Answer:
See explanation
Step-by-step explanation:
William has 24 cans of fruit and 60 cans of vegetables that he will be putting into bags for a food drive.
Factor number 24 and 60:
Find the greatest common factor
Hence, the greatest number of bags William can make is 12.
Each of these bags will have 
cans of fruit and 
cans of vegetables.
If he made fewer bags, 6 bags, each of these bags will have 
cans of fruit and 
cans of vegetables.
If he made fewer bags, 4 bags, each of these bags will have 
cans of fruit and 
cans of vegetables.
If he made fewer bags, 3 bags, each of these bags will have 
cans of fruit and 
cans of vegetables.
If he made fewer bags, 2 bags, each of these bags will have 
cans of fruit and 
cans of vegetables.
Answer:
Part A: 14x + 2
Part B: CJ's order is $30, and Cameron's order is $58.
Step-by-step explanation:
<u>Part A:</u> The total cost of the order would be $14 multiplied by the amount of canteens, plus the shipping price of $2. This is what my expression represents.
<u>Part B:</u>
CJ's Order:
14(2) + 2 =
28 + 2
$30
Cameron's Order:
14(4) + 2 =
56 + 2 =
$58
Hope this helps! :D
Answer:
-8.8c + 2.2 OR SOLVED c = 0.25
Step-by-step explanation:
<u>Step 1: Combine like terms</u>
- 5.8c + 4.2 - 3.1 + 1.4c − 5.8c + 4.2 − 3.1 + 1.4c
<em>-8.8c + 2.2</em>
<em />
<u>Step 2: If needed solve for c</u>
-8.8c + 2.2 - 2.2 = 0 - 2.2
-8.8c / -8.8 = -2.2 / -8.8
<em>c = 0.25</em>
<em />
Answer: -8.8c + 2.2 OR SOLVED c = 0.25
Answer:
x = 2.66666667
Step-by-step explanation:
3(x + 1) = 11
We multiply both x and 1 by 3 and get
3x + 3 = 11
Now we subtract 3 from both sides
3x + 3 - 3 = 11 - 3
3x = 8
Now divide both sides by 3
3x/3 = 8/3
x = 2.66666667
(っ◔◡◔)っ ♥ Hope It Helps ♥
Answer:
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
The margin of error of the interval is given by:
In this problem, we have that:
99.5% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?
This is n when M = 0.07. So
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
