okay im not sure but maybe try to find a number that can be = to 18 and that might be y
So for this, we will be using synthetic division. To set it up, have the equation so that the divisor is -10 (since that is the solution of k + 10 = 0) and the dividend are the coefficients. Our equation will look as such:
<em>(Note that synthetic division can only be used when the divisor is a 1st degree binomial)</em>
- -10 | 1 + 2 - 82 - 28
- ---------------------------
Now firstly, drop the 1:
- -10 | 1 + 2 - 82 - 28
- ↓
- -------------------------
- 1
Next, you are going to multiply -10 and 1, and then combine the product with 2.
- -10 | 1 + 2 - 82 - 28
- ↓ - 10
- -------------------------
- 1 - 8
Next, multiply -10 and -8, then combine the product with -82:
- -10 | 1 + 2 - 82 - 28
- ↓ -10 + 80
- -------------------------
- 1 - 8 - 2
Next, multiply -10 and -2, then combine the product with -28:
- -10 | 1 + 2 - 82 - 28
- ↓ -10 + 80 + 20
- -------------------------
- 1 - 8 - 2 - 8
Now, since we know that the degree of the dividend is 3, this means that the degree of the quotient is 2. Using this, the first 3 terms are k^2, k, and the constant, or in this case k² - 8k - 2. Now what about the last coefficient -8? Well this is our remainder, and will be written as -8/(k + 10).
<u>Putting it together, the quotient is
</u>
Answer:
(3x3) /9
Step-by-step explanation:
first you triple the three (3x3) then divide by 9
Answer:
a) 0.9
b) Mean = 1.58
Standard Deviation = 0.89
Step-by-step explanation:
We are given the following in the question:
A marketing firm is considering making up to three new hires.
Let X be the variable describing the number of hiring in the company.
Thus, x can take values 0,1 ,2 and 3.

a) P(firm will make at least one hire)

Also,


b) expected value and the standard deviation of the number of hires.
![E(x^2) = \displaystyle\sum x_i^2P(x_i)\\=0(0.1) + 1(0.4) + 4(0.32) +9(0.18) = 3.3\\V(x) = E(x^2)-[E(x)]^2 = 3.3-(1.58)^2 = 0.80\\\text{Standard Deviation} = \sqrt{V(x)} = \sqrt{0.8036} = 0.89](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20%5Cdisplaystyle%5Csum%20x_i%5E2P%28x_i%29%5C%5C%3D0%280.1%29%20%2B%201%280.4%29%20%2B%204%280.32%29%20%2B9%280.18%29%20%3D%203.3%5C%5CV%28x%29%20%3D%20E%28x%5E2%29-%5BE%28x%29%5D%5E2%20%3D%203.3-%281.58%29%5E2%20%3D%200.80%5C%5C%5Ctext%7BStandard%20Deviation%7D%20%3D%20%5Csqrt%7BV%28x%29%7D%20%3D%20%5Csqrt%7B0.8036%7D%20%3D%200.89)