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1 year ago

Owen can jog 660 yards in 3 minutes. how many yards would you expect him to jog in 10 minutes?

Mathematics

1 answer:

pshichka [43]1 year ago

6 0

Answer: 2200 yards

Step-by-step explanation:

First find the unit rate, 660/3 = 220

So, Owen can jog 220 yards in one minute.

220 times 10 to find how much Owen can jog in 10 minutes.

220*10 = 2200.

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Question 13

PtichkaEL [24]

Answer:

B. 30x

D. x(45-15)

Step-by-step explanation:

The distributive property lets you factor out the common factor of x. The sum of coefficients can be simplified to a single coefficient.

45x -15x = x(45 -15) = 30x

5 0

2 years ago

Avery received $80 in cash for mowing lawns. She did the following with the $80. Put 20% of the $80 into a bank account. Spent 1

PolarNik [594]

Answer:

$48

Step-by-step explanation:

To start off, we know that Avery put 20% of the $80 in her bank account. We can figure out how much 20% of 80 is by multiplying 80 by 20% and then subtracting that number from 80. Remember to switch 20% to its decimal form (20% -> .2).

Step 1) 80 x .2 = 16

Step 2) 80 - 16 = 64

Next, we have to figure out how much 1/4 of the remaining $64 would be. We just simply divide 64 by 4 to figure out how much Avery spent.

Step 3) 64 / 4 = 16

Lastly, since we know Avery spent 16 dollars, we can subtract the 16 from 64 to get our final answer.

Step 4) 64 - 16 = 48

8 0

3 years ago

The production department has installed a new spray machine to paint automobile doors. As is common with most spray guns, unsigh

marta [7]

Answer:

10000*0.6065=6065

A. about 6,065

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

$P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}$

And the parameter $\lambda$ represent the average ocurrence rate per unit of time.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=0.5$ , $Var(X)=\lambda=0.5$ , $Sd(X)=707$

Solution to the problem

We want "how many would have no blemishes" so first we need to find the probability that X=0, since X represent on this case the number of blemishes on each door. And if we use the mass function we got this:

$P(X=0) =0.5^0 \frac{e^{-0.5}}{0!}=0.6065$