Answer:1)5x^2-11x
2) x^2-3x
Step-by-step explanation:
1)(f+g)(x)
f(x)=3x-7
g(x)= 2x-4
[(3x-7)+(2x-4)](x)
3x^2-7x+2x6^2-4x
=<u>5x^2-11x</u>
2) (f-g)(x)
[(3x-7)-(2x-4)](x)
(3x-7-2x+4)(x)
3x^2-7x-2x^2+4x
x^2-3x
25.8%
First, determine how many standard deviations from the norm that 3 tons are. So:
(3 - 2.43) / 0.88 = 0.57/0.88 = 0.647727273
So 3 tons would be 0.647727273 deviations from the norm. Now using a standard normal table, lookup the value 0.65 (the table I'm using has z-values to only 2 decimal places, so I rounded the z-value I got from 0.647727273 to 0.65). The value I got is 0.24215. Now this value is the probability of getting a value between the mean and the z-score. What I want is the probability of getting that z-score and anything higher. So subtract the value from 0.5, so 0.5 - 0.24215 = 0.25785 = 25.785%
So the probability that more than 3 tons will be dumped in a week is 25.8%
Hello,
Assume Ax+By+Cz+D=0 the plane 's equation.
p=(0,0,0)==>A*0+B*0+C*0+D=0==>D=0
q=(0,1,0)==>A*0+B*1+C*0+0=0==>B=0
r=(1,2,3)==>A*1+0*2+C*3+0=0==>A=-3C
Let C=1==>A=-3
An equation of the plane is -3x+z=0
I would but I’m stuck on this problem too
The ans should be ASA because angle AVR is equal to angle EVN (opposite angles equal)
