Assuming m<PTQ is a right angle (measuring 90°) and angle TQP is already marked as 63° while knowing there are 180° in a triangle, you would take 180°- 63°- 90°= 27

8 0

3 years ago

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Which is the equivalent of 12.42 written in DMS form?

nika2105 [10]

12.42

12°+.42(60)

12°25.2'

12°25'+.2(60)

12°25'12"

3 0

3 years ago

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Assume the time required to complete a product is normally distributed with a mean 3.2 hours and standard deviation .4 hours. Ho

Alex787 [66]

Answer: 3.712 hours or more

Step-by-step explanation:

Let X be the random variable that denotes the time required to complete a product.

X is normally distributed.

$X\sim N(\mu=3.2\text{ hours},\ \sigma=0.4\text{ hours} )$

Let x be the times it takes to complete a random unit in order to be in the top 10% (right tail) of the time distribution.

Then, $P(\dfrac{X-\mu}{\sigma}>\dfrac{x-\mu}{\sigma})=0.10$

$P(z>\dfrac{x-3.2}{\sigma})=0.10\ \ \ [z=\dfrac{x-\mu}{\sigma}]$

As, $P(z>1.28)=0.10$ [By z-table]

Then,

$\dfrac{x-3.2}{0.4}=1.28\\\\\Rightarrow\ x=0.4\times1.28+3.2\\\\\Rightarrow\ x=3.712$

So, it will take 3.712 hours or more to complete a random unit in order to be in the top 10% (right tail) of the time distribution.

3 0

3 years ago

Brent has two barcode printing machines. Each machine can print 25 barcodes per minute. On day 1, only the first machine was use

Sonja [21]

I think it would be 95 + 25x + 25y

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3 years ago

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