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3 years ago

Four CDs and 4 DVDs cost $164. The cost of the 4 CDs is half the cost of the

Mathematics

1 answer:

pochemuha3 years ago

8 0

Answer:

Each CD cost $20.50

Step-by-step explanation:

You would divide the cost of the DVDs by 4: 164÷4=41

Now the CDs are half the cost of the DVDs so you would divide 41 by 2: 41÷2=20.50

Your answer would be $20.50

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There are 135 people in a sport centre. 73 people use the gym. 59 people use the swimming pool. 31 people use the track. 19 peop

wlad13 [49]

Answer:

Probability that the person doesn't use any facility = 0.0889

Step-by-step explanation:

Total number of people in a sports center = 135

Number of people using gym 'G' = 73

Number of people using swimming pool 'P' = 59

Number of people using track 'T' = 31

Number of people using gym and pool (G∩P) = 19

Number of people using pool and track (P∩T) = 9

Number of people using gym and track (G∩T) = 16

Number of people using all three facilities (G∩P∩T) = 4

Therefore, from the rule of compound probability,

P(GUPUT) = P(G) + P(P) + P(T) - P(G∩P) - P(G∩T) - P(P∩T) + p(G∩P∩T)

We know probability of an event = $\frac{\text{Favorable event}}{\text{Total outcomes}}$

We will plug in the values of probabilities of each event in the formula.

P(GUPUT) = $\frac{73}{135}+\frac{59}{135}+\frac{31}{135}-\frac{19}{135}-\frac{9}{135}-\frac{16}{135}+\frac{4}{135}$

= $\frac{1}{135}(73+59+31-19-9-16+4)$

= $\frac{123}{135}$

And probability that a person doesn't use any facility = 1 - P(GUPUT)

= 1 - $\frac{123}{135}$

= $\frac{135-123}{135}$

= $\frac{12}{135}$

= $\frac{4}{45}$ ≈ 0.0889

7 0

4 years ago

Read 2 more answers

Which descriptions from the list below accurately describe the relationship

yarga [219]

Answer:

QUT

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

(x2 + 3x + 1) + (2x2 + 2x)<br> HINT

viva [34]

Answer:

3x^2+5x+1

Step-by-step explanation:

(x^2 + 3x + 1) + (2x^2 + 2x)

Combine like terms

x^2 + 2x^2 + 3x +2x +1

3x^2+5x+1

4 0

3 years ago

Read 2 more answers

manufacturer tests 1200 computers and finds that 9 of them have defects. Find the probability that a computer chosen at random h

Kay [80]

Answer:

0.01.

113.

Step-by-step explanation:

Probability of a defect = 9/1200

= 0.0075

= 0.01 to the nearest hundredth.

Prediction of number of defects in 15,000 computers

= 15,000 * 0.0075

= 113.

5 0

3 years ago

Using the method of mathematical induction to prove that equalities are true for values of n indicated:

Dima020 [189]

$2^2+4^2+6^2+...(2n)^2=\frac{2n(n+1)(2n+1)}{3};\ n\geq1\\\\chek\ for\ n=1:\\L=2^2=4;\ R=\frac{2\cdot1(1+1)(2\cdot1+1)}{3}=\frac{2\cdot2\cdot3}{3}=4\\L=R\\-----------------------\\
assumption\ for\ n=k\\2^2+4^2+6^2+...+(2k)^2=\frac{2k(k+1)(2k+1)}{3}\\-----------------------\\thesis\ for\ n=k+1\\2^2+4^2+6^2+...+(2k)^2+[2(k+1)]^2=\frac{2(k+1)(k+1+1)[2(k+1)+1]}{3}\\-----------------------$
$proff:\\L=2^2+4^2+6^2+...+(2k)^2+(2k+2)^2=\frac{2k(k+1)(2k+1)}{3}+(2k+2)^2\\\\=\frac{(2k^2+2k)(2k+1)}{3}+\frac{3(2k+2)^2}{3}=\frac{4k^3+2k^2+4k^2+2k+3(4k^2+8k+4)}{3}\\\\=\frac{4k^3+6k^2+2k+12k^2+24k+12}{3}=\boxed{\frac{4k^3+18k^2+26k+12}{3}}\\\\R=\frac{2(k+1)(k+1+1)[2(k+1)+1]}{3}=\frac{(2k+2)(k+2)(2k+2+1)}{3}\\\\=\frac{(2k^2+4k+2k+4)(2k+3)}{3}=\frac{(2k^2+6k+4)(2k+3)}{3}=\frac{4k^3+6k^2+12k^2+18k+8k+12}{3}\\\\=\boxed{\frac{4k^3+18k^2+26k+12}{3}}\\\\L=R$

4 0

3 years ago