Which ratios are equivalent to the given ratio? 3 to 2 Select each correct answer. 23 4 : 12 1510 18 : 12 6 to 4
1 answer:
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Answer:
(a) S = {MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC}
(b) The probability that Roberto and John attend the conference is 0.10.
(c) The probability that Clarice attends the conference is 0.40.
(d) The probability that John stays home is 0.60.
Step-by-step explanation:
It is provided that :
Marco (<em>M</em>), Roberto (<em>R</em>), John (<em>J</em>), Dominique (<em>D</em>) and Clarice (<em>C</em>) works for the company.
The company selects two employees randomly to attend a statistics conference.
(a)
There are 5 employees from which the company has to select two employees to send to the conference.
So the total number of ways to select two employees is:
The 10 possible samples are:
MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC
(b)
The probability of the event <em>E</em> is:
Here,
n (E) = favorable outcomes
N = Total number of outcomes.
The variable representing the selection of Roberto and John is, <em>RJ</em>.
The favorable number of outcomes to select Roberto and John is, 1.
The total number of outcomes to select 2 employees is 10.
Compute the probability that Roberto and John attend the conference as follows:
Thus, the probability that Roberto and John attend the conference is 0.10.
(c)
The favorable outcomes of the event where Clarice attends the conference are:
n (C) = {MC, RC, JC and DC} = 4
Compute the probability that Clarice attends the conference as follows:
Thus, the probability that Clarice attends the conference is 0.40.
(d)
The favorable outcomes of the event where John does not attends the conference are:
n (J') = MR, MD, MC, RD, RC, DC
Compute the probability that John stays home as follows:
Thus, the probability that John stays home is 0.60.
We can see that revolving the region formed by intersecting 3 lines, we will get 2 cones that are connected their bases.
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50