Alright, let's do all of these (though this is a bit long).
1.
The constant is 1.8. All other values are coefficients to variables, which as the name implies will change.
2.
1 hour is 60 minutes, 1 minute is 60 seconds.
So, 4.2 *60 *60 = 15120 seconds.
3.
<span>−5x−4(x−6)=−3-5x-4(x-6)=-3
Let's move all x to one side, and all other numbers to another.
-5x-4(x-6)=-3-5x-4(x-6)=-3
x can be any value you want, if you actually solve this you'll only end up with -3 = -3, which is correct, of course.
Let me show you:
</span><span>−5x−4(x−6)=−3-5x-4(x-6)=-3
+5x +4(x-6) +5x +4(x-6)
-3 = -3
The value of x is irrelevant, then. X can be any real number.
4.
I'm going to assume it was an error in printing with this? If not please correct me.
m=a+2b(or b2)
subtract 2b from each
a=m-2b
(This question seems kind of odd. We should probably address this in the comments.)
5.
</span><span>5(x−2)<−3x+6
Move all x to one side, numbers to other.
5x-10<-3x+6
+3x +3x
+10 +10
8x<16
/8
<span>x < 2
</span>6.
y-3=3(x-5)
alright, to find zeros set one variable to zero and solve
x first
-3=3x-15
+15 +15
3x=12
/3
x=4
x-int is (4,0)
now y
</span>y-3=3(0-5)
y-3=-15
+3 +3
y=-12
so y-int is (0,-12)
i've got to sleep now so i'll do the rest tomorrow. Sorry for the incomplete answer.
Answer:
n=3.5
Step-by-step explanation:
2²=4
14÷4=3.5
3.5×4=14
3.5×2²=14
Answer:
The sample has not met the required specification.
Step-by-step explanation:
As the average of the sample suggests that the true average penetration of the sample could be greater than the 50 mils established, we formulate our hypothesis as follow
: The true average penetration is 50 mils
: The true average penetration is > 50 mils
Since we are trying to see if the true average is greater than 50, this is a right-tailed test.
If the <em>level of confidence</em> is α = 0.05 then the 
score against we are comparing with, is 1.64 (this is because the area under the normal curve N(0;1) to the right of 1.64 is 0.05)
The z-score associated with this test is
where
= <em>mean of the sample</em>
= <em>average established by the specification</em>
s = <em>standard deviation of the sample</em>
n = <em>size of the sample</em>
Computing this value of z we get z = 3.42
Since z > we can conclude that the sample has not met the required specification.
Answer:
-4
Step-by-step explanation:
I hope this helps
