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lora16 [44]
3 years ago
14

Which expression can be used to find the surface area of the following square pyramid?

Mathematics
1 answer:
telo118 [61]3 years ago
8 0

Answer:

Step-by-step explanation:

The slant height of one side of this pyramid is 5, and the base of this side is 4.  Thus, the area of one slant side is (1/2)(5)(4) = 10 units^2.

There are 4 such sides.  Thus, the total slant surface area is 4(10 units^2), or 40 units^2.  

If you also want to include the base area, the total would be

40 units^2 + 16 units^2 = 56 units^2.

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itzel noticed that the distance from her house to the library, which is 40 miles, was one fifth the distance from her house to h
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Answer:

200 miles

Step-by-step explanation:

40 x 5 = 200

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3 years ago
A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true
zloy xaker [14]

Answer:

The probability that the sample proportion will differ from the population proportion by greater than 0.03 is 0.009.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 \mu_{\hat p}=p

The standard deviation of this sampling distribution of sample proportion is:

 \sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

As the sample size is large, i.e. <em>n</em> = 492 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by the normal distribution.

The mean and standard deviation of the sampling distribution of sample proportion are:

\mu_{\hat p}=p=0.07\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.07(1-0.07)}{492}}=0.012

Compute the probability that the sample proportion will differ from the population proportion by greater than 0.03 as follows:

P(|\hat p-p|>0.03)=P(|\frac{\hat p-p}{\sigma_{\hat p}}|>\frac{0.03}{0.012})

                           =P(|Z|>2.61)\\\\=1-P(|Z|\leq 2.61)\\\\=1-P(-2.61\leq Z\leq 2.61)\\\\=1-[P(Z\leq 2.61)-P(Z\leq -2.61)]\\\\=1-0.9955+0.0045\\\\=0.0090

Thus, the probability that the sample proportion will differ from the population proportion by greater than 0.03 is 0.009.

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