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nadya68 [22]
3 years ago
11

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 63 months and a standard deviation of 8 months. Using the empirical rule (as presented in the book), what is the approximate percentage of cars that remain in service between 47 and 55 months?
Mathematics
2 answers:
nignag [31]3 years ago
6 0

Answer:

That probability is 0.1838 or 18.4%

Step-by-step explanation:

Sophie [7]3 years ago
4 0

Answer:

That probability is 0.1838 or 18.4% or enter 18.4

Step-by-step explanation:

mean 35 sd 5

20 is 3 sd s to the left of the mean

30 is 1 sd to the left.

The empirical rule has 68% within 1 sd or 34% on one side

It has 95% within 2 sd or 47.5% on one side

It has 99.7% within 3 sd or 49.85% on one side

Therefore between 3 sd and 1 sd on one side is 49.85-34=15.85% or enter 15.9

mean of 48 sd 7

between 48 and 55 is between -1 and 0 sd or 34% enter 34. The last one doesn't seem to post easily:z=(x-mean)/sd or z< (1217-1481)/293 or z <-264/293 or -0.901,

That probability is 0.1838 or 18.4% or enter 18.4

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Describe how the graph of y= x2 can be transformed to the graph of the given equation. y = x2 + 3 (5 points)
kari74 [83]

Answer:

x=-1 1/2,x=-1.5

Step-by-step explanation: x=-3/2

6 0
3 years ago
An insurance company sells automobile liability and collision insurance. Let X denote the percentage of liability policies that
qaws [65]

Answer:

-764.28

Step-by-step explanation:

Given the joint cumulative distribution of X and Y as

F(x,y) = \frac{xy(x+y)}{2000000}\ \ \ \, \ 0\leq x100, 0\leq y\leq 100

#First find F_x and probability distribution function ,f_x(x):

F_x(x)=F(x,100)\\\\\\=\frac{100x(x+100)}{2000000}\\\\\\\\=\frac{100x^2+10000x}{2000000}\\\\\\=>f_x(x)=\frac{x}{10000}+\frac{1}{200}

#Have determined the probability distribution unction ,f_x(x), we calculate the Expectation of the random variable X:

E(X)=\int\limits^{100}_0 \frac{x^2}{10000}+\frac{x}{200}  dx \\\\\\\\=|\frac{x^3}{30000}+\frac{x^2}{400}|\limits^{100}_0\\\\=58.33\\\\

#We then calculate E(X^2):

E(X^2)=\int\limits^{100}_0 \frac{x^3}{10000}+\frac{x^2}{200}\ dx\\\\=\frac{x^4}{40000}+\frac{x^3}{600}|\limits^{100}_0=4166.67\\\\Var(X)=E(X^2)-(E(X))^2=4166.67-58.33^2\\\\Var(X)=764.28

Hence, the Var(X) is 764.28  

4 0
3 years ago
What the missing term is
arlik [135]

Answer:

10x . . . . . (Note the sign of the middle term is negative. 10x goes in the box.)

Step-by-step explanation:

The sum of the two roots is ...

... (5 -3i) +(5 +3i) = 10

In a quadratic with leading coefficient 1, the coefficient of the 1st-degree term is the opposite of the sum of the roots. Here, that means the middle term is -10x. The minus sign is given, so the answer is 10x.

_____

<em>How Do We Know?</em>

When "a" and "b" are roots of a quadratic in x, it has factors (x -a)(x -b). The product of those two factors is ...

... (x -a)(x -b) = x² -(a+b)x +ab

Here, that means the product of the factors (5 -3i)(5 +3i) is ab = 34, which it is. Their sum is (a+b) = 10, so the x-term is -(a+b)x = -10x.

6 0
3 years ago
The sum of three numbers is 110. The third number is 4 times the second. The first number is 10 less than the second. What are t
Ede4ka [16]

Answer:

This is a bit confusing for me but here we go.

1x110

10x11 and

i don't really know the third

Step-by-step explanation:

5 0
3 years ago
True or False.<br><br> The expressions 4(y−3) and 4y−12 are equivalent?
pickupchik [31]

Answer:

true

Step-by-step explanation: its true 4 x 1y =4y then you multiply that 3 with the four bc you do the same for both sides it will look the same

7 0
3 years ago
Read 2 more answers
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