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3 years ago

10

Is a^-b = -a^b? PLEASE ANSWER!! 99 POINTS

2 answers:

natima [27]3 years ago

3 0

No, 1/a^b (a^-b) doesn't equal -a^b

Nat2105 [25]3 years ago

3 0

Yes a^-b is equal to -a^b

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A team won 9 of its first 12 games. At that rate, how many games will the team win if it plays 64 games in all?

Feliz [49]

9 out of 12 = 9/12 which can be simplified to 3/4. So the team wins 3/4 of its games. 3/4 of 64 = 64 divided by 4 = 16. 16 x 3 = 48.

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3 years ago

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The value of the expression 28+32–3x4–52-5+(7x4)is

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Amy wants to rent a bicycle for the day. The company charges $12 to rent the bike, plus $6 per hour. If Amy has a budget of $42,

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Using the equation y=6x+12, where x is the number of hours and y is the total cost for the given value of x, you can plug in 42 for y and solve:

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4 years ago

Sarah tried to evaluate 21 / 7 * 8 step by step.

vladimir2022 [97]

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24.

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First, use the Order of Operations:

PEMDAS.

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4 years ago

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 32% of all homeowners are ins

Bumek [7]

Answer:

$P(X = 0) = 0.2138$

$P(X = 1) = 0.4025$

$P(X = 2) = 0.2841$

$P(X = 3) = 0.0891$

$P(X = 4) = 0.0105$

Step-by-step explanation:

For each homeowner, there are only two possible outcomes. Either they have invested in earthquake insurance, or they have not. The probability of a home owner having invested in earthquake insurance is independent from other homeowners. So we use the binomial probability distribution 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.

32% of all homeowners are insured against earthquake damage.

This means that $p = 0.32$

Four homeowners are to be selected at random.

This means that $n = 4$

Find the probability distribution of X.

This is the probability of each outcome

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

$P(X = 0) = C_{4,0}.(0.32)^{0}.(0.68)^{4} = 0.2138$

$P(X = 1) = C_{4,1}.(0.32)^{1}.(0.68)^{3} = 0.4025$

$P(X = 2) = C_{4,2}.(0.32)^{2}.(0.68)^{2} = 0.2841$

$P(X = 3) = C_{4,3}.(0.32)^{3}.(0.68)^{1} = 0.0891$

$P(X = 4) = C_{4,4}.(0.32)^{4}.(0.68)^{0} = 0.0105$

6 0

3 years ago