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andreyandreev [35.5K]
3 years ago
15

Which type of sequence is shown? -2, 0, 2, 4, 6, . . .

Mathematics
1 answer:
Ivanshal [37]3 years ago
7 0

Answer:

The type of sequence that is show is an arithmetic sequence

Step-by-step explanation:

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What is the volume of a cylinder with base radius 2 and height 9?
Hatshy [7]
V- 113.1 is the answer
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3 years ago
Elsa is selling bracelets at craft shows to raise money for an animal shelter.
JulijaS [17]

Answer:

$258

Step-by-step explanation:

75% = 0.75 when we divide by 100 to convert a percentage to a decimal

75% of total sales = money donated

Total sales = sales from first show + sales from second show

                  = 24n+32 + (40n+56)

                  = 64n + 88

0.75 of (64n+88) = money donated

                            = 0.75 * (64n+88)

                            = 48n + 66

substitute 4 for n, as she charges $4 for each bracelet

money donated = 48*4 + 66 = 258

6 0
3 years ago
A fraction equivalent to 3 over -5
Burka [1]

Answer: -6/10 -9/15 those are equal to -3/5 but not simplified or if you want to eliminate your -3/5 you would use opposite reciprocal and it would be 5/-3

Step-by-step explanation:

3 0
3 years ago
At Factory Y, a worker's wages for a 40-hour week is $200. She is paid 10% of her regular weekly wages for every hour that she w
Leni [432]

Answer:

She worked 18 hours of overtime.

Step-by-step explanation:

If she works 40 hours per week and gets $200, subtract $380 - $200 = $180.

If you multiply $180 by 0.1 (which is 10% converted to a decimal), you get 18 which is your final answer.

3 0
3 years ago
Let $\mu$ and $\sigma^2$ denote the mean and variance of the random variable x. determine $e[(x-\mu)/\sigma]$ and $e{[((x-\mu)/\
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\mathbb E\left(\dfrac{X-\mu}{\sigma}\right)=\dfrac1\sigma\mathbb E(X)-\dfrac\mu\sigma=\dfrac{\mu-\mu}\sigma=0

\mathbb E\bigg(\left(\dfrac{X-\mu}\sigma\right)^2\bigg)=\dfrac1{\sigma^2}\mathbb E\left(X^2-2\mu X+\mu^2\right)=\dfrac{\mathbb E(X^2)-2\mu\mathbb E(X)+\mu^2}{\sigma^2}
=\dfrac{\mathbb E(X^2)-2\mu^2+\mu^2}{\sigma^2}=\dfrac{\mathbb E(X^2)-\mu^2}{\sigma^2}=\dfrac{\mathbb E(X^2)-\mathbb E(X)^2}{\sigma^2}
=\dfrac{\mathbb V(X)}{\sigma^2}=\dfrac{\sigma^2}{\sigma^2}=1
3 0
4 years ago
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