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Olin [163]
3 years ago
5

What number is 32% of 75?

Mathematics
2 answers:
BartSMP [9]3 years ago
7 0

Let

n-------> the number

we know that

75 represent the 100\%

so

by proportion

Find the 32\% of 75

\frac{75}{100\%}= \frac{n}{32\%} \\ \\100*n=75*32 \\ \\n= \frac{32}{100}*75 \\ \\ n=0.32*75

therefore

<u>the answer is</u>

the equation is equal to

n=0.32*(75)

vfiekz [6]3 years ago
3 0
1) 75/100 = 0.75 x 32 = 24
2) n= 0.32(75)
0.32x75 =24= n
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The equation tan(55 degree) equals StartFraction 15 Over b EndFraction can be used to find the length of Line segment A C.
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Answer:

AC is b

tan55 =  \frac{15}{b}

15/tan55 = b

tan 55 = 1.428

15/1.428 = 10.5 cm = AC

6 0
3 years ago
Express the exponential number 2.25 x 10-4 as an ordinary number. 22,500 0.000 022 5 225,000 0.000 225 none of the above
Evgesh-ka [11]
For this case we have the following number:
 2.25 x 10-4
 We observe that it is written in exponential notation.
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5 0
3 years ago
Last month you had 8 wins while playing fortnight. This month you had 5. What was your percent of change?
Dominik [7]

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%62.5

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3 0
3 years ago
12foot 11incher+11foot 3 incher​
mash [69]

Answer:

24feet 1 inches

Step-by-step explanation:

4 0
3 years ago
Help me with this!!!
PIT_PIT [208]

Answer:

Total  probability value is 1

P(green) = 0.2 , P(brown) = 0.54

Given P(red) = P(yellow) = x

Total Probability = P(green) + P(Brown) + x + x

1 = 0.2 + 0.54 + 2x

1 - 0.74 = 2x

x = 0.13

P(red) = P(yellow) = 0.13

Now number of times it landed on each color :

On \ red :\\P(red) = \frac{number \ of \ times \ on  \ red}{1200} \\\\0.13 *1200 = number \ of \ times \ on \ red\\number \ of \ times \ on \ red\\ = 156\\

Since P(red) = P(yellow) , the number of time on yellow and red are same.

On \ yellow :\\number \ of \ times \ on \ yellow = 156

On \ green :\\P(green ) = \frac{number \ of \ times \ on \ green}{1200}\\\\0.2 *1200 = number \ of \ times \ on \ green\\\\number \ of \ times \ on \ green = 240

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Answer :

Red      156

Green   240

Brown   648

Yellow   156

8 0
3 years ago
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