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1 answer:
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The selection of r objects out of n is done in
many ways.
The total number of selections 10 that we can make from 6+7=13 students is
thus, the sample space of the experiment is 286
A.
<span>"The probability that a randomly chosen team includes all 6 girls in the class."
total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.
</span>
<span>
P(all 6 girls chosen)=35/286=0.12
B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"
with the same logic as in A, the number of groups were all 7 boys are in, is
</span>
<span>
so the probability is 20/286=0.07
C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"
case 1: the team has 4 boys and 6 girls
this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.
there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.
</span>
<span>
the probability is 105/286=0.367
since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:
0.12+0.367=0.487 (approximately = 0.49)
D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"
selecting 5 boys and 5 girls can be done in
</span>
many ways,
so the probability is 126/286=0.44
The total number of selections 10 that we can make from 6+7=13 students is
thus, the sample space of the experiment is 286
A.
<span>"The probability that a randomly chosen team includes all 6 girls in the class."
total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.
</span>
<span>
P(all 6 girls chosen)=35/286=0.12
B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"
with the same logic as in A, the number of groups were all 7 boys are in, is
</span>
<span>
so the probability is 20/286=0.07
C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"
case 1: the team has 4 boys and 6 girls
this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.
there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.
</span>
<span>
the probability is 105/286=0.367
since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:
0.12+0.367=0.487 (approximately = 0.49)
D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"
selecting 5 boys and 5 girls can be done in
</span>
many ways,
so the probability is 126/286=0.44
Answer:
constant of variation = 3
Step-by-step explanation:
we know that b varies jointly with c and d
so:
b∝c∝d
and b varies inversely with e, so
b∝
and i will call the constant of variation k, this way we can make an equation for b in the following form:
this satisfy that b varies jointly with c and d (if b increases, c and d also increase) and inversely with e (if b increases, e decreases)
we know that when b is 18, c is 4, d is 9, and e is 6:
substituting this in our equation for b:
and we solve operations and clear for the constant of variation k:
the constant of variation is 3.