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Nataly_w [17]
4 years ago
7

Which expression is equivalent to (6i)^3

Mathematics
1 answer:
Rudiy274 years ago
6 0

Step-by-step explanation:

(6i)^3\qquad\text{use}\ (ab)^n=a^nb^n\\\\=6^3i^3\\\\\text{If}\ i=\sqrt{-1},\ \text{then}\\\\6^3i^3=216(-i)=-216i\\\\\text{else}\\\\6^3i^3=216i^3

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TIME REMAINING
mihalych1998 [28]

Answer:

The sum is 10s^2t-5st^2.  With two separate sections, it is a binomial, and since the sum of the powers of the variables is 2+1=3, it has a degree of 3.

4 0
3 years ago
8. Three groups of 10 kids. The average weight of a kid in each
sammy [17]

Answer: 385

Step-by-step explanation:

Mean the three groups of 10 kids=35

Total weight of 3 groups (10x35)=350

Mean weight of 3 groups and one boy  ( 35 + 350)=385

5 0
3 years ago
Find the scale factor please.
torisob [31]

Answer:

5/2

Step-by-step explanation:

The scale factor for linear measures, such as perimeter, is the square root of the scale factor for areas. The ratio of larger area to smaller is ...

... 75/12 = 25/4

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3 0
4 years ago
A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi
34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{X}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

\sigma_{X}=\sqrt{n}\sigma

It is provided that:

\mu=\$27\\\sigma=\$18\\n=310

As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let <em>s</em> denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding <em>z-</em>value is, -2.33.

Compute the value of <em>s</em> as follows:

z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

5 0
3 years ago
22x(4+16)-78x3
Deffense [45]

Answer:

what we solving for???? ill answer in the comments

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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