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Artemon [7]
2 years ago
13

514 divided by 9 as remainder

Mathematics
1 answer:
lesantik [10]2 years ago
6 0

Answer:

57 and a remainder of 1 or 0.11.

Step-by-step explanation:

514dividedby9=57.1

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A parabola and its focus are shown on the graph. The vertex of the parabola is at (0,0).
natulia [17]

Answer:

y = -3

Step-by-step explanation:

the length from directrix to vertex is equal to the length from focus to the vertex.

4 0
3 years ago
Read 2 more answers
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
Production times are evenly distributed between 8 and 15.8 minutes and production times are never outside of this interval. What
dezoksy [38]

Answer:

29.49% probability that a production time is between 9.7 and 12 minutes

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X between c and d, in which d is greater than c, is given by the following formula.

P(c \leq X \leq d) = \frac{d - c}{b-a}

Production times are evenly distributed between 8 and 15.8 minutes and production times are never outside of this interval.

This means that a = 8, b = 15.8

What is the probability that a production time is between 9.7 and 12 minutes?

d = 12, c = 9.7.

So

P(c \leq X \leq d) = \frac{d - c}{b-a}

P(9.7 \leq X \leq 12) = \frac{12 - 9.7}{15.8 - 8} = 0.2949

29.49% probability that a production time is between 9.7 and 12 minutes

5 0
3 years ago
(-2x^-4)^-4 * (2^2)^0 solve the integer equation
Tresset [83]

Answer:

= 1/16x^16

Step-by-step explanation:

6 0
3 years ago
Reverse percentage problem
Gala2k [10]

answer: you're missing the problem

8 0
3 years ago
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