Answer:
the probability that five randomly selected students will have a mean score that is greater than the mean achieved by the students = 0.0096
Step-by-step explanation:
From the five randomly selected students ; 160, 175, 163, 149, 153
mean average of the students = 160+175+163+149+153/5
= mean = x-bar = 800/5
mean x-bar = 160
from probability distribution, P(x-bar > 160) = P[ x-bar - miu / SD > 160 -150.8 /3.94]
P( Z>2.34) = from normal Z-distribution table
= 0.0096419
= 0.0096
hence the probability that five randomly selected students will have a mean score that is greater than the mean achieved by the students = 0.0096
where SD = standard deviation = 3.94 and Miu = 150.8
Answer:
The common ratio is 5 and the 5th term is 1250
Step-by-step explanation:
The common ratio is given by
r=U(2)/U (1)
=10/2
=5
The fifth term
U (n)=ar^n-1
U(5)=2(5)^5-1
=2×5^4
=2×625
=1250
In this problem, you must use SohCahToa which basically means:
Sin - opposite over hypotenuse
Cosine - adjacent over hypotenuse
Tangent - Opposite over hypotenuse
So in this picture we have 4 -which is on the opposite side of x - and a 5 -which is the hypotenuse of the triangle. This means we must use sin.
Using a calculator you put in:
Your answer is 53.13 degrees
Let the three gp be a, ar and ar^2
a + ar + ar^2 = 21 => a(1 + r + r^2) = 21 . . . (1)
a^2 + a^2r^2 + a^2r^4 = 189 => a^2(1 + r^2 + r^4) = 189 . . . (2)
squaring (1) gives
a^2(1 + r + r^2)^2 = 441 . . . (3)
(3) ÷ (2) => (1 + r + r^2)^2 / (1 + r^2 + r^4) = 441/189 = 7/3
3(1 + r + r^2)^2 = 7(1 + r^2 + r^4)
3(r^4 + 2r^3 + 3r^2 + 2r + 1) = 7(1 + r^2 + r^4)
3r^4 + 6r^3 + 9r^2 + 6r + 3 = 7 + 7r^2 + 7r^4
4r^4 - 6r^3 - 2r^2 - 6r + 4 = 0
r = 1/2 or r = 2
From (1), a = 21/(1 + r + r^2)
When r = 2:
a = 21/(1 + 2 + 4) = 21/7 = 3
Therefore, the numbers are 3, 6 and 12.
<span>$1025 * 0.06 * 5
</span>
= 333.125
= 333.13
<span>$1025 + $333.13 = $1,358.13
answer: </span><span>ending balance $1,358.13</span>
