(4-4x)(x-3)
4x-12-(4x^2)+12x
(-4x^2)+16x-12=(pxq)(x)
Answer:
f= -2/3
Step-by-step explanation:
Let's solve your equation step-by-step.
f(3)=−2
Step 1: Simplify both sides of the equation.
3f=−2
Step 2: Divide both sides by 3.
3f
3
=
−2
3
f=
−2
3
Answer:
f=
−2
3
Ok so what we need to find is 3 years of the simple interest so let’s get to it!
Ok so we found the principal price which is 250.00
And our rate is going to be 2.85%
A=P(1+rt)
Therefore the answer is going to be A=$271.38, therefore from the simple interest on a principal of $250.00 at a rate of 2.85% per year for 3 years is $271.38.
Answer:
x = 3.85
Step-by-step explanation:
Given equation:
6ˣ = 1,000
now,
on taking log both sides, we get
⇒ log(6ˣ) = log(1,000)
or
⇒ log(6ˣ) = log(10³)
now we know the property of log function that
log(aᵇ) = b × log(a)
thus, applying the above property, we get
⇒ x × log(6) = 3log(10)
or
⇒ x × log(2×3) = 3log(10)
now,
we have another property of log function as":
log(A) = log(A) + log(B)
therefore,
x × [log(2) + log(3)] = 3log(10)
also,
log(10) = 1
log(2) = 0.3010
log(3) = 0.4771
Thus,
⇒ x × [0.3010 + 0.4771 ] = 3 × 1
or
⇒ x × 0.7781 = 3
or
⇒ x = 3.85
Answer:
a) 
And replacing we got:

b) ![E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148](https://tex.z-dn.net/?f=%20E%2880Y%5E2%29%20%3D80%5B%200%5E2%2A0.45%20%2B1%5E2%2A0.2%20%2B2%5E2%2A0.3%20%2B3%5E2%2A0.05%5D%3D%20148)
Step-by-step explanation:
Previous concepts
In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".
The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).
And the standard deviation of a random variable X is just the square root of the variance.
Solution to the problem
Part a
We have the following distribution function:
Y 0 1 2 3
P(Y) 0.45 0.2 0.3 0.05
And we can calculate the expected value with the following formula:

And replacing we got:

Part b
For this case the new expected value would be given by:

And replacing we got
![E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148](https://tex.z-dn.net/?f=%20E%2880Y%5E2%29%20%3D80%5B%200%5E2%2A0.45%20%2B1%5E2%2A0.2%20%2B2%5E2%2A0.3%20%2B3%5E2%2A0.05%5D%3D%20148)