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2 years ago

How much wrapping paper is needed to cover a box (rectangular prism) that measures 6 feet by 2.5 feet by 3 feet?

Mathematics

1 answer:

AleksAgata [21]2 years ago

8 0

Answer:

81 feet ²

Step-by-step explanation:

Given that:

Dimension of rectangular prism :

6 feets by 2.5 feets by 3 feets

To obtain the amount of wrapping paper required to cover the box ; we find the area of the box ;

Area of box = 2(wl + hl + hw)

Area of box = 2(6*2.5 + 6*3 + 2.5*3)

Area of box = 2(15 + 18 + 7.5)

Area of box = 81 feet²

You might be interested in

8y + 7y <br> in standard form

dybincka [34]

Answer:

15y

Step-by-step explanation:

8y + 7y

first, we add the numbers which is 8+7 = 15

now we add the ys that = y+y= y ( when you add any alphabet letter the answer will be itself .)

then we add 15 + y but we can not add these but we can write it as 15y

= 15y

have a wonderful day

4 0

3 years ago

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

klio [65]

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$