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pishuonlain [190]

3 years ago

14

one circle has a circumference of 12 pi . another circle has a circumference of 32 pi . what is the ratio of the radius of the s

maller circle to the radius of the larger circle?

1 answer:

SOVA2 [1]3 years ago

6 0

<span>circumference=pi * 2 radius=12pi, radius(small)=6;
</span>circumference=pi * 2 <span>radius=32pi, radius(small)=16;</span>
6:16=3:8

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Use rounding or compatible numbers to estimate each product or quotient.

zmey [24]

Part 1

<span>Using rounding or compatible numbers, we estimate the product

$5\times2 \frac{3}{5}$
to
$5\times3$

Therefore, the estimate of the product </span><span> $5\times2 \frac{3}{5}$ , using rounding and compatible numbers is about 15.</span>

Part 2

<span>Using rounding or compatible numbers, we estimate the quotient

$25\frac{7}{8}\div2 \frac{8}{9}$
to
$27\div3$

Therefore, the estimate of the quotient </span><span> $25\frac{7}{8}\div2 \frac{8}{9}$ , using rounding and compatible numbers is about 9.</span>

Part 3

Given that Mr. Garcia's class ate $3\frac{2}{3}$ pizzas, Mrs Rizzoli's class ate <span> $4\frac{1}{8}$ pizzas, and Mrs. Tindal's class ate </span><span> $3\frac{1}{4}$ pizzas.

Using benchmarks, we estimate the total number of pizzas eaten by all three classes as follows: </span> $3\frac{3}{4}$ , $4$ and <span> $3\frac{1}{4}$

Thus, the </span><span>total number of pizzas eaten by all three classes is given by

$3\frac{3}{4}+4+3\frac{1}{4}=11$

Therefore, </span>the <span>total number of pizzas eaten by all three classes is about 11.</span>

3 0

3 years ago

Roulette is a very popular game in many American casinos. In roulette, a ball spins on a circular wheel that is divided into 38

ludmilkaskok [199]

Answer:

a. P(AUB) = 0.74

b. P(A∩C) = 0.24

c. P(BUC) = 0.71

d. P(Bc) = 0.55

e. P(A∩B∩C) = 0.11

Step-by-step explanation:

n(U) = Total number of elements in the set = number of elements in the universal set = 38

A: (Outcome is an odd number (00 and 0 are considered neither odd nor even)]

An odd number referred to any integer, that is not a fraction, which is not possible to be divided exactly by 2. Therefore, we have:

A: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35

B: 2, 4, 6, 8, 10, 11, 13, 15, 17, 20, 22, 24, 26, 28, 29, 31, 33, 35

C: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18

a. Calculate the probability of AUB

AUB picks not more than one each of the elements in both A and B without repetition. Therefore, we have:

AUB = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 15, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 33, 35

n(AUB) = number of elements in AUB = 28

P(AUB) = probability of AUB = n(AUB) / n(U) = 28 / 38 = 0.736842105263158 = 0.74

b. Calculate the probability of A ∩ C

A∩C picks only the elements that are common to both A and C. Therefore, we have:

A∩C: 1, 3, 5, 7, 9, 11, 13, 15, 17

n(A∩C) = number of elements in A∩C = 9

P(A∩C) = probability of A∩C = n(A∩C) / n(U) = 9 / 38 = 0.236842105263158 = 0.24

c. Calculate the probability of BUC

BUC picks not more than one each of the elements in both B and C without repetition. Therefore, we have:

BUC: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 22, 24, 26, 28, 29, 31, 33, 35

n(BUC) = number of elements in BUC = 27

P(BUC) = probability of A∩C = n(BUC) / n(U) = 27 / 38 = 0.710526315789474 = 0.71

d. Calculate the probability of Bc

Bc indicates B component it represents all the elements in the universal set excluding the elements in B. Therefore, we have:

Bc: 00, 0, 1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36

n(Bc) = number of elements in Bc = 20

P(Bc) = probability of Bc = n(Bc) / n(U) = 20 / 38 = 0.526315789473684 = 0.55

e. Calculate the probability of A∩B∩C

A∩B∩C picks only the elements that are common to A, B and C. Therefore, we have:

A∩B∩C: 11, 13, 15, 17

n(Bc) = number of elements in A∩B∩C = 4

P(A∩B∩C) = probability of A∩B∩C = n(A∩B∩C) / n(U) = 4 / 38 = 0.105263157894737 = 0.11

Note: All the answers are rounded to 2 decimal places.

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