Answer:
9 = -2/-1 +b
9 = 2 + b
b = 9-2
b = 7
Step-by-step explanation:
Answer:
The probability that the average time 100 random students on campus will spend more than 5 hours on the internet is 0.5
Step-by-step explanation:
We are given that . At Johnson University, the mean time is 5 hrs with a standard deviation of 1.2 hrs.
Mean = 
Standard deviation = 
We are supposed to find the probability that the average time 100 random students on campus will spend more than 5 hours on the internet i.e. P(X>5)


Z=0
P(X>5)=1-P(X<5)=1-P(Z<0)=1-0.5=0.5
Hence the probability that the average time 100 random students on campus will spend more than 5 hours on the internet is 0.5
<em><u>{(-1, -2), (1, 3), (1, 5), (3, 5)}</u></em><em><u>.</u></em>
<em>[</em><em>that's</em><em> </em><em>it]</em><em>:</em><em>)</em>
Answer:
10.5 or 10 1/2
Step-by-step explanation:
You would do 4 2/3 times 2 1/4.Which would equal 10.5 or 10 1/2
Answer:
<u>The measure of the arc CD = 64°</u>
Step-by-step explanation:
It is required to find the measure of the arc CD in degrees.
So, as shown at the graph
BE and AD are are diameters of circle P
And ∠APE is a right angle ⇒ ∠APE = 90°
So, BE⊥AD
And so, ∠BPE = 90° ⇒(1)
But it is given: ∠BPE = (33k-9)° ⇒(2)
From (1) and (2)
∴ 33k - 9 = 90
∴ 33k = 90 + 9 = 99
∴ k = 99/33 = 3
The measure of the arc CD = ∠CPD = 20k + 4
By substitution with k
<u>∴ The measure of the arc CD = 20*3 + 4 = 60 + 4 = 64°</u>