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alina1380 [7]
2 years ago
15

A company makes wax candles in the shape of a solid sphere. Suppose each candle has a diameter of 15 cm. If

Mathematics
1 answer:
jekas [21]2 years ago
8 0

We have been given that a company makes wax candles in the shape of a solid sphere. Each candle has a diameter of 15 cm. We are asked to find the number of candles that company can make from 70,650 cubic cm of wax.

To solve our given problem, we will divide total volume of wax by volume of one candle.

Volume of each candle will be equal to volume of sphere.

V=\frac{4}{3}\pi r^3, where r represents radius of sphere.

We know that radius is half the diameter, so radius of each candle will be \frac{15}{2}=7.5 cm.

\text{Volume of one candle}=\frac{4}{3}\cdot 3.14\cdot (7.5\text{ cm})^3

\text{Volume of one candle}=\frac{4}{3}\cdot 3.14\cdot 421.875\text{ cm}^3

\text{Volume of one candle}=1766.25\text{ cm}^3

Now we will divide 70,650 cubic cm of wax by volume of one candle.

\text{Number of candles}=\frac{70,650\text{ cm}^3}{1766.25\text{ cm}^3}

\text{Number of candles}=\frac{70,650}{1766.25}

\text{Number of candles}=40

Therefore, 40 candles can be made from 70,650 cubic cm of wax.

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valentina_108 [34]

Solution:

f(x) = x^{10}-8x^8-8x^3+12x^2-5x-5

We have to find the remainder when f(x) is divided by x^2-1.

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f(1)=1^{10}-8(1)^8-8(1)^3+12(1)^2-5(1)-5=1-8-8+12-5-5=1-16+12-10=13-26=-13\\\\f(-1)=(-1)^{10}-8(-1)^8-8(-1)^3+12(-1)^2-5(-1)-5=1-8+8+12+5-5=1+12=13

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5 0
1 year ago
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laila [671]

Answer:

Probability that Adam and Ella will be selected:

\displaystyle \frac{1}{10}=0.1

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Probabilities

The probability of a random event E to occur is a real number between 0 and 1, both inclusive, where 0 indicates an impossible event and 1 a sure event. There are many techniques to compute probabilities depending on the particular situation and distribution.

This question will be solved by simple calculations and logic, given its simplicity. We know the middle school chess club has 5 members: Adam, Bradley, Carol, Dave, and Ella. Two of them are going to be selected at random to participate in the county chess tournament. We can calculate the number of different ways it can be done without any restriction. It's called the sample space.

The sample space of this event is the combination of 5 members regardless of their position. If {a,b,c,d,e} are the five members, then the possible combinations are {ab,ac,ad,ae,bc,bd,be,cd,ce,de}. Notice that there are only 10 possibilities because the combination ab is the same as ba since it's the same team for the tournament.

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Answer:

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