Total number of computers sold last week from Best Bargain = 340
Percentage of laptops sold last week from Best Bargain = 75%
These are the information's that we can find given in the question. Based on these information's the answer to the question can be easily found.
Number of laptops sold from Best Bargain last week = 340 * (75/100)
= 340 * (3/4)
= 85 * 3
= 255
So out of the total of 340 computers sold from Best Bargain last week, 255 were laptops. I hope this is the answer you were looking for.
-9x+2y = -36
when x=0 ; 2y = -36
so y = -18
then y-intercept(x=0) is -18
In the table states the frequency of the marbles. It says in the 40 tries she had a frequency of 24 marbles that are yellow and for green it states the frequency is 16 marbles for green.
Answer:
Step-by-step explanation:
Solution is attached below
Answer:
The number of business students that must be randomly selected to estimate the mean monthly earnings of business students at one college is 64.
Step-by-step explanation:
The (1 - <em>α</em>) % confidence interval for population mean is:

The margin of error for this interval is:

The information provided is:
<em>σ</em> = $569
MOE = $140
Confidence level = 95%
<em>α</em> = 5%
Compute the critical value of <em>z</em> for <em>α</em> = 5% as follows:

*Use a <em>z</em>-table.
Compute the sample size required as follows:
![n=[\frac{z_{\alpha/2}\times \sigma}{MOE}]^{2}](https://tex.z-dn.net/?f=n%3D%5B%5Cfrac%7Bz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Csigma%7D%7BMOE%7D%5D%5E%7B2%7D)
![=[\frac{1.96\times 569}{140}]^{2}\\\\=63.457156\\\\\approx 64](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B1.96%5Ctimes%20569%7D%7B140%7D%5D%5E%7B2%7D%5C%5C%5C%5C%3D63.457156%5C%5C%5C%5C%5Capprox%2064)
Thus, the number of business students that must be randomly selected to estimate the mean monthly earnings of business students at one college is 64.