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3 years ago

11

A rectangular solid has a base with length 2 centimeters and a width of 8 centimeters. If the volume of the solid is 160 centime

ters to the power of 3 find the height of the solid

1 answer:

Whitepunk [10]3 years ago

8 0

10 centimeters it is 160/8=20/2=10

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Evaluate the algebraic expression for the value given: -3x + X +9;=2*8)1

Strike441 [17]

y =-3x + x + 9

Find y when x = 2

Substitute the value of x= 2 into the equation

y = -3(2) + 2 + 9

y = - 6 + 2 + 9

y = -4 + 9

y = 5

The answer is 5

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8 months ago

Hello, can someone please help me on solving this kind of question?

jek_recluse [69]

Answer: so i think for part a , the first term is a subscrpit 0 which is 10. Then the formula for any term in the sequence is Ao x R ^n. So R =2 so

1st term = 10 ( 2^0)= 1

2nd term = 10 (2^1) =20

3rd term = 10 (2^2)= 40

4th = 80

5th = 160

Part B

The limit as n approaches infinity for a sub n when R is 0, 1 and 2. THE LIMIT FOR R of 0, 1 and 2 is infinity.

Part C...i will have to research...sorry

Step-by-step explanation:

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2 years ago

How do you find the maximum or minimum of g(x) = 2x^2 + 12x +14

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2 years ago

What is the solution of log3x − 5 16 = 2?

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2 years ago

According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years old or older are randomly se

belka [17]

Answer:

a) The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

Step-by-step explanation:

For each smoker, there are only two possible outcomes. Either they started smoking before 21 years old, or they did not. Smokers are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

50% of adult smokers started smoking before 21 years old.

This means that $p = 0.5$

5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

This means that $n = 5$ .

a) The probability that at least 2 of them started smoking before 21 years of age is

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125$

$P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.03125 + 0.15625 = 0.1875$

The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is

This is:

$P(X \leq 4) = 1 - P(X = 5)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125$

$P(X \leq 4) = 1 - P(X = 5) = 1 - 0.03125 = 0.96875$

The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is

This is P(X = 3). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125$

The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

4 0

1 year ago