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2 years ago

The length of a rectangular plot is 2 1/3 m. The area of the reclangle is5 1/3 m? What is the width of the plot?

Mathematics

1 answer:

blondinia [14]2 years ago

6 0

Answer:

Step-by-step explanation:

Width = Area÷ length

= 5 1/3 ÷ 2 1/3

= 16/3 ÷ 7/3

$= \dfrac{16}{3}*\dfrac{3}{7}\\\\\\=\dfrac{16}{7}\\\\=2\dfrac{2}{7}$

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Solve the inequality.<br> I need help please.<br> -11.6>6.7+4.3+m

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First, combine like terms:

(6.7+4.3= 11)

-11.6>11+m
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2 years ago

Add. Simplify. Please HELP!!!

Archy [21]

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2 years ago

Read 2 more answers

According to survey data, the distribution of arm spans for males is approximately Normal with a mean of 71.4 inches and a stan

Gekata [30.6K]

Answer:

a) 89.97% of men have arm spans between 66 and 76 inches.

b) The z-score for this person's arm span is 5.68. 0% of males have an arm span at least as long as this person

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 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.

Mean of 71.4 inches and a standard deviation of 3.1 inches.

This means that $\mu = 71.4, \sigma = 3.1$

a. What percentage of men have arm spans between 66 and 76 inches?

The proportion is the pvalue of Z when X = 76 subtracted by the pvalue of Z when X = 66. The percentage is the proportion multiplied by 100.

X = 76

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

$Z = \frac{76 - 71.4}{3.1}$

$Z = 1.48$

$Z = 1.48$ has a pvalue of 0.9306

X = 66

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

$Z = \frac{66 - 71.4}{3.1}$

$Z = -1.74$

$Z = -1.74$ has a pvalue of 0.0409

0.9306 - 0.0409 = 0.8997

0.8997*100% = 89.97%

89.97% of men have arm spans between 66 and 76 inches.

b. A particular professional basketball player has an arm span of almost 89 inches. Find the z-score for this person's arm span. What percentage of males have an arm span at least as long as this person?

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

$Z = \frac{89 - 71.4}{3.1}$