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Citrus2011 [14]

2 years ago

5

To make a bakery’s signature chocolate muffins, a baker needs 2.5 ounces of chocolate for each muffin.How many pounds of chocola

te are needed to make 48 signature chocolate muffins? (1 pound = 16 ounces)
a.7.5
b.10
c.50.5
d.120

1 answer:

Ann [662]2 years ago

8 0

Step-by-step explanation:

Simple first we will multiply

2.5 x 48 = 120 ounces

1 pound = 16 ounces

120 ounces = 120 ÷ 16 = 7.5 pounds

Option A is the correct answer

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2 years ago

What is square root of the product of the number z and itself<br>

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Answer:

$\large\boxed{z}$

Step-by-step explanation:

What is square root of the product of the number z and itself?

Break down into smaller parts

What is the product of the number z and itself?

Product = multiply

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2 years ago

A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that t

Alina [70]

Answer:

75.76% probability that there will be 10 or more customers at this bank in one hour.

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.

A bank gets an average of 12 customers per hour.

This means that $\mu = 12$

Find the probability that there will be 10 or more customers at this bank in one hour.

Either there are less than 10 customers, or there are 10 or more. The sum of the probabilities of these events is 1. Then

$P(X < 10) + P(X \geq 10) = 1$

We want $P(X \geq 10)$

Then

$P(X \geq 10) = 1 - P(X < 10)$

In which

$P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)$

So

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

$P(X = 0) = \frac{e^{-12}*12^{0}}{(0)!} \approx 0$

$P(X = 1) = \frac{e^{-12}*12^{1}}{(1)!} = 0.0001$

$P(X = 2) = \frac{e^{-12}*12^{2}}{(2)!} = 0.0004$

$P(X = 3) = \frac{e^{-12}*12^{3}}{(3)!} = 0.0018$

$P(X = 4) = \frac{e^{-12}*12^{4}}{(4)!} = 0.0053$

$P(X = 5) = \frac{e^{-12}*12^{5}}{(5)!} = 0.0127$

$P(X = 6) = \frac{e^{-12}*12^{6}}{(6)!} = 0.0255$

$P(X = 7) = \frac{e^{-12}*12^{7}}{(7)!} = 0.0437$

$P(X = 8) = \frac{e^{-12}*12^{8}}{(8)!} = 0.0655$

$P(X = 9) = \frac{e^{-12}*12^{9}}{(9)!} = 0.0874$

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$P(X \geq 10) = 1 - P(X < 10) = 1 - 0.2424 = 0.7576$

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