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natulia [17]
3 years ago
15

each day last week, Ms. wilson walked3/4 mile. what is the total distance, in miles, that Ms. Wilson walked in 4 days?

Mathematics
2 answers:
zloy xaker [14]3 years ago
6 0
Hey there! :D

Since she walks 3/4 a mile a day, we can multiply that by 4.

3/4= .75

4*.75= 3

She walked 3 miles in 4 days.

I hope this helps!
~kaikers
ad-work [718]3 years ago
4 0
The answer is 3 because 3/4 * 4 =3
———————————————————
3/4 * 4/1
3 times 4 equals 12 and 4 times 1 is 4 which means the answer is at this point 12/4 which can be simplified to 3 because 4 goes into 12, 3 times.


Hope this helps:)
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Naval intelligence reports that 99 enemy vessels in a fleet of 1818 are carrying nuclear weapons. If 99 vessels are randomly tar
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Answer:

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

Step-by-step explanation:

The vessels are destroyed and then not replaced, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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)!}

In this question:

Fleet of 18 means that N = 18

9 are carrying nuclear weapons, which means that k = 9

9 are destroyed, which means that n = 9

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

P(X \leq 1) = P(X = 0) + P(X = 1)

In which

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 0) = h(0,18,9,9) = \frac{C_{9,0}*C_{9,9}}{C_{18,9}} = 0.000021

P(X = 1) = h(1,18,9,9) = \frac{C_{9,1}*C_{9,8}}{C_{18,9}} = 0.001666

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.000021 + 0.001666 = 0.001687

0.001687 = 0.1687% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

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