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Mandarinka [93]
3 years ago
7

In 1997, there were 43.2 million people who used free weights. Assuming the use of free weights increases 6% annually, which equ

ation can be used to predict the number of people using free weights t years from 1997?
p = 43.2(0.06)t
p = 43.2(1.06)t
p = 43.2(0.94)t
p = 43.2(1.005)t
Mathematics
1 answer:
Yanka [14]3 years ago
5 0
<span>P=43.2(1.06)^t, Would Be The Answer.</span>
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Consider three boxes with numbered balls in them. Box A con- tains six balls numbered 1, . . . , 6. Box B contains twelve balls
murzikaleks [220]

Answer:

a) 0.73684

b) 2/3

Step-by-step explanation:

part a)

P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)}

Using conditional probability as above:

(A,B,C)

Cases for numerator when:

P( A is 1 and exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1)

= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4})  + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) = 0.048611111

Cases for denominator when:

P( Exactly two balls are 1) = P( 1, not 1, 1) + P(1, 1, not 1) + P(not 1, 1 , 1)

= (\frac{1}{6}* \frac{11}{12}*\frac{1}{4})  + (\frac{1}{6}*\frac{1}{12}*\frac{3}{4}) + (\frac{5}{6}*\frac{1}{12}*\frac{1}{4})= 0.0659722222

Hence,

P ( A is 1 / exactly two balls are 1) = \frac{P ( A is 1 and that exactly two balls are 1)}{P (Exactly two balls are one)} = \frac{0.048611111}{0.06597222} \\\\= 0.73684

Part b

P ( B = 12 / A+B+C = 21) = \frac{P ( B = 12 and A+B+C = 21)}{P (A+B+C = 21)}

Cases for denominator when:

P ( A + B + C = 21) = P(5,12,4) + P(6,11,4) + P(6,12,3)

= 3*P(5,12,4 ) =3* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{96}

Cases for numerator when:

P (B = 12 & A + B + C = 21) = P(5,12,4) + P(6,12,3)

= 2*P(5,12,4 ) =2* \frac{1}{6}*\frac{1}{12}*\frac{1}{4}\\\\= \frac{1}{144}

Hence,

P ( B = 12 / A+B+C = 21) = \frac{\frac{1}{144} }{\frac{1}{96} }\\\\= \frac{2}{3}

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Answer:

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Step-by-step explanation:

Use formula

I=P\cdot r\cdot t,

where

I = interest,

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P + I =$12,720, thus

12,720-P=P\cdot 0.06\cdot 1\\ \\12,720-P=0.06P\\ \\12,720=P+0.06P\\ \\1.06P=12,720\\ \\P=\dfrac{12,720}{1.06}\\ \\P=\$12,000\\ \\I=\$12,720-\$12,000=\$720

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