30in I know that because I just did this .this is EASY!
Assume
if |a|=b
a=b and -a=b
so
9|9-8x|=2x+3
divide both sides by 9
|9-8x|=2/9x+1/3
assume
9-8x=2/9x+1/3 and
-9+8x=2/9x+1/3
9-8x=2/9x+1/3
times 9 both sides
81-72x=2x+3
add 72x both sides
81=74x+3
minus 3
78=74x
divide both sides by 74
39/37=x
other one
-9+8x=2/9x+1/3
multiply both sides by 9
-81+72x=2x+3
minus 2x both sides
-81+70x=3
add 81 to both sides
70x=84
divide boh sides by 70
x=6/5
x=39/37 and 6/5
check
9|9-8(39/37)|=2(39/37)+3
2646/37=189/37
false
9|9-8(6/5)|=2(6/5)+3
351/5=27/5
false
no soltion evidently, weird
Y = -1/2x + 3
Why?
Looking at the line start where the line intercepts the y line, it goes down one over 2 and that’s your slope, the 3 is bc that’s where the line crosses the y line
Answer:
90
Step-by-step explanation:
84+90=174
174/2=87
Answer:
1. E(Y) = 50.54°F
2. SD(Y) = 11.34°F
Step-by-step explanation:
We are given that The daily high temperature X in degrees Celsius in Montreal during April has expected value E(X) = 10.3°C with a standard deviation SD(X) = 3.5°C.
The conversion of X into degrees Fahrenheit Y is Y = (9/5)X + 32.
(1) Y = (9/5)X + 32
E(Y) = E((9/5)X + 32) = E((9/5)X) + E(32)
= (9/5) * E(X) + 32 {
expectation of constant is constant}
= (9/5) * 10.3 + 32 = 50.54
Therefore, E(Y), the expected daily high in Montreal during April in degrees Fahrenheit is 50.54°F .
(2) Y = (9/5)X + 32
SD(Y) = SD((9/5)X + 32) = SD((9/5)X) + SD(32)
=
* SD(X) + 0 {
standard deviation of constant is zero}
=
* 3.5 = 11.34°F
Therefore, SD(Y), the standard deviation of the daily high temperature in Montreal during April in degrees Fahrenheit is 11.34°F .