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alukav5142 [94]
2 years ago
7

HELP PLEASE! What is the base area of the following cylinder

Mathematics
1 answer:
nataly862011 [7]2 years ago
8 0

Answer:

12

Step-by-step explanation:

thats it okatyyyyyyy

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System of equations y >4, and y> |x-1|
Paul [167]
Y>(+)x+1
Y>4

Y>(+)x
Y>3

X=3
6 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
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>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
6 0
3 years ago
Read 2 more answers
Hard time with these type of problems, need help
amid [387]
Not sure I have had idea of what it is but I don’t wanna give u the wrong answer
4 0
2 years ago
PLEASE HELP THIS IS VERY IMPORTANT I WILL GIVE U BRAIN THING IF ITS CORRECT AND IT SAYS WRITE YOUR REASON AS A DECIMAL SO PWEASE
lozanna [386]

Answer:

its, 4.4

Step-by-step explanation:

5 0
2 years ago
Find all solutions of the given system of equations (If the system is infinite many solution, express your answer in terms of x)
lisov135 [29]

Answer:

(a) The system of the equations \left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right. has no solution.

(b) The system of the equations \left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right. has many solutions y=\frac{2x}{3}-\frac{5}{3}

Step-by-step explanation:

(a) To find the solutions of the following system of equations \left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right. you must:

Multiply 2x-3y=3 by 2:

\begin{bmatrix}4x-6y=6\\ 4x-6y=3\end{bmatrix}

Subtract the equations

4x-6y=3\\-\\4x-6y=6\\------\\0=-3

0 = -3 is false, therefore the system of the equations has no solution.

(b) To find the solutions of the system \left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right. you must:

Isolate x for 4x-6y=10

x=\frac{5+3y}{2}

Substitute x=\frac{5+3y}{2} into the second equation

16\cdot \frac{5+3y}{2}-24y=40\\8\left(3y+5\right)-24y=40\\24y+40-24y=40\\40=40

The system has many solutions.

Isolate y for 4x-6y=10

y=\frac{2x}{3}-\frac{5}{3}

3 0
2 years ago
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