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Step2247 [10]
3 years ago
6

The area of a rectangular dog pen is 8 1/2 square feet. If the width is 3 2/5 feet, what is the length ,in feet ?

Mathematics
2 answers:
Anarel [89]3 years ago
5 0
Since area is length times width, you can divide 8 1/2 by 3 2/5 to get the length. 8.5/3.2=2.65625 feet
victus00 [196]3 years ago
4 0

Answer:

2 1/2 ft

Step-by-step explanation:

Area is given by the formula

A = l*w

We know the area is 8 1/2  

Changing this to an improper fraction  8 1/2 = (2*8 +1)/2 = 17/2

The width is 3 2/5  as an improper fraction  = (5*3+2)/5 = 17/5

A = l*w

17/2 = l* 17/5

Multiply each side by 5/17

17/2 * 5/17 = l* 17/5 * 5/17

5/2 = l

2 1/2 = l


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A)A random sample was used.

Step-by-step explanation:

Due to which the house with vehicles occured mostly and others with no vehicles were left unseen ..

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3 years ago
The manager of a new supermarket wished to estimate the likely expenditure of his customers. A sample of till slips from a simil
bagirrra123 [75]

Answer:

0.0228 = 2.28% probability that any shopper selected at random spends more than $80 per week.

88.54% of shoppers are expected to spend between $30 and 80 per week.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean of $50 and a standard deviation of $15.

This means that \mu = 50, \sigma = 15

Find the probability that any shopper selected at random spends more than $80 per week?

This is 1 subtracted by the p-value of Z when X = 80. So

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

Z = \frac{80 - 50}{15}

Z = 2

Z = 2 has a p-value of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that any shopper selected at random spends more than $80 per week.

Find the percentage of shoppers who are expected to spend between $30 and 80 per week?

The proportion is the p-value of Z when X = 80 subtracted by the p-value of Z when X = 30.

X = 80

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

Z = \frac{80 - 50}{15}

Z = 2

Z = 2 has a p-value of 0.9772

X = 30

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

Z = \frac{30 - 50}{15}

Z = -1.33

Z = -1.33 has a p-value of 0.0918

0.9772 - 0.0918 = 0.8854

0.8854*100% = 88.54%

88.54% of shoppers are expected to spend between $30 and 80 per week.

8 0
3 years ago
Manual ate 1/3 of the crackers on the plate. His brother ate 1/4 of the crackers on the plate. There were 5 crackers left on the
KATRIN_1 [288]

Answer:

there were 11 crackers to begin with.

Step-by-step explanation:

8 0
3 years ago
Is. 36yd2ft bigger than 114ft 2in
Zarrin [17]
NO! 
1 yd=3 ft
36*3=108
108+2=110
114>110, therefore no
5 0
4 years ago
Paul worked as a library volunteer for 8 ¼ hours. Harry worked for 3 ½ hours. How much longer did Paul work than Harry?
stiks02 [169]

Answer:

4 3/4

Step-by-step explanation:

= (8 - 3) + (1/4 - 1/2)

= 5 + 1/4 - (1 x 2 over 2 x 2)

= 5 + 1/4 - 2/4

= 5 and (1/2 over 4)

= 5 + -1/4

= 4 3/4

8 0
3 years ago
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