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Lerok [7]
3 years ago
10

40 divided by 28! HURRY!!

Mathematics
1 answer:
Schach [20]3 years ago
3 0

Answer:

fraction~ 10/7

decimal~ 1.428571

mixed number~ 1 3/7

Step-by-step explanation: for decimal make sure to put the line over the 428571 because its an on going number

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Elena drove a total of 203 miles last week. She drove 28 miles during the week to run errands and 35 miles each day to go to wor
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Answer:

35×7=56 is a AWNSER djndhei

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2 years ago
Read 2 more answers
Tay–Sachs Disease Tay–Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carri
BARSIC [14]

Answer:

a) 0.0156 = 1.56% probability that all children will develop the disease.

b) 0.4219 = 42.19% probability that only one child will develop the disease.

c) 0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they carry the disease, or they do not. The probability of a children carrying the disease is independent of any other children, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The probability that their offspring will develop the disease is approximately .25.

This means that p = 0.25

Three children:

This means that n = 3

Question a:

This is P(X = 3). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{3,3}.(0.25)^{3}.(0.75)^{0} = 0.0156

0.0156 = 1.56% probability that all children will develop the disease.

Question b:

This is P(X = 1). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{3,1}.(0.25)^{1}.(0.75)^{2} = 0.4219

0.4219 = 42.19% probability that only one child will develop the disease.

c. The third child will develop Tay–Sachs disease, given that the first two did not.

Third independent of the first two, so just multiply the probabilities.

First two do not develop, each with 0.75 probability.

Third develops, which 0.25 probability. So

p = 0.75*0.75*0.25 = 0.1406

0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

6 0
3 years ago
A park is 4 times as long as its wide. If the distance around the park is12.5kilometers, what is the are of the park
Lilit [14]
Let x be the width, length=4x
(4x+x)×2×2=12.5
20x=12.5
x=0.625
Length = 4×0.625=2.5
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Find the area. Round your answer to the nearest hundredth 15m
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Mgs mg2 ag2 aq aq ags
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What is the property of -8+3 = 3+ (-8)
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Answer:

12

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its not 12 im just trying to make an account

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