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2 years ago

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Gabriel is saving money by mowing lawns on the weekends. Last month, he saved $150. This month, saved 20% more than he saved las

t month. How much money did Gabriel save this month?

2 answers:

Katyanochek1 [597]2 years ago

3 0

Answer:

180

Step-by-step explanation:

so last month he saved 150 right first you find 20% of 150. that is 30. you add 150 and 30 and get 180.

nadya68 [22]2 years ago

3 0

Answer:

180

Step-by-step explanation:

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Step-by-step explanation:

Rachel earned $34 in 4 hours at her job today. It means we can find his earning in 1 hour.

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Remember
a²-b²=(a-b)(a+b)

8x²-50y²
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(2)([2x]²-[5y]²)
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A hole is drilled in a sheet-metal component, and then a shaft is inserted through the hole. The shaft clearance is equal to dif

VashaNatasha [74]

Answer:

(A)

$P(X \geq 0.8) = \int\limits_{0.8}^{\infty} f(x) \, dx = \int\limits_{0.8}^{1} 1.25(1-x^4) \, dx = 0.08192$

(B)

Then the cumulative function would be

$CF(x) = 1.25x - 0.25x^5$ if 0<x<1

0 otherwise.

Step-by-step explanation:

(A)

We are looking for the probability that the random variable X is greater than 0.8.

$P(X \geq 0.8) = \int\limits_{0.8}^{\infty} f(x) \, dx = \int\limits_{0.8}^{1} 1.25(1-x^4) \, dx = 0.08192$

(B)

For any $x$ you are looking for the probability $P(X \geq x)$ which is

$P(X \geq x) = \int\limits_{-\infty}^{x} 1.25(1-t^4) dt = \int\limits_{0}^{x} 1.25(1-t^4) dt = 1.25x - 0.25x^2$

Then the cumulative function would be

$CF(x) = 1.25x - 0.25x^5$ if 0<x<1

0 otherwise.

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2 years ago