Answer:
21°
Step-by-step explanation:
set it up
7x-1+3x-9=180
combine like terms
10x-10=180
take the ten to the other side
10x=190
divide both sides by 10
x=19
plug x in
3(10)-9
30-9
21
(Incase you need it the other angle is 159°)
As described in z-distribution the answers are given below:
a) The 95% confidence interval estimate for the population mean spending by a customer on visiting salon per month is given as follows: (747, 853).
b) The sampling error at 95% confidence level is of: $35.78.
What is a z-distribution ?
The normal distribution with a mean of 0 and a standard deviation of 1 is referred to as the standard normal distribution (also known as the Z distribution) (the green curves in the plots to the right). It is frequently referred to as the bell curve since the probability density graph resembles a bell.
solution:
The bounds of the confidence interval are given as follows:
In which:
is the sample mean.
z is the critical value.
n is the sample size. is the standard deviation for the population.
The parameters for this problem are given as follows:
Hence the lower bound of the interval is of:
800 - 200 x 1.96/square root of 55 = 747.
The upper bound of the interval is of:
800 + 200 x 1.96/square root of 55 = 853.
The sampling error for a sample size of 120 is calculated as follows:
200 x 1.96/square root of 120 = $35.78.
To learn more about the z-distribution from the given link
brainly.com/question/4079902
#SPJ1
Answer:
B. 30°
Step-by-step explanation:
Hello,
Let's place the last digit: it must be 2 or 4 or 8 (3 possibilities)
It remainds 4 digits and the number of permutations fo 4 numbers is 4!=4*3*2*1=24
Thus there are 3*24=72 possibilities.
Answer A
If you do'nt believe run this programm
DIM n(5) AS INTEGER, i1 AS INTEGER, i2 AS INTEGER, i3 AS INTEGER, i4 AS INTEGER, i5 AS INTEGER, nb AS LONG, tot AS LONG
tot = 0
n(1) = 1
n(2) = 2
n(3) = 4
n(4) = 7
n(5) = 8
FOR i1 = 1 TO 5
FOR i2 = 1 TO 5
IF i2 <> i1 THEN
FOR i3 = 1 TO 5
IF i3 <> i2 AND i3 <> i1 THEN
FOR i4 = 1 TO 5
IF i4 <> i3 AND i4 <> i2 AND i4 <> i1 THEN
FOR i5 = 1 TO 5
IF i5 <> i4 AND i5 <> i3 AND i5 <> i2 AND i5 <> i1 THEN
nb = ((((n(i1) * 10) + n(i2)) * 10 + n(i3)) * 10 + n(i4)) * 10 + n(i5)
IF nb MOD 2 = 0 THEN
tot = tot + 1
END IF
END IF
NEXT i5
END IF
NEXT i4
END IF
NEXT i3
END IF
NEXT i2
NEXT i1
PRINT "tot="; tot
END
