Answer:
x≤ −2/3y + −1/3
Step-by-step explanation:
because...
Step 1: Add -4y to both sides.
6x + 4y + −4y ≤ −2+−4y
6x ≤ −4y −2
Step 2: Divide both sides by 6.
6x/6 ≤ 
x ≤ −2/3y + −1/3
Therefor, the answer is x≤ −2/3y + −1/3
* Hopefully this helps:) Mark me the brainliest:)!!
Answer:
0.7698
Step-by-step explanation:
If you call your random variable
, then what you are looking for is

because you want the probability of
being <em>between 87 and 123.</em>
We need a table with of the normal distribution. But we can only find the table with
and
. Because of that, first we need to <em>normalize </em>our random variable:

(you can always normalize your variable following the same formula!)
now we can do something similar to our limits, to get a better expression:


And we transform our problem to a simpler one:
(see Figure 1)
From our table we can see that
(this is represented in figure 2).
Remember that the whole area below the curve is exactly 1. So we can conclude that
(because 0.8849 + 0.1151 = 1). We also know the normal distribution is symmetric, then
.
FINALLY:

Well if the two numbers are equal to 5x(8+3).
Lets say that these two numbers are X and Y,
So that would be X + Y = 5(8+3)
which means X + Y = 55
You could write that as a fuctions Y = 55 - X
Now by using the graph you get and the X values of [0,55] ( This means every value from 0 to 55 ), you will get a X and Y value that if you added up together will give you the value of 8(5+33)
An examples for the answers would be
(0,55) (1,54) (2,53) (3,52) (4,51) (5,50) (6,49) ... it goes on until x reaches 55.
Answer:0.1446
Step-by-step explanation:
Let p be the population proportion of adults say that cloning of humans should be allowed.
As per given , the appropriate set of hypothesis would be :-

Since
is left-tailed and sample size is large(n=1012) , so we perform left-tailed z-test.
Given : The test statistic for this test is -1.06.
P-value for left tailed test,
P(z<-1.06)=1-P(z<1.06) [∵P(Z<-z)=1-P(Z<z]
=
(Using z-value table.)
Hence, the p-value for this test statistic= 0.1446