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3 years ago

10

Sandra is making decorations for the school dance out of construction paper. Each decoration uses 223 2 2 3 sheets of paper. San

dra notices that she only has 1012 10 1 2 sheets left.

1 answer:

scoundrel [369]3 years ago

7 0

Answer:

Step-by-step explanation:

CACA POOP

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Use Euler’s Formula to find the missing number.<br> Edges: 37<br> Faces: 25<br> Vertices: ?

matrenka [14]

Answer:

Step-by-step explanation:

Comment

The formula that relates edges faces and vertices is F + V = E + 2

Givens

Edges (E): 37

Faces (F) = 25

Vertices: x

Solution

25 + x = 37 + 2 Subtract 25 from both sides.

25-25 +x= 37 - 25 + 2 Combine

x = 12 + 2

x = 14

Answer: The vertices =<u> 14</u>

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2 years ago

A laptop is 30% off the regular price of 920$ If a 6% sales tax is added to the cost what is total cost of the laptop

Lunna [17]

Answer

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2 years ago

Suppose that the sitting back-to-knee length for a group of adults has a normal distribution with a mean of mu equals 24.4 in.

solong [7]

Answer:

$P(X \geq 26.6) = 0.0336$ , which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 24.4, \sigma = 1.2$

In this problem, a value x is significantly high if:

$P(X \geq x) = 0.01$

Using these criteria, is a back-to-knee length of 26.6 in. significantly high?

We have to find the probability of the length being 26.6 in or more, which is 1 subtracted by the pvalue of Z when X = 26.6. So

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

$Z = \frac{26.6 - 24.4}{1.2}$

$Z = 1.83$

$Z = 1.83$ has a pvalue of 0.9664.

1 - 0.9664 = 0.0336

$P(X \geq 26.6) = 0.0336$ , which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

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3 years ago