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jarptica [38.1K]
3 years ago
13

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

other features. The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 41 months and a standard deviation of 5 months. Using the empirical rule (as presented in class), what is the approximate percentage of cars that remain in service between 46 and 51 months
Mathematics
1 answer:
erastova [34]3 years ago
3 0

Answer:

15.85%

Step-by-step explanation:

Empirical rule states that for a normal distribution, 68% of the data falls within one standard deviation, 95% falls within two standard deviation and 99.7% falls within three standard deviation.

Given mean (μ) = 41 months, standard deviation (σ) = 5 months

One standard deviation = μ ± σ = 41 ± 5 = (36, 46)

Therefore 68% falls within 36 months and 41 months

Two standard deviation = μ ± 2σ = 41 ± 2(5) = (31,51)

Therefore 95% falls within 31 months and 51 months

Three standard deviation = μ ± 3σ = 41 ± 3(5) = (26, 56)

Therefore 99.7% falls within 26 months and 56 months

The percentage of cars that remain in service between 46 and 51 months = 32%/2 - 0.3%/2 = 16% - 0.15% = 15.85%

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Fofino [41]
<h3>Answer:   14400</h3>

Chances are your teacher doesn't want you to enter any commas. You won't type in the dollar sign either.

============================================================

Explanation:

Nice work on selecting the correct equation to set up.

We'll cross multiply to help solve for x.

The idea of cross multiplication is that \frac{A}{B} = \frac{C}{D} turns into A*D = B*C

So,

\frac{5040}{280000} = \frac{x}{800000}\\\\5040*800000 = 280,000*x\\\\4,032,000,000 = 280,000x\\\\280000x = 4,032,000,000\\\\x = \frac{4,032,000,000}{280,000}\\\\x = 14,400\\\\

That means the person would pay $14,400 on a property worth $800,000.

4 0
3 years ago
In a basketball game, Team A defeated Team B with a score of 97 to 63. Team A won by scoring a combination of two-point baskets,
-BARSIC- [3]

Answer:

14 free throw baskets , 25 two point baskets and 11 three point baskets

Step-by-step explanation:

Let n₁ represent the number of free-throw baskets, n₂ represent the number of two point baskets and n₃ represent the number of three point baskets.

Now, from the question, the number of two point baskets, n₂ is greater than the free throw baskets by 11. This is written as n₂ = n₁ + 11. Also, the number of three point baskets n₃ is three less than the number of free point baskets. This is written as n₃ = n₂ - 3. Since our total number of points equals 97, it follows that, sum of number of points multiplied by each point equals 97. So, ∑(number of points × each point) = 97. Thus,

n₁ + 2n₂ + 3n₃ = 97. Substituting n₂ and n₃ from above, we have n₁ +2(n₁ + 11) + 3(n₁ - 3) = 97.

Expanding the brackets, we have, n₁ + 2n₁ + 22 + 3n₁ - 9 = 97

collecting like terms, we have 6n₁ + 13 = 97

6n₁ = 97 - 13

6n₁ = 84

dividing through by n₁ we have, n₁ = 84/6 =14

so n₁ our free throw baskets equals 14. Substituting this into n₂ our number of two point baskets equals n₂ = n₁ + 11 = 14 + 11 = 25. Our number of three point baskets n₃ = n₁ - 3. So, n₃ = 14 -3 = 11.

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3 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
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Answer:

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Step-by-step explanation:

Given that

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A School Organization sells t-shirts for a fundraiser. Youth-sizes sell for $10 dollars each, and adult sizes sell for $15 each.
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In my opinion there are an infinite amount of answers though what i got when i solver wasn’t right i would think you just have to try and solve on your own.
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4 years ago
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