Alguém me ajuda rapido por favor tou aki rachando a cuca e n consigo
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So in this case, we have to replace the known value.
y=3
y=-2x+3
3=-2x+3
Then we leave our unknown value alone.
= x
In this case, our x value would be 0.
We check it...
3=-2(0)+3
3=0+3
3=3
So y=3 x=0
For the second one we have...
y=3x+2
y=-3x-4
For this we substitute the y in any of the equation...
3x+2=-3x-4
We move the unknown values to one side and the ones without unlown values to the other side...
3x+3x=-4-2
Then we solve
6x=-6
Then we leave the unknown value alone.
x=
Then solve for x.
x= -1
Then for our y value we return to one of the original equations and substitute the x value.
y=3x+2
y=3(-1)+2
y=-3+2
y=-1
y=-3x-4
y=-3(-1)-4
y=3-4
y=-1
So in this case we got that x= -1 and y= -1
y=3
y=-2x+3
3=-2x+3
Then we leave our unknown value alone.
In this case, our x value would be 0.
We check it...
3=-2(0)+3
3=0+3
3=3
So y=3 x=0
For the second one we have...
y=3x+2
y=-3x-4
For this we substitute the y in any of the equation...
3x+2=-3x-4
We move the unknown values to one side and the ones without unlown values to the other side...
3x+3x=-4-2
Then we solve
6x=-6
Then we leave the unknown value alone.
x=
Then solve for x.
x= -1
Then for our y value we return to one of the original equations and substitute the x value.
y=3x+2
y=3(-1)+2
y=-3+2
y=-1
y=-3x-4
y=-3(-1)-4
y=3-4
y=-1
So in this case we got that x= -1 and y= -1
Answer:
The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,
And the standard deviation of the distribution of sample mean is given by,
The information provided is:
<em>μ</em> = 144 mm
<em>σ</em> = 7 mm
<em>n</em> = 50.
Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.
Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:
*Use a <em>z</em>-table for the probability.
Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.