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Vera_Pavlovna [14]

3 years ago

8

Jeremy mowed several lawns to earn money for camp. After he paid $17 for gas, he had $75 leftover to pay towards camp. Write and

solve an equation to find how much money Jeremy earned mowing lawns.

1 answer:

Alexxx [7]3 years ago

6 0

Money earned = ee=75+17=92

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Math pow i dont understand this problem

Marina CMI [18]

83220 is your answer

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4 years ago

Which one of the following statements is true? (5 points)

DanielleElmas [232]

Answer:

None of these statements are true.

Step-by-step explanation:

a) The derivative of (fg)(x) is f'g +fg' according to the product rule for derivatives.

__

b) The derivative of |x² +x| is a 3-part piecewise linear function equal to 2x+1 for |x+1/2| > 1/2, and equal to -2x-1 for |x+1/2| < 1/2. It is undefined for x=0 and x=1.

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c) for y = √f(x), y' = f'(x)/(2√f(x))

8 0

3 years ago

Fresh milk costs 12.50 per liter what is the cost of y liters of fresh milk?

victus00 [196]

Answer:

12.50y

Step-by-step explanation:

If 1 ltr=12.50
what about y ltrs

Therefore, 1 ltr= 12.50

y ltrs=?

you do cross multiplication

hence, 12.50 × y= 12.50y

total cost= sh.(12.50y)

8 0

3 years ago

If I go to sleep at 10:30, how many hours of sleep will I get if I wake up at 5:00 in the morning?

kkurt [141]

Six hours & Thirty minutes

4 0

4 years ago

Read 2 more answers

Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival ti

natima [27]

Answer:

D. 91%

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Less than 15 minutes.

Event B: Less than 10 minutes.

We are given the following probability distribution:

$f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5$

Simplifying:

$f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}$

Probability of arriving in less than 15 minutes:

Integral of the distribution from 5 to 15. So

$P(A) = \int_{5}^{15} = \frac{375}{t^4}$

Integral of $\frac{1}{t^4} = t^{-4}$ is $\frac{t^{-3}}{-3} = -\frac{1}{3t^3}$

Then

$\int \frac{375}{t^4} dt = -\frac{125}{t^3}$

Applying the limits, by the Fundamental Theorem of Calculus:

At $t = 15$ , $f(15) = -\frac{125}{15^3} = -\frac{1}{27}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}$

Probability of arriving in less than 15 minutes and less than 10 minutes.

The intersection of these events is less than 10 minutes, so:

$P(B) = \int_{5}^{10} = \frac{375}{t^4}$

We already have the integral, so just apply the limits:

At $t = 10$ , $f(10) = -\frac{125}{10^3} = -\frac{1}{8}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}$

If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087$

Thus 90.87%, approximately 91%, and the correct answer is given by option D.

3 0

3 years ago