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podryga [215]
2 years ago
11

Peter's salary is twice Anne's salary and half of David's salary. Then the average salary of Anne and David is _ of Peter's sala

ry​
Mathematics
1 answer:
satela [25.4K]2 years ago
6 0
<h3><u>Answer:</u></h3>

Average of salary of Anne and David is 1.25 of Peter's Salary

<h3><u>Explanation:</u></h3>

In the question, relations between salaries of Peter, Anne, and David is given. From the given relations, we need to make few equations and calculate the average of Anne and David in terms of Peter's Salary.

Let us assume Peter's Salary as P.

Then we can calculate Anne's salary A = \frac{P}{2}.

Also David's salary = D = 2 \times P

Average of Anne's and David's salary = \frac{A+D}{2}

Average of Anne's and David's salary  = \frac{\frac{P}{2} + 2 \times P}{2}

Average of Anne's and David's salary  = \frac{5 \times P}{4}

Average = 1.25 \times P

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Tesla wanted to determine the average miles per kWh that their vehicles get across all models and variations. They took a sample
LiRa [457]

Answer:

\mu_{\bar x} = \mu = E(X) =30KWh

\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{100}}=0.3

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".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we select a sample of n =100

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

So then the sample mean would be:

\mu_{\bar x} = \mu = E(X) =30KWh

And the standard deviation would be:

\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{100}}=0.3

6 0
1 year ago
cents.ivan, bella, and ian agreed to share a pizza which was cut into 101010 slices. ivan ate 40\@@, percent of the pizza.how ma
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6 0
2 years ago
Glenna wants to rent a car for a trip to key west for one weekshe calls two car rental companies to get prices. mr kotters renta
Korvikt [17]

Answer: Barbarino's rentals has a better deal.

She has to drive 887.5 miles to spend the same amount at either company.

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

<em>Mr.kotters rentals (A) </em>

  • <em>$99 PER WEEK </em>
  • <em>$0.11per mile over 100 miles </em>

<em>Barbarino's rentals (B) </em>

  • <em>$75 per week </em>
  • <em>$0.15 per mile over 150 miles </em>

For "A"

Cost = 0.11 (432-100) + 99 = $135.52

For "B"

Cost= 0.15 (432-150) +75 = $117.3

Barbarino's rentals has a better deal, since $117.3(B) < $135.52 (A)

To find how many miles would Glenna drive before she would be spending the same amount at either company:

A =B

0.11 (M-100) + 99 =0.15 (M-150) +75 = $117.3

Solving for M (miles)

0.11 M -11+99 = 0.15 M -22.5+75

-11 +99 +22.5 -75 =0.15M -0.11 M

35.5 = 0.04M

35.5/0.04 = M

887.5 =M

She has to drive 887.5 miles to spend the same amount at either company.

4 0
2 years ago
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