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Vikki [24]
2 years ago
5

I will rate brainly if you answer this The number of weekly social media posts varies directly with the square root of the poste

r’s age and inversely with the cube root of the poster’s income. If a 16-year-old person who earns $8,000 makes 64 posts in a week, what is the value of k?
Mathematics
2 answers:
juin [17]2 years ago
6 0
K=320.

If a=age, m=income, and n=number of weekly posts:
The relationship can be modeled by
n=k * sqrt(a) / cbrt(m). sqrt(a) is in the numerator because it is directly proportional to n and cbrt(m) is in the denominator because it is inversely proportional to n.
Plugging in the given values, n=64, a=16, m=8000, 64=k* sqrt(16) / cbrt(8000). sqrt(16)=4, and cbrt(8000)=20, so 64=4k/20=k/5. So k=64*5= 320.
ryzh [129]2 years ago
5 0

Answer:

\large \boxed{\sf \bf \ \ k=320 \ \ }

Step-by-step explanation:

Hello,

The number of weekly social media posts varies directly with the square root of the poster’s age and inversely with the cube root of the poster’s income.

If a 16-year-old person who earns $8,000 makes 64 posts in a week, what is the value of k?

64=\dfrac{\sqrt{16}}{\sqrt[3]{8000}}\cdot k=\dfrac{4}{20}\cdot k=\dfrac{1}{5}\cdot k=0.2\cdot k\\\\k=64*5=320

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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A meter is 3.37 inches more than a yard swimmer a swims 100 meters and swimmer b swims 100 yeards how many more feet did swimmer
olga_2 [115]
Swimmer a swims 100 meters, which is 100 1-meters, which is 100
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1 yard is 3 feet, so 100 yards are 300 feet.

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Swimmer b swims 100 yards, which is 300 feet

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4 0
2 years ago
A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d
tankabanditka [31]
The selection of r objects out of n is done in

C(n, r)= \frac{n!}{r!(n-r)!} many ways.

The total number of selections 10 that we can make from 6+7=13 students is 

C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}=  286
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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>C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35
<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>C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20
<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>C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105
<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>C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126

many ways,

so the probability is 126/286=0.44
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