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8_murik_8 [283]
3 years ago
5

A purchaser of electric relays buys from two suppliers, A and B. Supplier A supplies two of every three relays used by the compa

ny. If 60 relays are selected at random from those in use by the company, find the probability that at most 39 of these relays come from supplier A. Assume that the company uses a large number of relays. (Use the normal approximation. Round your answer to four decimal places.)
Mathematics
1 answer:
Reptile [31]3 years ago
7 0

Answer: Our required probability is 0.3162.

Step-by-step explanation:

Since we have given that

n = 60

Probability of success = \dfrac{2}{3}

Probability that at most 39 of these relays come from supplier A would be

P(X\leq 39)=binomcdf(60,\dfrac{2}{3},39)=0.3162

Hence, our required probability is 0.3162.

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Mr. Vara is designing post caps in the shape of a square pyramid for a fence that he just built in his backyard. The caps are so
Lyrx [107]

Answer:

Height of the square pyramid is 6.569 centimeter

Step-by-step explanation:

The volume of the square pyramid is 94.5 cubic centimeters

The height of the square pyramid will be equal to the lengths of the sides of the square.

The volume of the square pyramid  is given by

a^2 * \frac{h}{3}

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\frac{a^3}{3}  = 94.5\\a^3 = 94.5 *3\\a = \sqrt[3]{94.5*3} \\a = 6.569centimeter

Height of the square pyramid is 6.569 centimeter

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g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur
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Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

\displaystyle V = \pi r^2h

Substitute:

\displaystyle (300) = \pi r^2 h

Solve for <em>h: </em>

\displaystyle \frac{300}{\pi r^2} = h

Recall that the surface area of a cylinder is given by:

\displaystyle A = 2\pi r^2 + 2\pi rh

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for <em>h</em>.

\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r}  \end{aligned}

Find its derivative:

\displaystyle A' = 4\pi r - \frac{600}{r^2}

Solve for its zero(s):

\displaystyle \begin{aligned} (0) &= 4\pi r  - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

\displaystyle  \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2}  \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm}   \end{aligned}

In conclusion, the dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

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