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vivado [14]
2 years ago
9

Thomas invested $8,500 for one year. Part of the money was invested at6% and the rest at 9%. The total interest earned was $667.

50. How much did Thomas invest at the 6% rate?
Mathematics
1 answer:
oee [108]2 years ago
4 0
Okay so say that x=amount invested with 6% and y=amount invested with 9%
x+y=8,500 so > x=8500-y
6%=0.06        <span>9%=0.09</span>
0.06x +0.09y=667.5 (Substitute in x=8500-y so only numbers and y)
0.06(8500-y)+0.09y=667.5 (expand brackets)
510-0.06y+0.09y=667.5 (-510)
0.03y=117.5 (/0.03)
$3916.67=Y  ->9%
X=8500-Y 
x=$4583.33 ->6%
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Answer:

37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 12.6, \sigma = 2.5

What is the probability that randomly selected homework will require between 8 and 12 minutes to grade?

This is the pvalue of Z when X = 12 subtracted by the pvalue of Z when X = 8. So

X = 12

Z = \frac{X - \mu}{\sigma}

Z = \frac{12 - 12.6}{2.5}

Z = -0.24

Z = -0.24 has a pvalue of 0.4052

X = 8

Z = \frac{X - \mu}{\sigma}

Z = \frac{8 - 12.6}{2.5}

Z = -1.84

Z = -1.84 has a pvalue of 0.0329

0.4052 - 0.0329 = 0.3723

37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade

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Answer:

  • There are 10 different combinations

  • The list of different combinations is:

        (10p, 1p), (10p, 50p), (10p, 2p), (10p, 20p), (1p, 50p), (1p, 2p),

        (1p, 20p), (50p, 2p), (50p, 20p), (2p, 20p)

Explanation:

The possible combinations are:

1. Assuming the first coin is 10p:

  • (10p, 1p)
  • (10p, 50p)
  • (10p, 2p)
  • (10p, 20p)

2. Asuming the first coin is 1p

Do not count (1p, 10p) as it is the same combination as (10p, 1p)

  • (1p, 50p)
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3. Assuming the first coin is 50p:

Do not count (50p, 10p) nor (50p, 1p) as they are the same combinations (10p, 50p) and (1p, 50p) counted earlier:

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4. Assuming the first coin is 2p:

The only new combination is:

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5. All the combinations with 20p have already been listed.

Therefore:

  • There are 4 + 3 + 2 + 1 = 10 different combinations

  • The list of different combinations is:

        (10p, 1p), (10p, 50p), (10p, 2p), (10p, 20p), (1p, 50p), (1p, 2p),

        (1p, 20p), (50p, 2p), (50p, 20p), (2p, 20p)

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