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love history [14]
1 year ago
14

Which statement is true about the slope of the graphed line?

Mathematics
1 answer:
horsena [70]1 year ago
5 0

Answer:

c

Step-by-step explanation:

because I had a test on these

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7- square root of 3i
r-ruslan [8.4K]

Step-by-step explanation:

Any complex number is of the form: a + bi where a is real part and B is imaginary part.

Therefore, in given complex number: 7- \sqrt {3}i

7 is real part.

\sqrt {3} is imaginary part.

Hence, first two options are correct.

Last two options are incorrect because of the following reasons:

- \sqrt {3} is coefficient of i & not 7- \sqrt {3}

Given number is a difference of a real number and imaginary number & not a sum of a real number and imaginary number.

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3 years ago
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The answer is A. If there are 3 thirds (1/3s) per yard and there are 6 yards, 3*6=18. Because the last cut will make two 1/3-yard segments, the answer is 17 cuts and not 18.
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2 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.

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2 years ago
0.001y^12 as a monomial
Cerrena [4.2K]

Answer:

Step-by-step explanation:

0.001y^12 has only one term and is thus a monomial.  Are you sure you've shared the entire question?

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2 years ago
A person has a bag containing quarters and dimes. There are a total of 61 coins in the bag, and the total value of the coins is
77julia77 [94]

Answer:

Q+D = 65, multiply both sides by 2525Q + 25D = 162525Q + 10D = 950  subtract to get            15D = 675               D = 675/15 = 135/3 = 45 dimes               Q = 65-45 = 20 quarters25(20) + 10(45) = 500+450 = 950

6 0
2 years ago
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