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Leona [35]
3 years ago
9

How to solve this problem

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
8 0
You're going to use cross multiplication.
1 gal x 21 miles = 25miles x x
21 miles = 25x
Divide each side by 25 to get x alone
21/25 = 0.84, which is your x value.
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4500/30000 =0.15

32000 * 0.15 = 4800

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Complete the third step to solve the equation.
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Answer:

x= 1/8 or 0.125

Step-by-step explanation:

8x=1

x=1/8

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2 years ago
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Find the area of the shaded region:
Deffense [45]

Answer:

34560

Step-by-step explanation:

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Hope I'm right...

Hope this helps plz like and brainly :D

6 0
3 years ago
If the heights of 300 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many studen
Lunna [17]
Given:
μ = 68 in, population mean
σ = 3 in, population standard deviation

Calculate z-scores for the following random variable and determine their probabilities from standard tables.

x = 72 in:
z = (x-μ)/σ = (72-68)/3 = 1.333
P(x) = 0.9088

x = 64 in:
z = (64 -38)/3 = -1.333
P(x) = 0.0912

x = 65 in
z = (65 - 68)/3 = -1
P(x) = 0.1587

x = 71:
z = (71-68)/3 = 1
P(x) = 0.8413

Part (a)
For x > 72 in, obtain
300 - 300*0.9088 = 27.36

Answer: 27

Part (b)
For x ≤ 64 in, obtain
300*0.0912 = 27.36

Answer: 27

Part (c)
For 65 ≤ x ≤ 71, obtain
300*(0.8413 - 0.1587) = 204.78

Answer: 204

Part (d)
For x = 68 in, obtain
z = 0
P(x) = 0.5
The number of students is
300*0.5 = 150

Answer: 150

3 0
3 years ago
An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of mov- ing violatio
olga2289 [7]

Answer:

a) E(Y) = \sum_{i=1}^n Y_i P(Y_i)

And replacing we got:

E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60

b) E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1

And then the expected value would be:

E(100Y^2) = 100*1.1= 110

Step-by-step explanation:

We assume the following distribution given:

Y       0       1        2        3

P(Y) 0.60 0.25  0.10  0.05

Part a

We can find the expected value with this formula:

E(Y) = \sum_{i=1}^n Y_i P(Y_i)

And replacing we got:

E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60

Part b

If we want to find the expected value of 100 Y^2 we need to find the expected value of Y^2 and we have:

E(Y^2) = \sum_{i=1}^n Y^2_i P(Y_i)

And replacing we got:

E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1

And then the expected value would be:

E(100Y^2) = 100*1.1= 110

4 0
3 years ago
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