Answer:
<A = 115
<C = 50
Step-by-step explanation:
<A + <B + <C = 180
SO
(4X -13) +15 +(X+18) =180
5X +20 =180
5X = 160
X= 32
LOG IN TO <A = 4*32 -13 = 115
<C = 32+18 = 50
3x - 17 = 9x + 7
Rearrange on same sides
-17 - 7 = 9x - 3x
-24 = 6x
Divide by 6 on either sides to isolate 6
![-\frac{24}{6} = \frac{6x}{6}](https://tex.z-dn.net/?f=%20-%5Cfrac%7B24%7D%7B6%7D%20%3D%20%5Cfrac%7B6x%7D%7B6%7D%20)
6 and 6 cancels out
-4 = x
Answer is:
B. x = -4
PROBABILITY = 11 letters
VOWELS + T = A, E, I, O, U, Y, T = 7
P=7/11
Answer:
Infinite many solutions. Any x-value can satisfy the equation.
Step-by-step explanation:
Let's work on simplifying the equation a little to investigate which x-values satisfy it. Start by combining like terms on the left side (6x +4x=10x),
then distribute the factor "10" into the binomial (x+10), obtaining 10x +30.
Now we have the same expression on the left and the right of the equal sign:
10x +30=10x+30. We may subtract 30 from both sides, and obtain 10x=10x, and at this point divide by 10 both sides, and we obtain: x=x
The process is shown below.
![6x+30+4x=10(x+3)\\6x+4x+30=10x+30\\10x+30=10x+30\\10x=10x\\x=x](https://tex.z-dn.net/?f=6x%2B30%2B4x%3D10%28x%2B3%29%5C%5C6x%2B4x%2B30%3D10x%2B30%5C%5C10x%2B30%3D10x%2B30%5C%5C10x%3D10x%5C%5Cx%3Dx)
x=x is an equation that is verified by absolutely ANY x value on the number line, and there are infinite x-values in the number line.
Therefore there are infinite many solutions to this equation (any x-value will satisfy it).
Answer: The mean of this distribution = 61.95
The standard deviation of this sampling distribution (i.e., the standard error= 0.048
Step-by-step explanation:
Given : Data from the U.S. Department of Education indicates that 59% of business graduate students from private universities had student loans.
i.e. proportion of business graduate students from private universities had student loans : p=0.59
sample size : n=105
Then , the mean of the distribution is given by :-
![\mu= np = (105)(0.59)=61.95](https://tex.z-dn.net/?f=%5Cmu%3D%20np%20%3D%20%28105%29%280.59%29%3D61.95)
∴The mean of this distribution = 61.95
Then standard deviation of this sampling distribution is given by :-
![\sqrt{\dfrac{p(1-p)}{n}}\\\\=\sqrt{\dfrac{(0.59)(1-0.59)}{105}}\\\\=\sqrt{0.00230381}\\\\=0.047998015832\approx0.048](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%5C%5C%5C%5C%3D%5Csqrt%7B%5Cdfrac%7B%280.59%29%281-0.59%29%7D%7B105%7D%7D%5C%5C%5C%5C%3D%5Csqrt%7B0.00230381%7D%5C%5C%5C%5C%3D0.047998015832%5Capprox0.048)
∴The standard deviation of this sampling distribution (i.e., the standard error= 0.048