Answer:
249.6
Step-by-step explanation:
I think it’s $34 but i’m not sure
Answer:
(9 x 7)/3 + 4
Step-by-step explanation:
9 x 7 = 63
63 divided by 3 is 21.
21 + 4 = 25
To find the expected value of the distribution, we multiply each outcome by it's probability. Doing this, we get that the expected value of defects on a skateboard is of 
.
Outcomes and probabilities:
0 defects, 9/10 probability
1 defect, 1/20 probability
2 defects, 1/25 probability
3 defects, 1/100 probability.
Expected value:
Dividing both numerator and denominator by 4:
Thus, the expected value of defects on a skateboard is of .
A similar problem is given at: brainly.com/question/23156292.
2x2-5x-18=0
Two solutions were found :
x = -2
x = 9/2 = 4.500
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(2x2 - 5x) - 18 = 0
Step 2 :
Trying to factor by splitting the middle term
2.1 Factoring 2x2-5x-18
The first term is, 2x2 its coefficient is 2 .
The middle term is, -5x its coefficient is -5 .
The last term, "the constant", is -18
Step-1 : Multiply the coefficient of the first term by the constant 2 • -18 = -36
Step-2 : Find two factors of -36 whose sum equals the coefficient of the middle term, which is -5 .
-36 + 1 = -35
-18 + 2 = -16
-12 + 3 = -9
-9 + 4 = -5 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -9 and 4
2x2 - 9x + 4x - 18
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (2x-9)
Add up the last 2 terms, pulling out common factors :
2 • (2x-9)
Step-5 : Add up the four terms of step 4 :
(x+2) • (2x-9)
Which is the desired factorization
Equation at the end of step 2 :
(2x - 9) • (x + 2) = 0
Step 3 :
Theory - Roots of a product :
3.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
