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anyanavicka [17]
3 years ago
6

PLS HELP I NEED THIS IN 3 HOURS PLS DO NOT GIVE ME A LINK FOR AN ANSWER

Mathematics
1 answer:
bezimeni [28]3 years ago
3 0

Answer:

Step-by-step explanation:

as Area of triangle is=1/2 base*height

Given

area=70

height=h=8

base lenght=?

put values in above formula for unkown

70=1/2 (8*b)

70=4*b

b=70/4

b=17.5 units

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2x – 3 &gt; -4x + 2<br> What is the solution to the inequality
bagirrra123 [75]

Answer:

x>\frac{5}{6}

Step-by-step explanation:

2x-3>-4x+2

Add 3 to both sides

2x-3+3>-4x+2+3

Simplify

2x>-4x+5

Add 4x to both sides

2x+4x>-4x+5+4x

Simplify

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Divide both sides by 6

\frac{6x}{6} >\frac{5}{6}

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4 0
3 years ago
Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 46 pounds and standar
Alik [6]

Answer:

85.99% of airline passengers incur this fee.

Step-by-step explanation:

When the distribution is normal, we use 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 46 pounds and standard deviation 3.7 pounds.

This means that \mu = 46, \sigma = 3.7

Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee.

As a proportion, this is 1 subtracted by the pvalue of Z when X = 50. So

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

Z = \frac{50 - 46}{3.7}

Z = 1.08

Z = 1.08 has a pvalue of 0.8599

0.8599*100% = 85.99%

85.99% of airline passengers incur this fee.

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Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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