What is the midpoint of a segment whose endpoints are (–4,

There are 2.54 centimeters to 1 inch. How many are in 15 inches

Elden [556K]

There are 38.1 centimeters in 15 inches. All you have to do is multiply 2.54 by 15 to get your answer.

8 0

3 years ago

A bag contains 4 red balls and 6 blue balls. If 3 balls are selected at random, find the probability of selecting 3 red balls?

Paha777 [63]

The probability is 4⋅3⋅2⋅1/10•9•8•7=(4/4)(6/0)/(10/4)=1/210 so the probability of selecting 3 red balls would be 1/210

3 0

2 years ago

I'LL MAKR U BRAINLIEST AND GIVE YOU 50 POINTS IF YOU ANSWER THIS ASAP PLEASEEE! THANK U SO MUCH! PIC IS INCLUDED.

solong [7]

Answer:

15 containers

Step-by-step explanation:

If the circus sells an average of 28,200 grams of peanuts you can expect that it would be close to that number again, it is equal to 28.2 kilograms. Since it only ships in 2 kilogram containers then you would want to buy 15 of them since 15*2=30, which is above what we need.

4 0

3 years ago

Read 2 more answers

Abby, Billy and Cathy fire one shot each at a target. The probability that Abby will hit the target is 1/5. The probability that

Tems11 [23]

Answer:

1. 0.0167

2. 0.2

3. 0.6

Step-by-step explanation:

Probabilities on Independent Events

If A and B are independent events (the occurrence of A doesn't affect the occurrence of B and vice-versa), then the probability that both events occur is

$\displaystyle P(A\bigcap B)= \ P(A).P(B)$

Being P(A) and P(B) the individual probability of each independent event

The probability that A does not occur is

$\displaystyle P(\bar{A})=1-P(A)$

The probability that B does not occur is

$\displaystyle P(\bar{B})=1-P(B)$

The probability that C does not occur is

$\displaystyle P(\bar{C})=1-P(C)$

We have 3 independent events. We know that because they fire together, no mutual affectation can happen

.

The probability that Abby will hit the target is 1/5.

$\displaystyle P(A)=\frac{1}{5}$

The probability that Billy will hit the target is 1/4.

$\displaystyle P(B)=\frac{1}{4}$

The probability that Cathy will hit the target is 1/3

$\displaystyle P(C)=\frac{1}{3}$

Part 1.

The probability that all three shots hit the target is

$\displaystyle P(A\bigcap B\bigcap C)=\frac{1}{5}.\frac{1}{4}.\frac{1}{3}$

$\displaystyle P(A\bigcap B\bigcap C)=\frac{1}{60}$

The probability that all three shots hit the target is

$\displaystyle P= \frac{1}{60}=0.0167$

Part 2.

The probability that only Cathy's shot hits the target is computed assuming Abby and Billy won't succeed

.

$\displaystyle P(\bar{A}\bigcap \bar{B}\bigcap C)=P(\bar{A})\ P(\bar{B})\ P(C)=(1-\frac{1}{5})\ (1-\frac{1}{4})\ (\frac{1}{3})=\frac{4}{5}.\frac{3}{4}.\frac{1}{3}=\frac{1}{5}=0.2$

3.

The probability that at least one shot hits the target is when one of them succeeds, two of them succed or all of them succeed

$\displaystyle P(A\bigcap \bar{B}\bigcap \bar{C})+P(\bar{A}\bigcap {B}\bigcap \bar{C})+P(\bar{A}\bigcap\bar{B}\bigcap C)+P(A\bigcap B\bigcap \bar{C})+P(A\bigcap \bar{B}\bigcap C)+P(\bar{A}\bigcap B\bigcap C)+P(A\bigcap B\bigcap C)$