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olchik [2.2K]
3 years ago
11

There are 750 toothpicks in a regular sized box. If a jumbo box is made by doubling all the dimensions of the regular sized box,

how many toothpicks will the jumbo box hold? (With work pls)
Mathematics
2 answers:
Serga [27]3 years ago
7 0

To answer this, let the 3 sides of the regular sized rectangular box be x, y, and z.

When we double all the sides, the new sides have lengths 2x, 2y, and 2z, respectively.

Since the regular box (with dimensions x,y, and z) have 750 toothpicks, we can write

xyz=750

Similarly we can write for the jumbo box as,

(2x)(2y)(2z)

which is equal to 8(xyz)

So this is 8 times the original of xyz. Hence the jumbo box will have 8 times 750 number of toothpicks.

ANSWER: 8*750=6000 toothpicks

stellarik [79]3 years ago
4 0

Answer:

6000

Step-by-step explanation:

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The heights of North American women are normally distributed with a mean of 64 inches and a standard deviation of 2inches.a. Wha
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Answer:

a)P(X>66)=P(Z>1)=1-P(Z

b)P(\bar X >66)=P(Z>2)=1-P(Z

c) P(\bar X >66)=P(Z>10)=1-P(Z

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

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2) Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

X \sim N(64,2)  

Where \mu=64 and \sigma=2

We are interested on this probability

P(X>66)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-\mu}{\sigma})=P(Z>\frac{66-64}{2})=P(Z>1)

And we can find this probability on this way:

P(Z>1)=1-P(Z

3) Part b

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

P(\bar X >66)=P(Z>\frac{66-64}{\frac{2}{\sqrt{4}}}=2)

And using a calculator, excel or the normal standard table we have that:

P(Z>2)=1-P(Z

4) Part c

P(\bar X >66)=P(Z>\frac{66-64}{\frac{2}{\sqrt{100}}}=10)

And using a calculator, excel or the normal standard table we have that:

P(Z>10)=1-P(Z

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