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GenaCL600 [577]
3 years ago
12

Simplify 6(-5) Simplify 6(-5)

Mathematics
2 answers:
Vadim26 [7]3 years ago
6 0

Answer:

-30

Step-by-step explanation:

DerKrebs [107]3 years ago
5 0

Answer:

The answer is -30

Step-by-step explanation:

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2000 multiplied by 3/8
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Type into a calculator 2000*(3/8)

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The following figure is made of six semicircles with the same diameter, and of four congruent equilateral triangles represented
Oxana [17]

6 semicircles = 3 circles

The formula of the area of a equilateral triangle:

A_\triangle=\dfrac{a^2\sqrt3}{4}

We have a = |AE| = 3cm. Substitute:

A_\triangle=\dfrac{3^2\sqrt3}{4}=\dfrac{9\sqrt3}{4}\ cm^2

The formula of the area of a circle:

A_O=\pi r^2

We have 2r = |AE| =3cm → r = 1.5cm. Substitute:

A_O=\pi\cdot1.5^2=2.25\pi\ cm^2

The area of the figure:

A=4A_\triangle+3A_O\\\\A=4\cdot\dfrac{9\sqrt3}{4}+3\cdot2.25\pi=9\sqrt3+6.75\pi=6\dfrac{3}{4}\pi+9\sqrt3\\\\=\dfrac{27}{4}\pi+\dfrac{36\sqrt3}{4}=\dfrac{9}{4}\cdot3\pi+\dfrac{9}{4}\cdot4\sqrt3=\dfrac{9}{4}(3\pi+4\sqrt3)\ cm^2

7 0
3 years ago
The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approxi
VikaD [51]

Answer:

$512.90 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05

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.

Normally distributed with mean $480 and standard deviation $20.

This means that \mu = 480, \sigma = 20

How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05?

This is the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95, so X when Z = 1.645.

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

1.645 = \frac{X - 480}{20}

X - 480 = 1.645*20

X = 512.9

$512.90 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05

5 0
3 years ago
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