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wlad13 [49]
3 years ago
5

Aretha can type 143 words per minute.If she types at the same rate,how many words can she type in 35 minuets?

Mathematics
2 answers:
TiliK225 [7]3 years ago
5 0
Hello!

Multiply the rate per minute by the amount of minutes:

143 × 35 = 5,005

Aretha can type 5,005 words in 35 minutes.
Reptile [31]3 years ago
4 0
The answer is 5,005. i hope this helps you :)
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1/6 is the answer. Good luck!
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3 years ago
The first term of a geometric sequence is 6, and the common ratio is 3. What is the 7th term of the sequence?
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An = a1 r^(n-1)

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The awnser is (3,0)
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3 years ago
If the distribution is really (5.43,0.54)
defon

Answer:

0.7486 = 74.86% observations would be less than 5.79

Step-by-step explanation:

I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)

Problems of normally distributed samples can be 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.

The general format of the normal distribution is:

N(mean, standard deviation)

Which means that:

\mu = 5.43, \sigma = 0.54

What proportion of observations would be less than 5.79?

This is the pvalue of Z when X = 5.79. So

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

Z = \frac{5.79 - 5.43}{0.54}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486

0.7486 = 74.86% observations would be less than 5.79

7 0
3 years ago
The 173 workers at a factory together produce 21, 452 items per day. What is the daily rate of items per worker?
mafiozo [28]

Answer:

124 items per worker per day

Step-by-step explanation:

21,452/173=124

6 0
2 years ago
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