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shtirl [24]
3 years ago
14

A company manufactures ball bearings for precision machines. The average diameter of a certain type of ball bearing should be 6.

0 mm. To check that the average diameter is correct, the company formulates a statistical test. a. What is the sampling distribution of the mean score x of a sample of 50 ball bearings given the standard deviation is 0.7 mm. b. State the null and alternative hypothesis. c. Suppose that one sample data gives x
Mathematics
1 answer:
Sindrei [870]3 years ago
6 0

Answer:

(a) =210

(b) = 0.11

Step-by-step explanation:

Sampling distribution of the mean = 50 × 6 × 0.7= 210

Now to get the null divide 0.7 by 6

This gives 0.11

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I’m really struggling! Please help me!!
Andrew [12]

Answer:

8.0

Step-by-step explanation:

The triangle is a right triangle, so we are able to use trigonometric functions. Relative to the angle of 37°, we have the opposite side which is 6 and the adjacent side which is x. The trig function that uses the opposite and adjacent is tangent(SohCahToa). We can set up the following equation:

tan(37) = \frac{opposite}{adjacent}

tan(37) = \frac{6}{x}

tan(37) evaluates to about , so we can plug it in and solve for x:

0.7536 = \frac{6}{x} \\0.7536x = 6\\x = 7.962

To the nearest tenth, x rounds to 8.0.

4 0
2 years ago
I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.
masya89 [10]

Answer:

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

4 0
2 years ago
The diagram shows four graphs match the graphs to the equations
cluponka [151]

Answer:

y=x^2 is graph D

y=1/x is graph A

otherwise, you're right

7 0
2 years ago
If f(x) =7x+5/3-x, then show that fof power -1(x) is an identity function ​
Irina-Kira [14]

If f(X)= 7x +5/3 - x =6x +5/3

y= 6x +5/3

6x= y-5/3

f^-1(y)=x=(y-5/3)/6

So we can write

f(f^-1(y))= f((y-5/3)/6) = 6 (y-5/3)/6 +5/3= y-5/3+5/3= y

The same result can be obtained for f^-1(f(x))=x

If the function is f(x)= 7x +5/(3-x) then it is not invertible since it is not injective (pictures)

4 0
3 years ago
How is the words enter and victor alike
pickupchik [31]
These words are represented as homophones as they sound alike but have the different meanings.
6 0
3 years ago
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