#2 is 5
#3 is 18.9
#4 is 14.1
#14 is yes it does
Mel should use the least common multiple to solve the problem
<u>Solution:</u>
Given, Mel has to put the greatest number of bolts and nuts in each box so each box has the same number of bolts and the same number of nuts.
We have to find that should Mel use the greatest common factor or the least common multiple to solve the problem?
He should use least common multiple.
Let us see an example, suppose 12 bolts and nuts are to be fit in 6 boxes.
Then, if we took H.C.F of 12 and 6, it is 6, which means 6 bolts and nuts in each box, but, after filling 2 boxes with 6 bolts and nuts, there will be nothing left, which is wrong as remaining boxes are empty.
So the remaining method to choose is L.C.M.
Hence, he should use L.C.M method.
30 %
3/10 * 10 = 30/100
30/100= 30%
Answer:
(a) The sample sizes are 6787.
(b) The sample sizes are 6666.
Step-by-step explanation:
(a)
The information provided is:
Confidence level = 98%
MOE = 0.02
n₁ = n₂ = n

Compute the sample sizes as follows:



Thus, the sample sizes are 6787.
(b)
Now it is provided that:

Compute the sample size as follows:

![n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%28z_%7B%5Calpha%2F2%7D%29%5E%7B2%7D%5Ctimes%20%5B%5Chat%20p_%7B1%7D%281-%5Chat%20p_%7B1%7D%29%2B%5Chat%20p_%7B2%7D%281-%5Chat%20p_%7B2%7D%29%5D%7D%7BMOE%5E%7B2%7D%7D)
![=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2.33%5E%7B2%7D%5Ctimes%20%5B0.45%281-0.45%29%2B0.58%281-0.58%29%5D%7D%7B0.02%5E%7B2%7D%7D%5C%5C%5C%5C%3D6665.331975%5C%5C%5C%5C%5Capprox%206666)
Thus, the sample sizes are 6666.
<span>We cannot deduce about the exact location of P between J and K. But we can conclude: segment JP + segment PK = line JK.
</span><span>JP + PK = JK.
</span><span>Substitute first each.
(8z - 17) + (5z + 37) = 17z - 4
Combine like terms.
13z + 20 = 17z - 4
Isolate the variable z.
4z = 24
z = 6
The value of the variable z is then 6 units.</span>