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3 years ago

What is 9^3/9^1 in expanded form?

Mathematics

2 answers:

amm18123 years ago

7 0

Answer:

9*9*9/9*1

Step-by-step explanation:

9^3 = 9*9*9

9^1 = 9

expanded form would be 9*9*9/9*1

9*9*9/9*1

Alenkinab [10]3 years ago

4 0

Answer:

9*9

Step-by-step explanation:

Expanded form is writing out the exponent with multiplication.

9^3/9^1=9^2, or 9*9

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London was offered a job that paid a salary of \$93,000$93,000 in its first year. The salary was set to increase by 2% per year

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Answer: Amount = $143776

Step-by-step explanation:

Given that London was offered a job that paid a salary of $93,000 in its first year. The salary was set to increase by 2% per year every year.

Let P = 93000

Rate R = 2%

The amount of salary he will receive after 22 years can be calculated by using exponential equation

A = P(1 + 2%)^t

Where t = 22 years

A = 93000(1.02)^22

A = 93000 × 1.545979

A = 143776.11 dollars

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3 years ago

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3 years ago

Standard deviation of 19,29,21,22,23

MaRussiya [10]

Answer:

3.370459909

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3 years ago

Is 12a-a simplified?

7nadin3 [17]

12a−a

=12a−1a

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3 years ago

Of all customers purchasing automatic garage-door openers, 75% purchase Swedish model. Let X = the number among the next 15 purc

Lelechka [254]

Answer:

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = 0.2737

d) μ = 15*0.75 = 11.25. σ² = 11.25*0.25 = 2.8125

Step-by-step explanation:

X is a binomial random variable with parameters n = 15, p = 0.75. Therefore

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

P(X>10) = P(X=11) + P(X=12)+ P(X=13)+P(X=14)+P(x=15)

$P(X=11) = {15 \choose 11} * 0.75^{11} * 0.25^4 = 0.2252$

$P(X=12) = {15 \choose 12} * 0.75^{12} * 0.25^3 = 0.2252$

$P(X=13) = {15 \choose 13} * 0.75^{13} * 0.25^2 = 0.1559$

$P(X=14) = {15 \choose 14} * 0.75^{14} * 0.25 = 0.0668$

$P(X=15) = {15 \choose 15} * 0.75^{15} = 0.0134$

Thus,

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X=9) + P(X=10)

$P(X=6) = {15 \choose 6} * 0.75^{6} * 0.25^9 = 0.0034$

$P(X=7) = {15 \choose 7} * 0.75^{7} * 0.25^8 = 0.0131$

$P(X=8) = {15 \choose 8} * 0.75^{8} * 0.25^7 = 0.0393$

$P(X=9) = {15 \choose 9} * 0.75^{9} * 0.25^6 = 0.0918$

$P(X=10) = {15 \choose 10} * 0.75^{10} * 0.25^{5} = 0.1652$

Thereofre,

$P(6 \leq X \leq 10) = 0.0034 + 0.0134 + 0.0393 + 0.0918 + 0.1652 = 0.2737$

d) μ = n*p = 15*0.75 = 11.25

σ² = np(1-p) = 11.25*0.25 = 2.8125

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3 years ago