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2 years ago

Someone please help i will rate! :(

Mathematics

1 answer:

Angelina_Jolie [31]2 years ago

7 0

What you will want to do is replace x with the amount of customers so it would be 42 squared divided by 3025-55x42 which should hopefully give you your answer which is 2.339 but you may have to round it as you can’t have 1/3 of a person etc hope this helps, let me know if I’m wrong

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» What is the distance between -25 and -12 on<br> a number line?

solmaris [256]

Answer: 13

Step-by-step explanation: All you have to do is subtract -12 from -25 and you get -13. If it was -25 from 12 you would just add the two together.

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2 years ago

Rewrite the equations in the form ax^2+bx+c =0:<br> -5x (x +6)=4(x-3) -10<br> 4x^2-2x(3x+1)=5

Rainbow [258]

Step-by-step explanation:

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3 years ago

Combine like terms in the expression below:<br> 9x + 4.05xy + 15y + 30y - 11x + 9.46

aev [14]

Answer: 4.05 x y − 2 x + 45 y + 9.46

Step-by-step explanation:

9x + 4.05xy + 15y + 30y - 11x + 9.46

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2 years ago

Consider a Poisson distribution with μ = 6.

bearhunter [10]

Answer:

a) $P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b) f(2) = 0.04462

c) f(1) = 0.01487

d) $P(X \geq 2) = 0.93803$

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 successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In this question:

$\mu = 6$

a. Write the appropriate Poisson probability function.

Considering $\mu = 6$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b. Compute f (2).

This is P(X = 2). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

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

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

$P(X \geq 2) = 0.93803$

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2 years ago

At a local Dunkin’ Donuts, a survey showed that out of 1,300 customers eating lunch, 520 ordered coffee with their meal. What pe

k0ka [10]

First subtract the number that did, to find the number that didn't.

1300 - 520 = 780 customers did not order coffee.

Now divide the number who didn't order coffee by total number of customers:

780 / 1300 = 0.6

Multiply by 100 to get the percent:

0.6 x 100 = 60% did not order coffee.

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2 years ago