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Vera_Pavlovna [14]
3 years ago
11

If Jerry Adams would normally pay $875 for bodily injury

Mathematics
1 answer:
netineya [11]3 years ago
3 0

Answer:

Jerry Adams normally pays $875 for bodily injury and property damage insurance. His insurance company increases premiums by 150% for 1 accident, 200% for 2-3 accidents, and 250% for 4 accidents. Find ... If the probability that he will live through the year is 0.9989, what is the expected value for the insurance policy?

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Please help on b (left of c) and c !!!
vesna_86 [32]
Alrighty


remember
(ab)^c=(a^c)(b^c)
and
x^\frac{m}{n}=\sqrt[n]{x^m}
and
(x^m)^n=x^{mn} and [tex]x^0=1 for all real numbers x
and
x^{-m}=\frac{1}{x^m}


b.
(x^5y^4)^\frac{1}{2}=((x^5)^\frac{1}{2})((y^4)^\frac{1}{2})=
(x^\frac{5}{2})(y^\frac{4}{2})=(\sqrt{x^5})(\sqrt{y^4})=x^2y^2\sqrt{x}

c.
x^0=1
so
that (x^-3y)^0=1
because exponents first in pemdas
so we are left with
x^2y^-1
x^2y^{-1}=(x^2)(y^{-1})=(x^2)(\frac{1}{y^1})=\frac{x^2}{y}
3 0
2 years ago
Read 2 more answers
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2-cm and a standard dev
Varvara68 [4.7K]

Answer:

The probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is 0.65173.

Step-by-step explanation:

We are given that a company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2 cm and a standard deviation of 2.1 cm. For shipment, 17 steel rods are bundled together.

Let \bar X = <u><em>the average length of rods in a randomly selected bundle of steel rods</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean length of rods = 259.2 cm

           \sigma = standard deviaton = 2.1 cm

           n = sample of steel rods = 17

Now, the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is given by = P(\bar X > 259 cm)

 

     P(\bar X > 259 cm) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{259-259.2}{\frac{2.1}{\sqrt{17} } } ) = P(Z > -0.39) = P(Z < 0.39)

                                                                = <u>0.65173</u>

The above probability is calculated by looking at the value of x = 0.39 in the z table which has an area of 0.65173.

8 0
3 years ago
In a bag of small balls 1/4 are green, 1/8 are blue, 1/12 are yellow and the remaining 26 white. How many balls are blue?
Darina [25.2K]
6/ 24 green 
3/24  blue 
2/24 yellow 
total 11/24 
                
13/24 left which is white 
13 * 2 = 26 white 
blue 4 *2 = 8



8 0
3 years ago
A man sells a type of nut for $7 per pound and a different one for $4.20 per pound, how much of each type should be used to make
qwelly [4]

Answer:

a type nut is 10 pounds

a different one is 14 pounds

Step-by-step explanation:

let a type of the nut be represented by t

Let a different one be represented by d

a type of nut cost $7 per pound

a different one cost $4.20 per pound

The cost of the mixture for 24 pounds = 5.37 * 24

= $128.88

t + d = 24 ........(1)

7t + 4.2d = 128.88 ..........(2)

From equation (1), t = 24 - d

Put t = 24 - d in equation 2

7(24 - d) + 4.2d = 128.88

168 - 7d + 4.2d = 128.88

168 - 2.8d = 128.88

-2.8d = 128.88 - 168

-2.8d = -39.12

d = -39.12 / -2.8

d= 13.97

d = 14 pounds

t = 24 - d

t = 24 - 14

t = 10 pounds

A type nut is 10 pounds. A different one is 14 pounds

5 0
3 years ago
Read 2 more answers
Lena took a nap . she fell asleep at 11:37a.m and she woke up at 1:24p.m. how long did she sleep?
zalisa [80]
1 hour and 57 minutes
6 0
2 years ago
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