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Ede4ka [16]
3 years ago
7

Buck lodge israising money for hoops for heart in the spring. They raise $1000 in the fall and are asking the students to bring

in five dollars each. Write a function that represents how much money is raised for excellence. Then determine how much money will be raised 250 students participate

Mathematics
1 answer:
Wewaii [24]3 years ago
4 0

Answer: a) The function :

f(x) = 5x + 1000

b) f(x) = 2250 dollars

Step-by-step explanation: Please find the attached file for the solution

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What is this problem?<br><br>9+6÷(8-2)
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Hey there! Welcome to Brainly.

Let's solve using order of operations (PEMDAS), which is Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction.

First, we solve the parentheses.

9+6÷6

Now we take care of the division.

9+1

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Therefore, your answer is 10

I hope this helps!

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Please help<br> y∝1/√x If y=5<br> when x=36 find,<br> x when y=3
Fittoniya [83]

Answer:

<h3>x=60</h3>

Step-by-step explanation:

5=k×\frac{1}{36}

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3=180×\frac{1}{x}

x=60

6 0
2 years ago
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How to generate a Venn diagram to compare the following numbers 318,274 and 138,524
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3 0
3 years ago
The computer that controls a bank's automatic teller machine crashes a mean of 0.6 times per day. What is the probability that,
professor190 [17]

Answer:

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Mean of 0.6 times a day

7 day week, so \mu = 7*0.6 = 4.2

What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150

P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630

P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323

So

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

3 0
3 years ago
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