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2 years ago

Please help me!

2 answers:

7 0

To write -12/80 as a decimal you have to divide numerator by the denominator of the fraction.
We divide now 12 by 80 what we write down as 12/80 and we get -0.15

And finally we have:
-12/80 as a decimal equals -0.15

Hope this helps:3

n200080 [17]2 years ago

4 0

-0.15
I hope I'm correct

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3 years ago

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2 years ago

A chemist has three different acid solutions. The first acid solution contains 25 % acid, the second contains 45 % and the third

liq [111]

Answer:

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ALGEBRA WORD PROBLEM

Dorian S. asked • 06/28/17

THREE LINEAR EQUATIONS WITH THREE VARIABLES

A chemist has three different acid solutions. The first acid solution contains 15% acid, the second contains 25%,

and the third contains 70%. He wants to use all three solutions to obtain a mixture of 40 liters containing 45%

acid, using 2 times as much of the 70% solution as the 25% solution. How many liters of each solution should be used?

How to set up as three linear equations to find the answer?

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1 Expert Answer

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Best

Arthur D. answered • 06/28/17

TUTOR 4.9 (67)

Forty Year Educator: Classroom, Summer School, Substitute, Tutor

ABOUT THIS TUTOR ›

set up 2 equations

the first equation...

15%*x+25%*y+70%*2y=45%*40

0.15x+0.25y+1.4y=0.45*40

0.15x+1.65y=18

multiply all terms by 100

15x+165y=1800

divide all terms by 15

x+11y=120

the second equation...

x+y+2y=40

x+3y=40

x=40-3y

substitute into the first equation x+11y=120

40-3y+11y=120

40+8y=120

8y=120-40

8y=80

y=80/8

y=10 liters

x+3*10=40

x+30=40

x=40-30

x=10 liters

10 liters of 15%, 10 liters of 25%, and 20 liters of 70%

check:

10*15%=1.5

10*25%=2.5

20*70%=14

1.5+2.5+14=18

18/40=9/20=45/100=45%

45%*40=18

hope it helps

please mark brainliest

8 0

3 years ago

The number of tickets purchased by an individual for Beckham College's holiday music festival is a uniformly distributed random

Zanzabum

1) Mean: 6.5

2) Standard deviation: 3.2

Step-by-step explanation:

A uniform random distribution is the distribution of a variable that can assume n different values, and each of these values is equally likely to occur.

This means that the probability of occurring of each of these values is the same:

$p(x_1)=p(x_2)=...=p(x_n)$

For a uniform random distribution, the mean of the distribution is simply equal to the median value of the distribution, which is exactly at the center of the distribution itself. It can be calculated as

$\bar x = \frac{1}{2}(a+b)$