1. x= 11.7 μ = 7
2. z11.7= 1.3
3. Is 11.7 within a z-score of 3?
a. Yes because z11.7 < 3.
4. Which statement is true of z11.7?
b. z11.7 is between 1 and 2 standard deviations of the mean.
<span><span><u>Answer</u>
This explanation shows that the equation is solvable and there is not where in the steps we get a zero on the denominator.
</span><span><u>Explanation</u>
</span><span>The equation given in the statement is;
4/5+3/x=1/2
This equation can be solved as follows:
4/5+3/x=1/2
3/x=1/2-4/5=(5-8)/10
3/x= -3/10
Taking the reciprocal on both side of the equation;
x/3=-10/3
Multiplying by 3 both sides we get;
</span><span>x= -10
</span></span>
Answer:
The numbers of doors that will have no blemishes will be about 6065 doors
Step-by-step explanation:
Let the number of counts by the worker of each blemishes on the door be (X)
The distribution of blemishes followed the Poisson distribution with parameter
/ door
The probability mass function on of a poisson distribution Is:


The probability that no blemishes occur is :


P(X=0) = 0.6065
Assume the number of paints on the door by q = 10000
Hence; the number of doors that have no blemishes is = qp
=10,000(0.6065)
= 6065
Answer:
x is less-than-or-equal-to 2.25 (x ≤ 2.25)
Step-by-step explanation:
We can write down the inequality that represents the weight Li can add without going over the 50 pound limit:
47.75 + x ≤ 50
If we solve for x we have:
47.75 + x ≤ 50
x ≤ 50 - 47.75
x ≤ 2.25
Therefore, the weight Li can add to the suitcase is less-than-or-equal-to 2.25
Answer:
Pool 1 will be drained out just over a minute before pool 2.
Step-by-step explanation:
Pool 1
3700/31 = 119.35minutes
Pool 2
4228/42 = 100.67minutes
Pool 1 will be drained out just over a minute before pool 2