Deanna uses only fruit juice and sparkling water to make

xxTIMURxx [149]

1/18k^6 so uhhhh ya laalalallaa

4 0

3 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. n = 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 s denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding z-value is, -2.33.

Compute the value of s 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

Whats the most accurate estimation of 65+77

Bogdan [553]

You could round 77 to 80 and 65 to 70 and then add 70 and 80 and you will get 150

6 0

2 years ago

Read 2 more answers

What kind of problem do we need to use “indefinite integral” to solve? Use a real life example to explain.

anastassius [24]

Answer:

Indefinite integration acts as a tool to solve many physical problems.

There are many type of problems that require an indefinite integral to solve.

Basically indefinite integration is required when we deal with quantities that vary spatially or temporally.

As an example consider the following example:

Suppose that we need to calculate the total force on a object placed in a non- uniform field.

As an example let us consider a rod of length L that posses an charge 'q' per meter length and suppose that we place it in a non uniform electric field which is given by

$E(x)=\frac{E_{o}}{e^{kx}}$

Now in order to find the total force on the rod we cannot use the similar procedure as we can see that the force on the rod varies with the position of the rod.

But if w consider an element 'dx' of the rod at a distance 'x' from the origin the force on this element will be given by

$dF=E(x)\times qdx\\\\dF=\frac{qE_{o}}{e^{kx}}dx$