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3 years ago

What is the place value of 5 x 70 = 5 x _ tens

Mathematics

2 answers:

Andrej [43]3 years ago

8 0

7 because 7x10 is 70

Nataly [62]3 years ago

4 0

I think it would be 7 because 7x10=70

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Which of the following word problems can be solved using the equation 12 ÷ 4 = a?

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Answer:

Step-by-step explanation:

A) 12 is divided by 4 since the twelve cookies are shared equally between four children.

B) 12 is multiplied by 4 since each band uniform has 12 buttons and there are four band uniforms.

C) 12 is substracted by 4 since four boats will not participate in the race out of a total of 12 boats.

D) 4 is added to 12 since there are already 12 monkeys but now 4 more will arrive.

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4 years ago

Given h(x)=(3x-4), f(x)=(-7x+2), g(x)=(9x+1), which of the following is equivalent to 2x+3?

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3 years ago

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3 years ago

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5.53 An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day. What

Licemer1 [7]

Answer:

a) 38.4% probability that on a given day this item is requested more than 5 times.

b) 0.67% probability that on a given day this item is not requested at all.

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.

An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day.

This means that $\mu = 5$

What is the probability that on a given day this item is requested

(a) more than 5 times?

Either it is requested 5 times or less, or it is requested more than 5 times. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067$

$P(X = 1) = \frac{e^{-5}*(5)^{1}}{(1)!} = 0.0337$

$P(X = 2) = \frac{e^{-5}*(5)^{2}}{(2)!} = 0.0842$

$P(X = 3) = \frac{e^{-5}*(5)^{3}}{(3)!} = 0.1404$

$P(X = 4) = \frac{e^{-5}*(5)^{4}}{(4)!} = 0.1755$

$P(X = 5) = \frac{e^{-5}*(5)^{5}}{(5)!} = 0.1755$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755 + 0.1755 = 0.616$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.616 = 0.384$

38.4% probability that on a given day this item is requested more than 5 times.

(b) not at all?

$P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067$

0.67% probability that on a given day this item is not requested at all.

3 0

3 years ago