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Dvinal [7]
3 years ago
5

How many times does 85 go into 100

Mathematics
2 answers:
Aleks04 [339]3 years ago
5 0
85 goes into 100 one time
yanalaym [24]3 years ago
3 0
1 time.
You divide 100(divide sign)85
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Which of the following methods would be the easiest to use to solve 12x2 – 48 = 0?
Katarina [22]

\bf \stackrel{\textit{solving for \underline{x}}}{12x^2-48=0\implies 12x^2=48\implies x^2=\cfrac{48}{12}\implies x^2=4} \\\\\\ x=\pm\sqrt{4}\implies x=\pm 2

7 0
3 years ago
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Write a real-world problem involving a percent that can be solved by using the proportion 3/b= 60/100. Then solve the proportion
inessss [21]
Dax goes to the store to buy milk. There is a sale for milk, and it now costs 3 dollars since it was 60% off. What was the original price of the milk?
3 - 60
x - 100
Cross multiply
300=60x
5=x

The milk original cost 5 dollars.
8 0
2 years ago
PLS HELP I HAVE TO DO THIS BY TONIGHTT
gtnhenbr [62]

I got you !

Step-by-step explanation:

45 bulbs planted in first year.

65 bulbs the next year.

increase = 20 bulbs

Number of increase of planted tulip bulb is N = 65 - 45 = 20

( 20/45)  x 100% = 44.4 % to the nearest tenths

3 0
3 years ago
Assume that by continuing your education, you increased your yearly earning potential from $21,484 to $39,746. If the additional
Ede4ka [16]

Answer:

By continuing my education I increased my earning potential from $21,484 to $39,746 a year. That's a difference of $18262 a year.

If the additional education costs $18,000, then in one year it will pay for itself.

7 0
3 years ago
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Five distinct numbers are randomly distributed to players numbers 1 through 5. Whenever two playerscompare their numbers, the on
ad-work [718]

Answer:

Observe that the first and the second player have equal probability to get any of number. Using the principle of that symmetry, we have that  

P(X=0)=\frac{1}{2}

<u>Event X = 1 </u>means that the first player has got greater number than the second player, but not than the third player. So, choose any three numbers out of five of them and say that the minimal number out of these three goes to the second player, mean number to the first one and the largest to the third one. Permute remaining two numbers on remaining two people. Hence

P(X=1)=\frac{1}{6}

<u>Event X = 2</u> means that the first player has got greater number than the second and the third player, but not than the fourth player. So, choose any four numbers out of five of them and say that the minimal number and the next minimal out of these four go to the second and the third player (and permute them), third number to the first one and the largest to the fourth player. Give remaining number to the last person. Hence

P(X=2)=\frac{1}{12}

<u>Event X = 3</u> means that the first player has got greater number than the second, the third, and the forth player, but not than the fifth player. So, permute these five numbers as follows: give the highest to the last person, the second highest to the first, and permute remaining numbers on the remaining people. Hence  

P(X=3)=\frac{1}{20}

<u>Event X = 4</u> means basically the first player has won all the battles i.e, he has got the greatest number. Hence

P(X=4)=\frac{1}{5}

7 0
3 years ago
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