What is 18/25 as a decimal

Rebotar Inc.makes basketballs.Their fixed costs are $3,450.

melomori [17]

Answer:

Since the profit is positive, Rebotar not only broke even, they had earnings.

Step-by-step explanation:

Function Modeling

The costs, incomes, and profits of Rebotar Inc. can be modeled by means of the appropriate function according to known conditions of the market.

It's known their fixed costs are $3,450 and their variable costs are $12 per basketball produced and sold. Thus, the total cost of Rebotar is:

C(x) = 12x + 3,450

Where x is the number of basketballs sold.

It's also known each basketball is sold at $25, thus the revenue (income) function is:

R(x) = 25x

The profit function is the difference between the costs and revenue:

P(x) = 25x - (12x + 3,450)

Operating:

P(x) = 25x - 12x - 3,450

P(x) = 13x - 3,450

If x=300 basketballs are sold, the profits are:

P(300) = 13(300) - 3,450

P(300) = 3,900 - 3,450

P(300) = 450

Since the profit is positive, Rebotar not only broke even, they had earnings.

8 0

2 years ago

The food bank has 1774.4 oz of chicken.If 12.6 oz is 2 servings,how many servings of chicken does this provide?

diamong [38]

Answer:

281 servings

Step-by-step explanation:

From the table attached,

We will apply the unitary method to solve this problem.

∵ 12.6 oz of chicken represents number of servings = 2

∴ 1 oz of chicken will represents number of servings = $\frac{2}{12.6}$

∴ 1774.4 oz of chicken will make the total servings = $\frac{2}{12.6}\times 1774.4$

= 281.65

≈ 281 servings

Therefore, total number of servings provided = 281

6 0

3 years ago

A Starbucks coffee shop serves an average of 3500 customers per day, with a standard deviation of 250. What is the probability t

castortr0y [4]

Answer:

$P(3300$

And we can find this probability with this difference:

$P(-0.8$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the amount of cofee shops of a population, and for this case we know the distribution for X is given by:

$X \sim N(3500,250)$

Where $\mu=3500$ and $\sigma=250$

We are interested on this probability

$P(3300$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(3300$

And we can find this probability with this difference:

$P(-0.8$

6 0

3 years ago

Read 2 more answers

36.4 14 14 Find the area

weeeeeb [17]

Answer:7134.4

Step-by-step explanation: 36.4*14 is 509.6 and times 14 again is 7134.4

5 0

3 years ago

Read 2 more answers

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

3 years ago