Answer:
The Alans's average for the course is 84.5
Step-by-step explanation:
We are given
There are 4 tests, A Term paper and a Final examination.
Score = 92, 78, 82, 90.
Term paper score = 80
Final examination score = 86
Weighted mean = ∑ w.f/∑w
also the sum of all the weights is 100% = 1
weighted mean = 15%*92 + 15%*78+15%*82+15%*90+20%*80+20%*86/1
= 51.3 + 33.2
= 84.5
Therefore the Alans's average for the course is 84.5
Answer: 8.4; 14.
Step-by-step explanation:
To find how much ribbon we used to tie presents, we need to multiply 9 and 0.4.
9 x 0.4 = 3.6.
Diego used 3.6 meters of yard to tie some presents. We can subtract that from 12 to find how much ribbon we have left.
12 - 3.6 = 8.4.
We have 8.4 meters of ribbon to make wreaths.
Now we have to find how many wreaths we can make wiht 8.4 meters of ribbon. We know that each wreath needs 0.6, so we should divide 8.4 by 0.6 to find how many wreaths we can make.
8.4 ÷ 0.6 = 14.
Therefore, we can make 14 wreaths with the 8.4 meters of leftover ribbon.
(Part A's answer is 8.4, and Part B's answer is 14)
Answer:
A, C are true . B is not true.
Step-by-step explanation:
Mean of a discrete random variable can be interpreted as the average outcome if the experiment is repeated many times. Expected value or average of the distribution is analogous to mean of the distribution.
The mean can be found using summation from nothing to nothing x times Upper P (x) , i.e ∑x•P(x).
Example : If two outcomes 100 & 50 occur with probabilities 0.5 each. Expected value (Average) (Mean) : ∑x•P(x) = (0.5)(100) + (0.5)(50) = 50 + 25 = 75
The mean may not be a possible value of the random variable.
Example : Mean of possible no.s on a die = ( 1 + 2 + 3 + 4 + 5 + 6 ) / 6 = 21/6 = 3.5, which is not a possible value of the random variable 'no. on a die'
Answer:
Linear function
<h3>
</h3>
Step-by-step explanation:
<h2>
</h2><h3>Linear function</h3>
A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degree of variable y is 1 and the degree of variable x is 1.
<h2>
</h2><h3>Not linear function</h3>
A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degree of variable y is 1
, the degrees of the variables in the equation violate the linear equation definition, which means that the equation is not a linear equation.
<h2>
</h2><h3>Not linear function</h3>
A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degrees of the variables in the equation violate the linear equation definition, which means that the equation is not a linear equation.
<h2>
</h2><h3>Not linear function</h3>
A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degree of variable y is 1 and the degree of variable x is 2.
<h3>Hope it is helpful...</h3>
Using polynomial long division, we get
3x^3+6x^2+11x
_____________
(x+2) | 3x^4-x^2+cx-2
-(3x^4+6x^3)
____________
6x^3-x^2+cx-2
- (6x^3+12x^2)
_____________
11x^2+cx-2
-(11x^2+22x)
__________
(22+c)x-2.
If you're wondering how I did the long division, what I essentially did was get the first value (at the start, it was 3x^4) and divided it by the first value of the divisor (which in x+2 was x) to get 3x^3 in our example. I then subtracted the polynomial by the whole divisor multiplied by, for example, 3x^3 and repeated the process.
For this to be a perfect factor, (x+2)*something must be equal to (22+c)x-2. Focusing on how to cancel out the 2, we have to add 2 to it. To add 2 to it, we have to multiply (x+2) by 1. However, there's a catch, which is that we subtract whatever we multiply (x+2) by, so we have to multiply it by -1 instead. We still need to cross out (22+c)x. Multiplying (x+2) by -1, we get
(-x-2) but by subtracting the whole thing from something means that we have to add -(-x-2)=x+2 to something to get 0. x+2-x-2=0, xo (22+c)x-2 must equal -x-2, meaning that (22+c)=-1 and c=-23
