Answer:
(c) 43.
Step-by-step explanation:
Given:
Margin of error (E) = 2
Sample variance
= 44
Confidence level = 95%
The z- value at 95% confidence level is 1.96.
The sample using the provided information can be calculated as:
![n=(\frac{z(s)}{E})^{2} \\n=\frac{z^{2} s^{2} }{E^{2} } \\n=\frac{1.96^{2} (44) }{2^{2} }\\n= 42.52\\n=43](https://tex.z-dn.net/?f=n%3D%28%5Cfrac%7Bz%28s%29%7D%7BE%7D%29%5E%7B2%7D%20%5C%5Cn%3D%5Cfrac%7Bz%5E%7B2%7D%20s%5E%7B2%7D%20%7D%7BE%5E%7B2%7D%20%7D%20%5C%5Cn%3D%5Cfrac%7B1.96%5E%7B2%7D%20%2844%29%20%7D%7B2%5E%7B2%7D%20%7D%5C%5Cn%3D%2042.52%5C%5Cn%3D43)
Hence, the correct option is (c) 43.
You have to distribute the minus sign to the second parenthesis, changing the sign of every term:
![-3y^2-8-(5y^2+1) = -3y^2-8-5y^2-1 = -8y^2-9](https://tex.z-dn.net/?f=%20-3y%5E2-8-%285y%5E2%2B1%29%20%3D%20-3y%5E2-8-5y%5E2-1%20%3D%20-8y%5E2-9%20)
The average speed of Joshua during that time is 2500 m/h.
Explanation:
It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.
The total time taken by Joshua from 5:15 pm to 8:09 pm is
![2 { h } 54 \text { min }=2 \mathrm{h}+\frac{54}{60} {h}](https://tex.z-dn.net/?f=2%20%7B%20h%20%7D%2054%20%5Ctext%20%7B%20min%20%7D%3D2%20%5Cmathrm%7Bh%7D%2B%5Cfrac%7B54%7D%7B60%7D%20%7Bh%7D)
Dividing we get,
![2 { h } 54 \text { min }=2 \mathrm{h}+0.9 {h}](https://tex.z-dn.net/?f=2%20%7B%20h%20%7D%2054%20%5Ctext%20%7B%20min%20%7D%3D2%20%5Cmathrm%7Bh%7D%2B0.9%20%7Bh%7D)
Adding, we have,
![2 { h } 54 \text { min }=2.9 {h}](https://tex.z-dn.net/?f=2%20%7B%20h%20%7D%2054%20%5Ctext%20%7B%20min%20%7D%3D2.9%20%7Bh%7D)
Thus, the total time taken by Joshua is ![2.9 {h}](https://tex.z-dn.net/?f=2.9%20%7Bh%7D)
To determine the average speed we use the formula,
![speed=\frac{distance}{time}](https://tex.z-dn.net/?f=speed%3D%5Cfrac%7Bdistance%7D%7Btime%7D)
where
and ![time=2.9](https://tex.z-dn.net/?f=time%3D2.9)
Hence, substituting the values we have,
![speed=\frac{7250}{2.9}](https://tex.z-dn.net/?f=speed%3D%5Cfrac%7B7250%7D%7B2.9%7D)
Dividing, we get,
![speed=2500](https://tex.z-dn.net/?f=speed%3D2500)
Thus, the average speed of Joshua during that time is 2500 m/h.
Joel's work makes $9.50/ hr. During the 2 Day, 11 hrs shift, he earns a total of $217.55 but spends a cash of $9 on his bus. Therefore, the money left is $208.55. While on his 4 day, 6 hrs shift, he earns a total of $228 but spends $18, with a total money of $210. The option that he should take to make the most money is the 4 days for 6 hrs shift.
I think the answer is the one in the bottom right.