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Lelu [443]
4 years ago
11

three students gambled with the amount of 7:6:5 and finished with sums in the ratio 1:4:7 . if one of them lost $22, then what w

ould the total amount at the start of the game be?
Mathematics
1 answer:
agasfer [191]4 years ago
8 0
Make the ratios have a common "total."

7:6:5 is the start
change 1:4:7 to 1.5:6:10.5 so that there are 18 parts.

the only person who lost, lost 5.5 parts which is $22

each part is $4

there are 18 parts, they started out with $72
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Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.
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Answer:

0.1994 is the required probability.      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling =

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851

P(sample of 20 boxers is more than 167 pounds)

P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)

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Calculation the value from standard normal z table, we have,  

P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

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3 years ago
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Setler [38]

Answer:

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Step-by-step explanation:

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3 years ago
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Igoryamba

Answer:

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Step-by-step explanation:

6 3/5 ÷ 0.8

6 3/5 = 33/5

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6 3/5 ÷ 0.8

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The angles across from the legs are called base angles

Step-by-step explanation:

we know that

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Step-by-step explanation:

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