Answer:
2,365.31
Step-by-step explanation:
since she earns $750 a week you subtract 750 from 1,076.25 which leaves you with 326.25 so you take the 326.25 an multiply it with 0.0725 and you get 2,365.3125 but you round it to 2,365.31, so that number is how much money worth of stuff she sold last week
The answer is about 12.5 (when rounded to the nearest 10th).
This can be found using the distance formula between each point. One distance is 5, one is
and one is
. Added together gives you the answer above.
Answer:
a. z = 2.00
Step-by-step explanation:
Hello!
The study variable is "Points per game of a high school team"
The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)
The hypothesis is:
H₀: μ ≤ 99
H₁: μ > 99
α: 0.01
There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.
The statistic value under the null hypothesis is:
Z= X[bar] - μ = 101 - 99 = 2
σ/√n 6/√36
I don't have σ, but since this is an approximation I can use the value of S instead.
I hope it helps!
Answer:
If a+b+c=1,
a
2
+
b
2
+
c
2
=
2
,
a
3
+
b
3
+
c
3
=
3
then find the value of
a
4
+
b
4
+
c
4
=
?
we know
2
(
a
b
+
b
c
+
c
a
)
=
(
a
+
b
+
c
)
2
−
(
a
2
+
b
2
+
c
2
)
⇒
2
(
a
b
+
b
c
+
c
a
)
=
1
2
−
2
=
−
1
⇒
a
b
+
b
c
+
c
a
=
−
1
2
given
a
3
+
b
3
+
c
3
=
3
⇒
a
3
+
b
3
+
c
3
−
3
a
b
c
+
3
a
b
c
=
3
⇒
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
a
b
−
b
c
−
c
a
)
+
3
a
b
c
=
3
⇒
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
−
(
a
b
+
b
c
+
c
a
)
+
3
a
b
c
=
3
⇒
(
1
×
(
2
−
(
−
1
2
)
+
3
a
b
c
)
)
=
3
⇒
(
2
+
1
2
)
+
3
a
b
c
=
3
⇒
3
a
b
c
=
3
−
5
2
=
1
2
⇒
a
b
c
=
1
6
Now
(
a
2
b
2
+
b
2
c
2
+
c
2
a
2
)
=
(
a
b
+
b
c
+
c
a
)
2
−
2
a
b
2
c
−
2
b
c
2
a
−
2
c
a
2
b
=
(
a
b
+
b
c
+
c
a
)
2
−
2
a
b
c
(
b
+
c
+
a
)
=
(
−
1
2
)
2
−
2
×
1
6
×
1
=
1
4
−
1
3
=
−
1
12
Now
a
4
+
b
4
+
c
4
=
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
2
b
2
+
b
2
c
2
+
c
2
a
2
)
=
2
2
−
2
×
(
−
1
12
)
=
4
+
1
6
=
4
1
6
Extension
a
5
+
b
5
+
c
5
=
(
a
3
+
b
3
+
c
3
)
(
a
2
+
b
2
+
c
2
)
−
[
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
c
2
)
]
=
3
⋅
2
−
[
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
b
2
)
]
Now
a
3
(
b
2
+
c
2
)
+
b
3
(
c
2
+
a
2
)
+
c
3
(
a
2
+
b
2
)
=
a
2
b
2
(
a
+
b
)
+
b
2
c
2
(
b
+
c
)
+
c
2
a
2
(
a
+
c
)
=
a
2
b
2
(
1
−
c
)
+
b
2
c
2
(
1
−
a
)
+
c
2
a
2
(
1
−
b
)
=
a
2
b
2
+
b
2
c
2
+
c
2
a
2
−
(
a
2
b
2
c
+
b
2
c
2
a
+
c
2
a
2
b
)
=
−
1
12
−
a
b
c
(
a
b
+
b
c
+
c
a
)
=
−
1
12
−
1
6
⋅
(
−
1
2
)
=
0
So
a
5
+
b
5
+
c
5
=
6
−
0
=
6
Step-by-step explanation:
