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ICE Princess25 [194]
4 years ago
13

The figure below shows the net of a triangular pyramid. The given height is rounded to the nearest hundredth. If all the triangl

es are equilateral, what is the surface area of the pyramid in square centimeters? HEIGHT: 4.33 cm BASE: 5 cm. *The net shows 4 triangles but I think one of them is the base of the triangle.*
Mathematics
1 answer:
Nataly [62]4 years ago
4 0

Answer: 43.3 square inches

Step-by-step explanation:

Formula for Triangle: A = bh/2

Key:

* = multiple

/ = divide

Fill in the formula with what you know:

A = bh/2

A = 5 * 4.33 / 2

A= 21.65 / 2 = 10.825 square inches

However, 10.825 is not our answer. 10.825 is the area of only one triangle. Since all four of the triangles are congruent (or the same) we can:

Add: 10.825 + 10.825 + 10.825 + 10.825 = 43.3 square inches

OR

Multiple: 10.825 * 4 = 43.3 square inches

Thus, the surface area of the pyramid is 43.3 square inches.

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Step-by-step explanation:

Option A is the correct answer

f(x)=8x+2

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3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7
Zinaida [17]

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

    The probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable <em>X</em> ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 2.7 and standard deviation, <em>σ</em> = 0.75.

(a)

A random sample of <em>n</em> = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

P(\bar X\leq 3.00)=P(\bar X

                    =P(\bar X

                    =P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

                    =P(Z

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

P(2.70

                                =P(0

*Use a <em>z</em>-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

P(\bar X\leq 3.00)\geq 0.99

P(\bar X

P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

P(Z

The value of <em>z</em> for the probability above is

<em>z</em> ≥ 2.33

Compute the value of <em>n</em> as follows:

\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33

\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}

\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77

Thus, the sample size must be at least 77.

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Answer:

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