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Archy [21]
3 years ago
8

There are 100 gumdrops. 78 of them are large, and the rest are small. 33 of them are red, and the rest are other colors. 16 of t

he gumdrops are large red ones. You choose 1 gumdrop at random, and it is that yummy red color. What is the probability that the gumdrop is large, given that it is red?
Write fractions using the slash ( / ) key. Reduce fractions to their lowest terms.
Mathematics
1 answer:
g100num [7]3 years ago
3 0
Large red ones= 16/78
33-16=17 small gumdrops

22 gumdrops are small
small red ones=17/22

Probability of large and small red ones:
prob. of large red ones:     16/78+22/22= 38/100=38%
prob. of small red ones:    17/22+78/78= 95/100=95%
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3 years ago
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IRINA_888 [86]

Answer:

<em>The approximate percentage of cars that remain in service between 73 and 84 months</em>

<em>P( 73 < x < 84 ) = 0.02145 = 2.1 %</em>

<u>Step-by-step explanation</u>:

<u><em>Explanation</em></u>:-

Given mean of the Population 'μ ' = 51 months

Standard deviation of the Population 'σ' = 11 months

Let 'X' be the random variable of Normal distribution

<em>Let    'X'  = 73</em>

<em></em>Z = \frac{x-mean}{S.D} = \frac{73-51}{11} = 2<em></em>

<em>Let  'X' = 84</em>

<em></em>Z = \frac{84-51}{11} = 3<em></em>

<em>The approximate percentage of cars that remain in service between 73 and 84 months</em>

<em>P( 73 < x < 84 )      = P( 2 < Z < 3)</em>

<em>                               = P( Z<3) - P( Z <2)</em>

<em>                              =  0.5 + A(3) - ( 0.5 + A(2))</em>

<em>                             = A(3) - A( 2)</em>

<em>                             = 0.49865 - 0.4772     ( From Normal table)</em>

<em>                             = 0.02145</em>

<em>  P( 73 < x < 84 ) = 0.02145</em>

<em>The approximate percentage of cars that remain in service between 73 and 84 months</em>

<em>P( 73 < x < 84 ) = 0.02145 = 2.1 %</em>

7 0
3 years ago
Please help me with this question
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Answer: 275 cm

volume = length x width x height

v = 6.25 x 8 x 5.5

v = 275

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