Answer:
<u>The probability that a randomly selected boy in school can run the mile in less than 348 seconds is 1.1%.</u>
Step-by-step explanation:
1. Let's review the information provided to us to answer the question correctly:
μ of the time a group of boys run the mile in its secondary- school fitness test = 440 seconds
σ of the time a group of boys run the mile in its secondary- school fitness test = 40 seconds
2. Find the probability that a randomly selected boy in school can run the mile in less than 348 seconds.
Let's find out the z-score, this way:
z-score = (348 - 440)/40
z-score = -92/40 = -2.3
Now let's find out the probability of z-score = -2.3, using the table:
p (-2.3) = 0.0107
p (-2.3) = 0.0107 * 100
p (-2.3) = 1.1% (rounding to the next tenth)
<u>The probability that a randomly selected boy in school can run the mile in less than 348 seconds is 1.1%.</u>
Answer:
Should be 530.55 total that week
And 80.55 in commissions
Replace f(x) with 0 and swith the equation around to solve for x
x^4 - 32x^2 - 144 = 0
Factor the left side:
(x+6) (x-6) (x^2+4)
so far we have x+6 = 0, x = -6
x -6 = 0, x = 6
solve x^2 +4 = 0
x^2 = -4
x = sqrt(-4)
x = 2i, -21
The answer is B) 2i, 6, -6
-5/8c = 20....multiply both sides by -8/5 (this cancels out the -5/8 on the left
c = 20(-8/5)
c = -160/5
c = - 32
check..
-5/8(-32) = 20
160/8 = 20
20 = 20 (correct)
so c = -32
The z-score of the information given is:
z=(x-μ)/σ
where
μ-mean
σ- standard deviation
from the information given:
x=93.02
μ=89
σ=√625=25
hence
z=(93.02-89)/25
z=0.1608
thus
P(X≥x)=1-0.5636=0.4364=43.64%
