The radius of a circle is decreasing at a constant rate of 0.1 centimeter per second. In terms of the circumference C, what is t
1 answer:
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Answer:
a. The average number of hashtags used in a social media post from a marketing agency is different than 7 hashtags.
Step-by-step explanation:
A social media platform states that a social media post from a marketing agency has 7 hashtags, on average.
This means that at the null hypothesis, we test if the mean is 7, that is:
A digital marketing specialist studying social media advertising believes the average number of hashtags used in a post from a marketing agency is different than the number stated by the social media platform.
Keyword is different, so at the null hypothesis, we test if the mean is different of 7, that is:
Thus, the correct answer is given by option a.
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