcome is three boys then a girl. Each of these outcomes has a 50% probability. So

The probability that they have 5 children, that is 4 boys before a girl is:

The desired outcome is four boys then a girl. Each of these outcomes has a 50% probability. So

The probability that they have 6 children, that is 5 boys before a girl is:

The desired outcome is five boys then a girl. Each of these outcomes has a 50% probability. So

Which of these is most likely?

The largest probability is the one of them having three children, that is, two boys before a girl.

So the correct answer is:

The couple having 3 children

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Answer:

1/2

Step-by-step explanation:

the rate of change is = the change in y/ the change in x

if the points are (2, 11) and (8, 14)  

Rate of change = (14-11) / (8-2) = 3 / 6 = 1/2

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2 years ago
Write the algebraic expression that matches each graph? PLS HELP I NEED HELP!!!
Vladimir [108]

Looks like y = |x| because the slopes of the lines are both 1

You shift it to the right making it |x-2| and you shift it down making y = -2

combine to get <em>y = |x - 2| - 2</em>.

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3 years ago
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a scanner scanned 72 photos in 8 minutes if it scans photos at a constant rate it can scan ___ photos in 23 minutes
Leni [432]
If it can scan 72 photos in 8 minutes, it can scan 72/8 = 9 photos per minute.
So in 23 minutes, it can scan 9x23 = 207 photos 
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2 years ago
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At the beach, Pancho and his sister both built sandcastles and then measured their heights. Pancho's sandcastle was 3/5 of a foo
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1/5 of a foot

Simply subtract the height of Pancho’s sandcastle (3/5 of a foot) minus the height of his sister’s sandcastle (2/5 of a foot) to find a difference of 1/5 of a foot, meaning Pancho’s sandcastle was 1/5 of a foot taller than his sister’s sandcastle.

7 0
3 years ago
A market research team thinks that their new ad campaign is better than their old one. They used to be able to sell to 50% of th
andreev551 [17]

Answer:

a. The test statistic is 2 and we conclude that the new ad campaign is not signficantly better.

Step-by-step explanation:

They used to be able to sell to 50% of those who saw their ads. Test if the new campaign is better.

At the null hypothesis, we test is it is the same, that is, the proportion is the same.

H_0: p = 0.5

At the alternate hypothesis, we test if it is significantly better, that is, the proportion is above 50%.

H_1: p > 0.5

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.5 is tested at the null hypothesis:

This means that \mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5

They take a random sample of 100 potential buyers and find that they convinced 60 of these people to buy their product.

This means that n = 100, X = \frac{60}{100} = 0.6

Test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.6 - 0.5}{\frac{0.5}{\sqrt{100}}}

z = 2

The test statistic is 2.

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion above 0.6, which is 1 subtracted the by p-value of z = 2.

Looking at the z-table, z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

The p-value of the test is 0.0228 > 0.01, which means that we cannot conclude that the new ad campaign is signficantly better, so the correct answer is given by option A.

5 0
3 years ago
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