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3 years ago

8

A box with a square base and no top is to be made from a square piece of carboard by cutting 4 in. squares from each corner and

folding up the sides. The box is to hold 1444 in3. How big a piece of cardboard is needed

1 answer:

Lunna [17]3 years ago

7 0

Answer:

$C=27inch\ by\ 27inch$

Step-by-step explanation:

Squares $h=4inch$

Volume $v=1444in^3$

Generally the equation for Volume of box is mathematically given by

$V=l^2h$

$1444=l^2*4$

$l^2=361$

$l=19in$

Since

Length of cardboard is

$l_c=19+4+4$

$l_c=27in$

Therefore

Dimensions of the piece of cardboard is

$C=27inch\ by\ 27inch$

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A process is producing a particular part where the thickness of the part is following a normal distribution with a µ = 50 mm and

Hitman42 [59]

Answer:

0.13% probability that this selected sample has an average thickness greater than 53

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 normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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 $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem, we have that:

$\mu = 50, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1$

What is the probability that this selected sample has an average thickness greater than 53?

This is 1 subtracted by the pvalue of Z when X = 53. So

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

By the Central Limit Theorem

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

$Z = \frac{53 - 50}{1}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.13% probability that this selected sample has an average thickness greater than 53

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Ryan scored 87, 93, 96, and 89 on her first four history quizzes. What score does Ryan need to get on her fifth quiz to have an

hichkok12 [17]

Answer: For Ryan to get an average of exactly 91, she needs to get a 90 on her fifth quiz.

Step-by-step explanation:

You can find the average of a set of numbers by adding them together and dividing the sum by the amount of numbers added.

Ryan has five total quizzes. Only four of them are known, so we can represent her fifth quiz as X.

$\frac{87 + 93 + 96 + 89 + x}{5}$ , dividing the five quizzes' sum by 5

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A random sample of 401 student were recent survey regarding their class standing freshmen software junior senior and their major

SOVA2 [1]

the Solution

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To ascertain whether Sophomore and Non-stem are dependent, we have to test the following:

$\begin{gathered} If\text{ Pr(sophomore or Non-stem) = Pr(sophomore and Non-stem),} \\ \text{then we conclude that both events are Independent,} \\ \text{otherwise, they are dependent.} \end{gathered}$ $\begin{gathered} Pr(\text{sophomore or Non-stem)=Pr(sophomore)+Pr(Non-stem)} \\ -Pr(\text{sophomore and Non-stem)} \end{gathered}$ $Pr(\text{Sophomore or Non-stem)=}\frac{87}{401}+\frac{243}{401}-(\frac{87}{401}\times\frac{234}{401})$ $=\frac{330}{401}-\frac{21141}{160801}=0.82294-0.13147=0.69147$

Cleary, we have that sophomore and non-stem events are dependent events since 0.69147 is not the same as 0.13147.

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