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Slav-nsk [51]
3 years ago
14

What percent of 120 is 90

Mathematics
1 answer:
Alex777 [14]3 years ago
7 0

Answer:

75%

Step-by-step explanation:

90/ 120 is the same as 75 or 3/4 so  75% of 120 is 90

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For which pair A function is the exponential constantly growing at a faster rate than a quadratic over the interval zero less th
VashaNatasha [74]

Answer:

Here is the complete question:

For which pair of functions is the exponential consistently growing at a faster rate than the quadratic over the interval 0<=X<=5.

Answer is C (the third option)

Step-by-step explanation:

Basically in exponential growth a quantity may increase over time. When a quantity increases of decreases by equal or same percent over equal period of times this means that the quantity increases or decreases exponentially. I have attached the image of the correct option.

4 0
3 years ago
N x 3/8 = 1 What is N?
Dominik [7]

Answer:

N=\frac{8}{3}

Step-by-step explanation:

The image below shows step-by-step on how to solve it.

Hope this helps! :)

7 0
3 years ago
PLEASE SOLVE. Students are selling candy bars. They earn $0.50 for each one they sell. They want to earn at least $455. Write an
kari74 [83]
So each candy bar is .50 and they want to reach 455.00

we see how many .50's can go into 455.00

we notice that .50 times 2=1
so if we multiply the 455  1's by 2 we get how many candy bars to sell or
910.00
6 0
3 years ago
Read 2 more answers
USA Today reported that approximately 25% of all state prison inmates released on parole become repeat offenders while on parole
Zepler [3.9K]

Answer:

\mu =E(X) = 0*0.207+1*0.367+ 2*0.227+ 3*0.162+ 4*0.036+ 5*0.001= 1.456

For the variance we need to calculate first the second moment given by:

E(X) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 0^2*0.207+1^2*0.367+ 2^2*0.227+ 3^2*0.162+ 4^2*0.036+ 5^2*0.001= 3.33

And the variance is given by:

\sigma^2 = E(X^2) - [E(X)]^2 = 3.334 -[1.456]^2 =1.21

And the deviation is:

\sigma = \sqrt{1.214} = 1.10

Step-by-step explanation:

For thi case we have the following distribution given:

X         0          1         2         3         4            5

P(X)  0.207  0.367 0.227  0.162  0.036    0.001

For this case the expected value is given by:

E(X) = \sum_{i=1}^n X_i P(X_i)

And replacing we got:

\mu =E(X) = 0*0.207+1*0.367+ 2*0.227+ 3*0.162+ 4*0.036+ 5*0.001= 1.456

For the variance we need to calculate first the second moment given by:

E(X) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 0^2*0.207+1^2*0.367+ 2^2*0.227+ 3^2*0.162+ 4^2*0.036+ 5^2*0.001= 3.33

And the variance is given by:

\sigma^2 = E(X^2) - [E(X)]^2 = 3.334 -[1.456]^2 =1.21

And the deviation is:

\sigma = \sqrt{1.214} = 1.10

3 0
4 years ago
PART A
krok68 [10]

im guessing its about $10

6 0
3 years ago
Read 2 more answers
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