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ladessa [460]
3 years ago
9

At her last party, Miracle decorated with window stickers. For this party, she wants to use 4 times as many stickers. Write an e

xpression for the number of stickers Miracle will use. (Use the variable s to represent the number of stickers she used at her last party).
Mathematics
1 answer:
mixas84 [53]3 years ago
6 0

Answer:

s x 4 or 4s

s being the variable

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3/5 divided by 1 1/4
Kipish [7]

Answer:

\frac{12}{55}

Step-by-step explanation:

To divide fractions, here are your steps:

1) Invert the fraction you are dividing by.

2) Multiply the numerators and denominators.

3) Simplify if necessary.

1)

\frac{3}{5} / \frac{4}{11}

2)

\frac{3 * 4}{5 * 11} = \frac{12}{55}

3)

Cannot be further simplified.

6 0
2 years ago
Which expression is equivalent to StartFraction RootIndex 7 StartRoot x squared EndRoot Over RootIndex 5 StartRoot y cubed EndRo
Softa [21]

Answer:

D (x^7/2)(y^-5/3)

Step-by-step explanation:

6 0
2 years ago
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I need help!! Describe how to determine the average rate of change between x = 1 and x = 3 for the function f(x) = 3x^3 + 1. Inc
Lostsunrise [7]

Answer:

39

Step-by-step explanation:

6 0
3 years ago
9.647 • 10 to the 7 power
DIA [1.3K]

Answer:

96470000


Step-by-step explanation:

Exact Form:

9.647 ⋅ 10^7

6 0
2 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
2 years ago
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