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3 years ago

Which of the following expressions has the greatest value when x=5? show how you arrived at your choice.

Mathematics

1 answer:

Nitella [24]3 years ago

6 0

Answer:

2x^2+7

Step-by-step explanation:

Let x=5

2x^2+7

2 ( 5)^2 +7 = 2*25 +7 = 50+7 = 57

(x^3-5)/3

(5^3 -5)/3 = (125-5)/3 = 120/3 = 40

(10x -2)/ (x-3)

(10*5-2)/(5-3) = (50-2)/(2) = 48/2 = 24

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ABC Auto Insurance classifies drivers as good, medium, or poor risks. Drivers who apply to them for insurance fall into these th

Misha Larkins [42]

Answer:

a. $P(E_1/A)=0.0789$

b. $P(E_2/A)=0.395$ \

c. $P(E_3/A)=0.526$

Step-by-step explanation:

Let $E_1,E_2,E_3$ are the events that denotes the good drive, medium drive and poor risk driver.

$P(E_1)=0.30,P(E_2)=0.50,P(E_3)=0.20$

Let A be the event that denotes an accident.

$P(A/E_1)=0.01$

$P(A/E_2=0.03$

$P(A/E_3)=0.10$

The company sells Mr. Brophyan insurance policy and he has an accident.

a.We have to find the probability Mr.Brophy is a good driver

Bayes theorem, $P(E_i/A)=\frac{P(A/E_i)\cdot P(E_1)}{\sum_{i=1}^{i=n}P(A/E_i)\cdot P(E_i)}$

We have to find $P(E_1/A)$

Using the Bayes theorem

$P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}$

Substitute the values then we get

$P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}$

$P(E_1/A)=0.0789$

b.We have to find the probability Mr.Brophy is a medium driver

$P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395$

c.We have to find the probability Mr.Brophy is a poor driver

$P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526$

7 0

3 years ago

What is the domain and range please?

Stella [2.4K]

The second graph: y = 3
Domain: all real number
Range: y = 3

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3 years ago

A right rectangular pyramid is sliced parallel to the base, as shown What is the area of the resulting two-dimensional cross sec

omeli [17]

Answer:

15 cm^2

Step-by-step explanation:

Simply multiply the base times the height of the cross section. So 3 x 5 is 15

8 0

3 years ago

I already solved this problem. I just need help with the HINT part.

frutty [35]

$x+y=25 \\ y=25-x~~Other~number~in~terms~of~x \\ \\ x^2+y^2=313 \\ x^2+2xy+y^2=313+2xy \\ (x+y)^2=313+2x(25-x) \\ \frac{25^2-313}{2} =25x-x^2 \\ x^2-25x+156 \\ \\ \Delta =(-25)^2-4\times 1\times 156 \\ \Delta =625-624 \\ \sqrt{\Delta}=1 \\ \\ x= \frac{25 \pm 1}{2} \\ x= \left \{ {{x_1=13} \atop {x_2=12}} \right.$

The number "x" is given by:
$y= \left \{ {{y_1=25-13 \rightarrow~y_1=12} \atop {y_1=25-12 \rightarrow ~y_1=13}} \right.$

Therefore, we have:

$\boxed {S=({12,13})} \rightarrow~The~curly~braces~does~not~want~appear~here.$

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3 years ago

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Goshia [24]

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3 years ago