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Bumek [7]
3 years ago
10

Please help fast!!

Mathematics
2 answers:
Natasha2012 [34]3 years ago
8 0
G (h(-8)) = 118

Hope this helps
nevsk [136]3 years ago
6 0

Answer:

g(h(- 8)) = 119

Step-by-step explanation:

Evaluate h(- 8), then substitute the result obtained into g(x), that is

h(- 8) = (- 8)² - 2 = 64 - 2 = 62, then

g(62) = 2(62) - 5 = 124 - 5 = 119

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A company manager for a tire manufacturer is in charge of making sure there is the least amount of defective tires. If the tire
Anestetic [448]

Answer:

0.347% of the total tires will be rejected as underweight.

Step-by-step explanation:

For a standard normal distribution, (with mean 0 and standard deviation 1), the lower and upper quartiles are located at -0.67448 and +0.67448 respectively. Thus the interquartile range (IQR) is 1.34896.

And the manager decides to reject a tire as underweight if it falls more than 1.5 interquartile ranges below the lower quartile of the specified shipment of tires.

1.5 of the Interquartile range = 1.5 × 1.34896 = 2.02344

1.5 of the interquartile range below the lower quartile = (lower quartile) - (1.5 of Interquartile range) = -0.67448 - 2.02344 = -2.69792

The proportion of tires that will fall 1.5 of the interquartile range below the lower quartile = P(x < -2.69792) ≈ P(x < -2.70)

Using data from the normal distribution table

P(x < -2.70) = 0.00347 = 0.347% of the total tires will be rejected as underweight

Hope this Helps!!!

6 0
3 years ago
The exponent portion of a scientific number can be less than 1 true or false​
WITCHER [35]

Answer:

True

have a nice day!!

4 0
3 years ago
Im so confused all i know is that it isn't C
hodyreva [135]
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A nursery sells English, tea, and miniature rosebushes in red, pink, yellow, and white. How many
rusak2 [61]

is this year 3 math? Its 4

8 0
3 years ago
is it possible to find a side length that would be perfect for a square with an area of 45 square units?
Cerrena [4.2K]

If a square has an area of 45 square units its side has a length of

s=\sqrt{45} = 3 \sqrt{5}

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5 0
3 years ago
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