Answer:
girll how is this college math im in 8th grade and n learned this in 1st grade loll
Step-by-step explanation:
its 15 m I THINK DONT TAKE MY WORD ON IT but tytyty for the points
A = l * w
66 = l * (l - 5)
66 =

-5l
0 =

-5l - 66
l = 11 or 6 check which works.
A = 11 * 6, which fits.
length = 11 miles
width = 5 miles
<h2>
Answer:</h2>
17
<h2>
Step-by-step explanation:</h2>
Cost per Packet * Number of Packets + Shipping = Amount to Spend
0.90 * b + 2.50 = A
You can spend a maximum of $18.50
A < 18.50
Rearrange the formula to find b
Number of Packets = Amount to Spend - Shipping / Cost per Packet
18.50 - 2.50 = 16.00 / 0.9 = 17.7...
You can't have 0.7 of a packet so you have to round down to 17.
Now if you put it into the original formula:
0.90 * 17 + 2.50 = 17.80
And 17.80 is less than 18.50.
Answer:
(8, 10, 12, 14, 16) is the set which describes the set of even integers from 8 to 16
Step-by-step explanation:
To find the set of even integers from 8 to 16 as below :
There are 8,10,12,14,16 even integers in between the numbers 8 and 16
Therefore (8, 10, 12, 14, 16) is the set which describes the set of even integers from 8 to 16
Answer:
Now we can find the p value using the alternative hypothesis with this probability:
Since the p value is large enough, we have evidence to conclude that the true proportion for this case is NOT significanctly higher than 0.75 since we FAIL to reject the null hypothesis at any significance level lower than 30%
Step-by-step explanation:
Information provided
n=100 represent the random sample selected
estimated proportion of students that are satisfied
is the value that we want to test
z would represent the statistic
represent the p value
System of hypothesis
We want to verify if more than 75 percent of his customers are very satisfied with the service they receive, then the system of hypothesis is.:
Null hypothesis:
Alternative hypothesis:
The statistic is given by:
(1)
Replacing the info given we got:
Now we can find the p value using the alternative hypothesis with this probability:
Since the p value is large enough we have evidence to conclude that the true proportion for this case is NOT significanctly higher than 0.75 since we FAIL to reject the null hypothesis at any significance level lower than 30%