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4 years ago

Help me with this please!

Mathematics

2 answers:

DedPeter [7]4 years ago

8 0

Answer:

Area of triangle = 20√2

Step-by-step explanation:

Formula:

Area of triangle = bh/2

b - base of triangle

h - height of triangle

Distance formula

d = √(x₂ - x₁)² +(y₂ - y₁)²

From the figure we can see that, the given triangle is right angled triangle.

Base = AC and Height = AB

To find AB and AC

A(1,3), B(4,7) and C(9, -5)

AB = √(4 - 1)² +(7 - 3)² = √(3² + 4²) = 5

AC = √(9 - 1)² +(-5 - 3)² = √(8² + -8²) = 8√2

To find the are of triangle

Area = bh/2 = (5* 8√2)/2 = 20√2

Therefore the area of triangle = 20√2 square units

svetoff [14.1K]4 years ago

8 0

Answer:

Area = 20√2 is the area of triangle.

Step-by-step explanation:

We have given the triangle in the graph.

We have to find the area of the triangle.

The formula for the area of triangle is :

Area = 1/2(base×height)

In the graph, the base is represented by AC, and height is represented

by AB.

First, we use distance formula to find AC and AB.

Base = AC =√(9-1)² + (-5-3)² = 8√2

Height = AB=√(4-1)² +(7-3)² = 5

Put this values of height and base in the formula we get,

Area = 1/2 (8√2 ×5)

Area = 20√2 is the area of triangle.

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Step-by-step explanation:

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3 years ago

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3 years ago

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Kryger [21]

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Step-by-step explanation:

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3 years ago

Sick-leave time used by employees of a firm in a course of one month has approximately normal distribution, with a mean of 200 h

Usimov [2.4K]

Answer:

a)0.62% probability that total sick leave for next month will be less than 150 hours.

b) 225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 200, \sigma = \sqrt{400} = 20$

a.Find the probability that total sick leave for next month will be less than 150 hours.

This probability is the pvalue of Z when X = 150. So:

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

$Z = \frac{150 - 200}{20}$

$Z = -2.5$

$Z = -2.5$ has a pvalue of 0.0062.

So there is a 0.62% probability that total sick leave for next month will be less than 150 hours.

b.In planning schedules for next month, how much time should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

This is the value of X when Z has a pvalue of 0.90. So $Z = 1.28$