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Maru [420]
3 years ago
7

-2n-13= - 3n - 5 pls solve!

Mathematics
2 answers:
Elis [28]3 years ago
8 0

Answer:

n = 8

Step-by-step explanation:

-2n - 13 = -3n - 5

-2n + 3n = -5 + 13

n = 8

Hope this helps. Have a nice day!

Yuri [45]3 years ago
8 0
Hey Your answer is n=8
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Surface area of a rectangular prism if W = 6in L = 9in and H = 2in
Delicious77 [7]

Answer:

Step-by-step explanation:

Two faces are 6” by 9”. Two faces are 6” by 2”. Two faces are 9” by 2”.

Surface are = 2*6*9 + 2*6*2 + 2*9*2 = 108 + 24 + 36 = 168 square inches

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3 years ago
Which expressions are equivalent to cos2(105°) – sin2(105°)? Check all that apply.
IgorLugansk [536]

Answer:

1,2,5

Step-by-step explanation:

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3 years ago
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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
2 years ago
A doctor wants to know if her patients are happy with her. She decides to survey a sample of her clients. What method would be b
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8 0
3 years ago
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QUICK<br><br> Find the area of the shaded region. Use 3.14 for pi.
ivolga24 [154]

Answer:

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Step-by-step explanation:

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2 years ago
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