Which expression is equivalent to

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.3. (Round your ans

Maurinko [17]

Answer:

0.0039 is the probability that the sample mean hardness for a random sample of 12 pins is at least 51.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 50

Standard Deviation, σ = 1.3

Sample size, n = 12

We are given that the distribution of hardness of pins is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{1.3}{\sqrt{12}} = 0.3753$

P(sample mean hardness for a random sample of 12 pins is at least 51)

$P( x \geq 51) = P( z \geq \displaystyle\frac{51 - 50}{0.3753}) = P(z \geq 2.6645)$

$= 1 - P(z < 2.6645)$

Calculation the value from standard normal z table, we have,

$P(x \geq 51) = 1 - 0.9961= 0.0039$

0.0039 is the probability that the sample mean hardness for a random sample of 12 pins is at least 51.

3 0

3 years ago

I need help!! I don’t know which one it is. Which point is closest to point A? Point__

aleksandrvk [35]

Answer:

Point D

Step-by-step explanation:

The length from Point A to Point D is 6 units, while the length from point A to Point C is 7 units, and the length from Point A to Point B is also 7 units. Because to reach Point C, you have to make one turn, adding one more unit. Same with Point B.

4 0

2 years ago

Read 2 more answers

PLZ HELP on THIS QUESTION

Greeley [361]

Its a rectangle cut the corners

8 0

3 years ago

Help me guys thx so much

navik [9.2K]

Answer:

Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Step-by-step explanation:

Step 1: Multiply

$(\sqrt{10x^4} -x\sqrt{5x^2} )*(2\sqrt{15x^4} + \sqrt{3x^3})\\ (\sqrt{10 * x^2 * x^2} -x\sqrt{5 * x^2} ) * (2\sqrt{15 * x^2 * x^2} +\sqrt{3 * x^2 * x})\\(x^2\sqrt{10} -x^2\sqrt{5} )*(2x^2\sqrt{15} +x\sqrt{3x}) \\\\$

$(x^2\sqrt{10}*2x^2\sqrt{15} )+(x^2\sqrt{10}*x\sqrt{3x} ) + (-x^2\sqrt{5} *2x^2\sqrt{15}) + (-x^2\sqrt{5} *x\sqrt{3x}$

$(2x^4\sqrt{150} ) + (x^3\sqrt{30x}) + (-2x^4\sqrt{75}) + (-x^3\sqrt{15x} )$

$2x^4\sqrt{5^2*6} + x^3\sqrt{30x} -2x^4\sqrt{5^2*3} -x^3\sqrt{15x}$

$10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Answer: Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

6 0

3 years ago

Read 2 more answers

A ship leaves port and travels due west for 30 knots, then changes course to S 30° W and travels 50 more knots. Find the bearing

leva [86]

Answer:

232°

Step-by-step explanation:

There are a couple of ways to find the desired direction. Perhaps the most straightforward is to add up the coordinates of the travel vectors.

30∠270° +50∠210° = 30(cos(270°), sin(270°)) +50(cos(210°), sin(210°))

= (0, -30) +(-43.301, -25) = (-43.301, -55)

Then the angle from port is ...

arctan(-55/-43.301) ≈ 231.79° . . . . . . . 3rd quadrant angle

The bearing of the ship from port is about 232°.

_____

Comment on the problem statement

The term "knot" is conventionally used to indicate a measure of speed (nautical mile per hour), not distance. It is derived from the use of a knotted rope to estimate speed. Knots on the rope were typically 47 ft 3 inches apart. As a measure of distance 30 knots is about 1417.5 feet.

5 0

3 years ago