Answer: 0.4302
Step-by-step explanation:
Given : Mean : 
Standard deviation : 
Sample size : 
Also, these distances are normally distributed.
Then , the formula to calculate the z-score is given by :-
For x=32.5
For x=40.5
The p-value = 
Hence, the required probability :-0.4302
Answer:
$4500 in interest.
Step-by-step explanation:
9% of 10,000 is 900, 900 x 5 is 4500
The angle that is coterminal to 425° is the last one:
B = 425° + n*1,440°
<h3>Which measure is of an angle that is coterminal with a 425° angle?</h3>
By definition, for any angle A, we can say that an angle B is coterminal to A if:
B = A + n*360°
where n can be any integer.
So, from the given options, we need to see which one is a multiple of 360°.
Of the given options, the only that meets this condition is the last one:
B = 425° + n*1,440°
Where:
1,440°/360° = 4
Then we conclude that:
425° + n*1,440° is coterminal to 425°.
If you want to learn more about coterminal angles:
brainly.com/question/3286526
#SPJ4
Answer:
Option B and D
Step-by-step explanation:
Multiply 80 by 1/10
Divide 80 by 10
We are choosing 2
2
r
shoes. How many ways are there to avoid a pair? The pairs represented in our sample can be chosen in (2)
(
n
2
r
)
ways. From each chosen pair, we can choose the left shoe or the right shoe. There are 22
2
2
r
ways to do this. So of the (22)
(
2
n
2
r
)
equally likely ways to choose 2
2
r
shoes, (2)22
(
n
2
r
)
2
2
r
are "favourable."
Another way: A perhaps more natural way to attack the problem is to imagine choosing the shoes one at a time. The probability that the second shoe chosen does not match the first is 2−22−1
2
n
−
2
2
n
−
1
. Given that this has happened, the probability the next shoe does not match either of the first two is 2−42−2
2
n
−
4
2
n
−
2
. Given that there is no match so far, the probability the next shoe does not match any of the first three is 2−62−3
2
n
−
6
2
n
−
3
. Continue. We get a product, which looks a little nicer if we start it with the term 22
2
n
2
n
. So an answer is
22⋅2−22−1⋅2−42−2⋅2−62−3⋯2−4+22−2+1.
2
n
2
n
⋅
2
n
−
2
2
n
−
1
⋅
2
n
−
4
2
n
−
2
⋅
2
n
−
6
2
n
−
3
⋯
2
n
−
4
r
+
2
2
n
−
2
r
+
1
.
This can be expressed more compactly in various ways.
