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Doss [256]
3 years ago
11

Help :( ima link it........

Mathematics
1 answer:
suter [353]3 years ago
8 0

Answer:

d

Step-by-step explanation:

he is DIVIDING it among his children so the equation will have division

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A production process is checked periodically by a quality control inspector. the inspector selects simple random samples of 30 f
777dan777 [17]

Answer:

Population Mean = 2.0

Population Standard deviation = 0.03

Step-by-step explanation:

We are given that the inspector selects simple random samples of 30 finished products and computes the sample mean product weight.

Also, test results over a long period of time show that 5% of the values are over 2.1 pounds and 5% are under 1.9 pounds.

Now, mean of the population is given the average of two extreme boundaries because mean lies exactly in the middle of the distribution.

So,   Mean, \mu = \frac{1.9+2.1}{2} = 2.0

Therefore, mean for the population of products produced with this process is 2.

Since, we are given that 5% of the values are under 1.9 pounds so we will calculate the z score value corresponding to a probability of 5% i.e.

             z = -1.6449 {from z % table}

We know that z formula is given by ;  

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

              -1.6449 = \frac{1.9 - 2.0}{\frac{\sigma}{\sqrt{n} } }     ⇒  \frac{\sigma}{\sqrt{n} }  = \frac{-0.1}{-1.6449}  

                                           ⇒ \sigma = 0.0608 * \sqrt{30}  {as sample size is given 30}

                                           ⇒ \sigma = 0.03 .

Therefore, Standard deviation for the population of products produced with this process is 0.0333.

7 0
3 years ago
A circle has centre (3,0) and radius 5. The line y = 2x + k intersects the circle in two points. Find the set
lara [203]

A circle with center (3,0) and radius 5 has equation

(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x  = 16

If we substitute y=2x+k in this equation, we have

x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

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The expression simplifies to

-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89

The solutions to the associated equation are

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So, the parabola is negative between the two solutions:

-6-5\sqrt{5}

8 0
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