Find the product of this

Laura takes very good care of her vehicles. She owns a blue van and a red truck. Although she bought them both new, she has owne

MrRissso [65]

The age of the van owned = 12 years

The age of the truck owned = 29 years

Step-by-step explanation:

Here, the given question is INCOMPLETE.

Laura takes very good care of her vehicles. She owns a blue van and a red truck. Although she bought them both new, she has owned the truck for 17 years longer than she has owned the van. If the sum of the ages of the vehicles is 41 years, how old is the van and how old is the truck?

Let us assume the number of years Laura has owned the van = S years

So, according to the question:

The number of years she has owned truck = S + 17 years

Now, Sum of the age of ( Truck + Van) = 41 years

⇒ (S + 17 years) + (S years) = 41 years

or, 2 S + 17 = 41

or, 2 S = 4 1 -1 7 = 24

or, S = 24/2 = 12 years

or, S = 12 years

Hence the age of the van owned = 12 years

The age of the truck owned = S + 17 = 12 + 17 = 29 years

6 0

3 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago

How do you do this? algebra II

Pavel [41]

$3^x=23\\
x=\log_323\approx2.85$

6 0

2 years ago

I really need help please :(

MA_775_DIABLO [31]

Answer: 125 mg per serving for orange juice. 90 mg per serving for cranberry

Step-by-step explanation:

6 0

2 years ago

Jake is building a fence around his property. He wants the perimeter to be no more than 100 feet. He also wants the length to be

emmasim [6.3K]

Answer:

C. 20 feet

Step-by-step explanation:

Let x = width

Let y = length

If the length is at least 10 feet longer than the width:

⇒ x + 10 ≤ y

⇒ x ≤ y - 10

If the perimeter is to be no more than 100 ft:

⇒ 2x + 2y ≤ 100

⇒ x + y ≤ 50

⇒ x ≤ 50 - y

Equate the equations for x and solve for y:

⇒ y - 10 = 50 - y

⇒ 2y = 60

⇒ y = 30

Substitute found value of y into one of the inequalities for x:

⇒ x ≤ 30 - 10

⇒ x ≤ 20

Therefore, the maximum possible width of the fence is 20 feet.

8 0

2 years ago

Find the product of this​

Find the product of this