Ask question

Ask question

All categories

3 years ago

9

if an athlete’s weight decreased from 160 pounds to 152 pounds, what was the percent decrease in the athlete’s weight? a) 16% b)

10% c) 8% d) 5% e) 4%

1 answer:

Eddi Din [679]3 years ago

7 0

Percent decrease=decrease/original

decrease=160-152=8
original=160

8/160=1/20=5/100=5%

5% decrease

D is answer

You might be interested in

If the simple interest on $7000 for 8 years is $4480, then what is the interest rate ?

OverLord2011 [107]

$560 for every year.

5 0

3 years ago

What is 1X 78945621331227894?

Salsk061 [2.6K]

Answer:

your answer would be 7.8945621e+16

Step-by-step explanation:

have a wonderful day

8 0

3 years ago

Read 2 more answers

A total of about 230 million turkeys were raised in the U.S. last year. Were more than one-quarter of them raised in Minesota if

pochemuha

No, 17.39% were raised in Minnesota.

You can find this by dividing 40,000,000 by 230,000,000 and then converting to a percent by multiplying by 100.

40,000,000/230,000,000=.1739
.1739 x 100= 17.39%

8 0

3 years ago

Simplify the expression 6+(12+x) <br> Answer: 18+x

dybincka [34]

Answer:

18+x. Are you kidding us?

8 0

3 years ago

Caleb's mother said they will drive 1 /4 hour to get there, and they will drive another 1 /4 hour back. Once they arrive, Caleb'

Nezavi [6.7K]

Answer: They are away for $2\dfrac{3}{4}\ hours$ from home.

Step-by-step explanation:

Since we have given that

Time to get there = $\dfrac{1}{4}$

Time to get back = $\dfrac{1}{4}$

Number of soccer games = 3

Time for each soccer games = $\dfrac{3}{4}$

So, total time for 3 games would be

$3\times \dfrac{3}{4}=\dfrac{9}{4}$

So, the time for which they will be away from home is given by

$\dfrac{1}{4}+\dfrac{9}{4}+\dfrac{1}4}=\dfrac{1+9+1}{4}=\dfrac{11}{4}=2\dfrac{3}{4}\ hours$

Hence, they are away for $2\dfrac{3}{4}\ hours$ from home.

3 0

4 years ago