What is the exact value of this?

Mathematics

1 answer:

Allushta [10]3 years ago

5 0

Answer:

Correct answer: second quadrant angle is 3π/4

fourth answer (2√2 + π) / 4

third quadrant angle is 5π/4

first answer (-2√2 + π) / 4

Step-by-step explanation:

Your question does not define the quadrant in which the angle is of the

cos⁻¹ (- √2/2).

If it is in the second quadrant then the angle is 3π/4 and if it is in the third quadrant then the angle is 5π/4

1) second quadrant angle is 3π/4

sin ( cos⁻¹ (- √2/2)) + tan⁻¹ ( sin ( π/2)) = sin ( 3π/4) + tan⁻¹ 1 =

= √2/2 + π/4 = (2√2 + π) / 4

2) third quadrant angle is 5π/4

sin ( cos⁻¹ (- √2/2)) + tan⁻¹ ( sin ( π/2)) = sin ( 5π/4) + tan⁻¹ 1 =

= - √2/2 + π/4 = (-2√2 + π) / 4

God is with you!!!

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Answer:

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4 years ago

Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 141 millimeters,

tiny-mole [99]

Answer:

Probability that the sample mean would be greater than 141.4 millimetres is 0.3594.

Step-by-step explanation:

We are given that Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 141 millimetres, and a standard deviation of 7.

A random sample of 39 steel bolts is selected.

Let $\bar X$ = sample mean diameter

The z score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } } }$ ~ N(0,1)

where, $\mu$ = population mean diameter = 141 millimetres

$\sigma$ = standard deviation = 7 millimetres

n = sample of steel bolts = 39

Now, Percentage the sample mean would be greater than 141.4 millimetres is given by = P( $\bar X$ > 141.4 millimetres)

P( $\bar X$ > 141.4) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } } }$ > $\frac{141.4-141}{\frac{7}{\sqrt{39} } } }$ ) = P(Z > 0.36) = 1 - P(Z $\leq$ 0.36)