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3 years ago

Find the values of x and y.

Mathematics

1 answer:

eimsori [14]3 years ago

3 0

Answer:

The answers are

x=12

y=60

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During registration at a university, 170 students took advantage of the online system to sign up for classes. Sixty-two students

Lina20 [59]

The correct answer is Choice B. The mean is about 31.

To find the mean we have to add up all the different wait times and then divide by the total.

Add them up:
62(20) + 87(35) + 21(50) = 5335

Divide it by the total:
5335 / (62 + 87 + 21) = 31.38

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3 years ago

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Answer:

C. 11/100

Step-by-step explanation:

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3 years ago

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x is directly proportional to y and inversely proportional to z. if x= 1/2 when y = 3/4 and z=2/3, find x when y= 7/8 and z = 7/

ipn [44]

Answer:

1/2

Step-by-step explanation:

x is directly proportional to y (this means y will go on top because of the directly part) and inversely proportional to z (this means z will go on bottom due to the inversely part).

There is a constant k such that:

$x=k \cdot \frac{y}{z}$ .

Contant means it will never change. It will not care what (x,y,z) you use, it will remain the same.

We will use the first point to find k and then that k will still be there no matter what (x,y,z) they give you.

We have (1/2 , 3/4 , 2/3) is on our graph of the equation:

$x=k \cdot \frac{y}{z}$ .

Insert the numbers:

$\frac{1}{2}=k \cdot \frac{\frac{3}{4}}{\frac{2}{3}}$

Multiply both sides by $\frac{2}{3}$ :

$\frac{1}{2}\cdot \frac{2}{3}=k \cdot \frac{3}{4}$

Simplify left hand side:

$\frac{1}{3}=k \cdot \frac{3}{4}$

Multiply both sides by 4:

$\frac{4}{3}=k \cdot 3$

Multiply both sides by 1/3 (or you can say divide by 3):

$\frac{4}{9}=k$

So k=4/9 no matter the (x,y,z).

$x=\frac{4}{9} \cdot \frac{y}{z}$

We are asked to find x given y=7/8 and z=7/9.

Input these numbers:

$x=\frac{4}{9} \cdot \frac{\frac{7}{8}}{\frac{7}{9}}$

Change the division to multiplication:

$x=\frac{4}{9} \cdot \frac{7}{8} \cdot \frac{9}{7}$

I see a 7 on top and bottom that I can cancel:

$x=\frac{4}{9} \cdot \frac{1}{8} \cdot \frac{9}{1}$

I see a 9 on top and bottom that I can cancel:

$x=\frac{4}{1} \cdot \frac{1}{8} \cdot \frac{1}{1}$

Let's go ahead and multiply and reduce more later if we can.

Multiply straight across on top.

Multiply straight across on bottom.

$x=\frac{4}{8}$

Divide top and bottom by 4:

$x=\frac{1}{2}$

4 0

3 years ago

Pat Statsdud is taking an economics course. Pat's exam strategy is to rely on luck for the next exam. The exam consists of 20 mu

Delicious77 [7]

Answer:

0.01386 or 1.386%

Step-by-step explanation:

Each question has a binomial distribution with probability of success p =0.25 (1 correct answer out of four alternatives).

The probability of 'k' successes in n trials is given by:

$P(x=k)=\frac{n!}{(n-k)!k!}*p^k*(1-p)^{n-k}$

Pat will pass the exam if x ≥ 10. The probability that Pat will pass is:

$P(pass)=P(x=10)+P(x=11)+P(x=12)+P(x=13)+P(x=14)+P(x=15)+P(x=16)+P(x=17)+P(x=18)+P(x=19)+P(x=20)$

The probability for each number of success is:

$P(x=10)=\frac{20!}{(20-10)!10!}*0.25^{10}*0.75^{10}=0.0099\\\\P(x=11)= \frac{20!}{(20-11)!11!}*0.25^{11}*0.75^{9}=0.0030\\\\P(x=12)=\frac{20!}{(20-12)!12!}*0.25^{12}*0.75^{8}=0.00075\\\\P(x=13)=\frac{20!}{(20-13)!13!}*0.25^{13}*0.75^{7}=0.00015\\\\P(x=14)=\frac{20!}{(20-14)!14!}*0.25^{14}*0.75^{6}=0.0000257\\\\P(x=15)=\frac{20!}{(20-15)!15!}*0.25^{15}*0.75^{5}=3.426*10^{-6}\\\\$

$P(x=16)=\frac{20!}{(20-16)!16!}*0.25^{16}*0.75^{4}=3.569*10^{-7}\\\\P(x=17)=\frac{20!}{(20-17)!17!}*0.25^{17}*0.75^{3}=2.799*10^{-8}\\\\P(x=18)=\frac{20!}{(20-18)!18!}*0.25^{18}*0.75^{2}=1.555*10^{-9}\\\\P(x=19)=\frac{20!}{(20-19)!19!}*0.25^{19}*0.75^{1}=5.457*10^{-11}\\\\P(x=20)=\frac{20!}{(20-20)!20!}*0.25^{20}*0.75^{0}=9.095*10^{-13}\\\\$

The probability that Pat will pass his exam is:

$P(pass)=0.01386$

4 0

3 years ago

Help me please!! thank you very much!!

Vinil7 [7]

B. 96 cubic inches.

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3 years ago