Answer: k = 55 degrees
Step-by-step explanation:
to find J, subtract 122 from 180, you get 58. The total number of degrees is 180. Since L is given as 67, and 67+58=125. Simply subtract 125 from 180 which is 55 degrees
Answer:
Wait From What I See Aren't You Just Supposed To Put A Name And Date?
Step-by-step explanation:
Answer:
Option B) 3
Step-by-step explanation:
we have the number
4162 0012 3456 789a
<u><em>Find the value of a (check number)</em></u>
step 1
Multiply every even-position digit (when counted from the right) in the number by two. If the result is a two digit number, then add these digits together to make a single digit
4162 0012 3456 789a
9(2)+7(2)+5(2)+3(2)+1(2)+0(2)+6(2)+4(2)
(18)+(14)+(10)+(6)+(2)+(0)+(12)+(8)
Remember that If the result is a two digit number, then add these digits together to make a single digit
so
(1+8)+(1+4)+(1+0)+(6)+(2)+(0)+(1+2)+(8)
(9)+(5)+(1)+(6)+(2)+(0)+(3)+(8)=34
step 2
add every odd-position digit
4162 00123456 789a
so
8+6+4+2+0+2+1=23
step 3
Adds the result in step 1 and the result in step 2
34+23=57
The check-digit is what number needs to be added to this total to make the next multiple of 10
In our case, we’d need to add 3 to make 60
therefore
The check number is 3
Answer:
71.123 mph ≤ μ ≤ 77.277 mph
Step-by-step explanation:
Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:
≤ μ ≤ 
Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and 
is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.
the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.
So, if we replace m by 74.2, s by 5.3083, n by 10 and 
by 1.8331, we get that the 90% confidence interval for the mean speed is:
≤ μ ≤ 
74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077
71.123 ≤ μ ≤ 77.277
