$1.00 for 4 cans of beans (Useing a unit rate)

The solid was created by connecting two congruent square pyramids to a rectangular prism.

AnnyKZ [126]

Answer:

1920 square inches

Step-by-step explanation:

For a rectangular prism, the lateral area can be found by ...

LA = Pl

where P is the perimeter, and l is the length.

For a square pyramid, the lateral area can be found by ...

LA = (1/2)Ph

where P is the perimeter of the base, and h is the slant height of the triangular faces.

For a figure with a square cross section of perimeter P "capped" by square pyramids on either end, the total surface area is the sum of the lateral areas of the three components:

SA = (Pl) + (1/2)Ph + (1/2)Ph

SA = P(l+h) = (4×15 in)(14 +18 in) = (60)(32) in²

SA = 1920 in²

The surface area of the solid seems to be 1920 square inches.

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<em>Caveat</em>

If the figure is something other than what we have tried to describe, your mileage may vary. A diagram would be helpful.

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4 years ago

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. The time required for a technician to machine a specific component is normally distributed with a mean of 2 hours and a standa

erma4kov [3.2K]

Answer:

a) There is a 3.84% probability that the technician can machine one component in 1.5 hours or less.

b) There is a 0.42% probability that the technician will require at least 2.75 hours to complete one component

Step-by-step explanation:

Problems of normally distributed samples can be 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

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

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.

In this problem, we have to be careful. The mean is in hours, while the standard deviation is in minutes. I am going to work with both in hours, as the problem states. 17 minutes is 0.283 hours, so:

$\mu = 2, \sigma = 0.283$

(a.) What is the probability that the technician can machine one component in 1.5 hours or less?

This probability is the pvalue of the Zscore when $X = 1.5$ . So:

$Z = \frac{1.5 - 2}{0.283}$

$Z = \frac{-0.5}{0.283}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

This means that there is a 3.84% probability that the technician can machine one component in 1.5 hours or less.

(b.) What is the probability that the technician will require at least 2.75 hours to complete one component?

The pvalue of the score of $X = 2.75$ is the probability that the technican will require less than 2.75 hours to complete one component. The probability that he will require at least 2.75 hours to complete one component is 1 subtracted by this pvalue. So: