Ask question

Ask question

All categories

2 years ago

12

Billy walks 15 miles every day to soccer practice, and catches a ride home from his grandma. Billy had 5 practices last week. Ho

w many yards did billy walk last week to practice.

1 answer:

kirza4 [7]2 years ago

8 0

Billy walked 75 yards last week

You might be interested in

Annie has two equal sized amounts of wax in two different colors she used 5/8 of the green wax and 1/3 of the blue wax to make s

lana66690 [7]

Find a common denominator that 8 and 3 can go into.. which is 24
5/8 = 1/3
15/24 - 8/24
7/24

Hope this helped

6 0

2 years ago

a 5 sided solid has the numbers 1,2,3,4, and 5. what is the probability of rolling two five-sided solids and getting a sum of ei

kipiarov [429]

So this is going to be alot of writing to show my thinking but ill bold the answer.

1,1
1,2
1,3
1,4
1,5

2,1
2,2
2,3
2,4
2,5

3,1
3,2
3,3
3,4
3,5

4,1
4,2
4,3
4,4
4,5

5,1
5,2
5,3
5,4
5,5

next ill mark all the ones that equal 4 or 8 when added together, with an x

1,1
1,2
x1,3
1,4
1,5

2,1
x2,2
2,3
2,4
2,5

x3,1
3,2
3,3
3,4
x3,5

4,1
4,2
4,3
x4,4
4,5

5,1
5,2
x5,3
5,4
5,5

that is 6 (that equal 4 or 8) out of 25

so your ratio would be 6:19

6 0

2 years ago

Ralph goes out to lunch and receives a bill listing a subtotal of $14.65 and tax of $1.47.Ralph leaves a 20% tip on the total am

Darina [25.2K]

The tip is greater than the amount charged

5 0

2 years ago

Milo walks 40 meters in 15 seconds and Mira walks 30 meters in 10 seconds. Who's walking rate is faster?

mestny [16]

Mira is walking faster because her pace is faster than Milo

7 0

2 years ago

Read 2 more answers

: A personal computer manufacturer buys 36% of its chips from Japan and the rest from the United St...

meriva

Answer: 0.9862

Step-by-step explanation:

Given : The probability that the chips belongs to Japan: P(J)= 0.36

The probability that the chips belongs to United States : P(U)= 1-0.36=0.64

The proportion of Japanese chips are defective : P(D|J)=0.017

The proportion of American chips are defective : P(D|U)=0.012

Using law of total probability , we have

$P(D)=P(D|J)\times P(J)+P(D|U)\times P(U)\\\\\Rightarrow\ P(D)=0.017\times0.36+0.012\times0.64=0.0138$

Thus , the probability that chip is defective = 0.0138

Then , the probability that a chip is defect-free= $1-0.0138=0.9862$

3 0

2 years ago