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Fittoniya [83]
3 years ago
11

What is the area of this object?

Mathematics
1 answer:
yarga [219]3 years ago
4 0

Answer:

280 is the surface area

Step-by-step explanation:

10+10+5+5=30 * 6 =180

180 + 2 * 50

280 in²

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The graph of an arithmetic sequence is shown.
GalinKa [24]

Answer:

The fifth term is 7

Step-by-step explanation:

Looking at the graph

we have the ordered pairs

(1,5),(2,5.5),(3,6),(4,6.5),(5,7)

so

Let

a_1=5\\a_2=5.5\\a_3=6\\a_4=6.6\\a_5=7

The common difference in this arithmetic sequence is 0.5

The value of the fifth term is a_5

therefore

The fifth term is 7

7 0
3 years ago
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Evaluate the function f(x) =-x/2 -7x -2 find f(-9)
NeX [460]

Answer:

f(-9)=\frac{131}{2}

Step-by-step explanation:

In order to evaluate a function at a given x-value, plug in that value into the actual function

f(x)=-\frac{x}{2} -7x-2\\\\f(-9)=-\frac{9}{2} -7(-9)-2\\\\f(-9) = \frac{131}{2}

3 0
2 years ago
What is the slope of a vertical line
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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
3 years ago
64,000 in<br> a fraction
earnstyle [38]

Answer:

undefined

Step-by-step explanation:

4 0
4 years ago
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