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makkiz [27]
3 years ago
5

Most people think babies are equally likely to come as either a boy or a girl. This is not true. Actually, about 51.3% of all ba

bies are boys. If a family has two children (not twins), what is the chance they have one boy and one girl
Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
4 0

Answer:

0.499662

Step-by-step explanation:

Now since 51.3% of all babies are boys, this means that the probability of having a baby boy P(b) = 51.3/100 = 0.513

Since there are only two options, you are either a boy or a girl. The Probability of having a girl child P(g) = 1 - P (b) = 1 - 0.513 = 0.487

Now, we proceed to calculate the probability.

The probability of having one boy and one girl = probability of one boy and one girl or probability of one girl and one boy

Or in probability means that we have to add the two possibilities together.

P = (0.513*0.487) + (0.487 * 0.513) = 0.499662

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3 years ago
Is 11/128 equal to a terminating decimal or a repeating decimal ? Explain how you know
Ostrovityanka [42]

We need to determine whether \frac{11}{128} is a terminating decimal or a repeating decimal.

Let's solve this question using the long division method

First, let's identify the divisor and dividend. The number to be divided is 11 hence this is the dividend, and it needs to be divided by 128 which is the divisor

Next, since the divisor (128) is greater than the dividend (11) it can not divide 11. Hence, we will introduce a decimal point in quotient, and append a 0 next to 11 and divide 110 by 128. Again, 128 is greater than 110 so we will introduce a 0 in the quotient, and append another 0 next to 110, and will divide 1100 by 128. We will see what multiple of 128 is less than or equal to 1100. That multiple is 8. So we write 8 in the quotient and multiply 128 with 8 and subtract the product (128*8 = 1024) from 1100. The remainder that we get is 76.

Next, we append a 0 to the remainder and divide 760 by 128. Now, we see what multiple of 128 is less than or equal to 760. That multiple is 5. So we write 5 next to the quotient and multiply 128 with 5 and subtract the product (640) from 760. Now, the remainder is 120.

Next, we append a 0 to the remainder and divide 1200 by 128. Now, we see what multiple of 128 is less than or equal to 1200. That multiple is 9. So we write 9 next to the quotient and multiply 128 with 9 and subtract the product (1152) from 1200. Now, the remainder is 48.

Next, we append a 0 to the remainder and divide 480 by 128. Now, we see what multiple of 128 is less than or equal to 480. That multiple is 3. So we write 3 next to the quotient and multiply 128 with 3 and subtract the product (384) from 480. Now, the remainder is 96.

Next, we append a 0 to the remainder and divide 960 by 128. Now, we see what multiple of 128 is less than or equal to 960. That multiple is 7. So we write 7 next to the quotient and multiply 128 with 7 and subtract the product (896) from 960. Now, the remainder is 64.

Next, we append a 0 to the remainder and divide 640 by 128. Now, we see what multiple of 128 is less than or equal to 640. That multiple is 5. So we write 5 next to the quotient and multiply 128 with 5 and subtract the product (640) from 640. Now, the remainder is 0.

Hence, we have solved the entire problem

Last, we look at the quotient i.e. 0.0859375, which is the solution to the problem. We see that the quotient has a definite number of digits in it, and terminates at 5. Hence, this is a terminating decimal.

A repeating decimal is one in which a particular pattern after the decimal point keeps re-occuring, which is not the case here. Hence, \frac{11}{128} is a terminating decimal.

Please refer to the attached image for visualization

3 0
3 years ago
Read 2 more answers
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