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2 years ago

Terri owns a computer repair shop. Her revenue at the end of last month was $8,000. What is Terri's Run Rate?

Mathematics

1 answer:

Ilia_Sergeevich [38]2 years ago

6 0

Based on the information given Terri's Run Rate is $96,000.

Using this formula

Run rate=Revenue for the Period × Number of months in a year

Where:

Revenue for the Period =$8,000

Number of months in a year=12 months

Let plug in the formula

Run rate=$8,000×12 months

Run rate=$96,000

Inconclusion Terri's Run Rate is $96,000.

Learn more about run rate here:brainly.com/question/16134508

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Two numbers total 31 and have a difference of 9. Find the two numbers.

Eva8 [605]

Answer:

20, 11

Step-by-step explanation:

x=first Number

x-9=second number

above two added together=31 or

x+(x-9)=31

2x=40

x=20 first number

x-9=second number

20-9=11 second number

20 (first number) + 11 (second number)=31

31=31

5 0

2 years ago

PLEASE HELP ASAP!!!!!

KATRIN_1 [288]

Answer:

Figure 3 because if u look at the figure it alines with the question its asking you

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2 years ago

Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by $Mean = 2.725$

Two deviations correspond to 5.540, so we want this probability:

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

$P(X$

8 0

3 years ago

if the perimeter of the blanket has to be at least 280 inches is there egnough room for all three of them to cuddle up under the

dezoksy [38]

Answer:

hope it helps

Step-by-step explanation:

angle < EDG = 90°

5x + 4x = 90

9x = 90

x = 90/9

x = 10°

which means

5x = 50°

4x = 40°

---------------------------

sice EF and DG are parallel lines

EHD = 40°

DHG is 90°

then HGD = 50°

-----------------------------------------

ED = 64 x2 (2 sides) = 128 inches

DG = 84 x2 (2 sides) = 168 inches

total perimeter = 128 + 168 = 296 inches

and yes. theres enough room for all 3 to cuddle

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2 years ago

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Dimas [21]

The answer is 10x - 1380m + z

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2 years ago