The geometric mean of two numbers is 25√. One of the numbers is 6. Find the other number.

The sum of two numbers is 12. The first number,x, is twice the second number ,y,

vlada-n [284]

Judging by the question at hand I generated this equation.
x+y=12
x=2y

I begin this question by plugging in the x=2y into the equation for x.
So the new equation should be 3y=12. I then divide the entire equation by 3 to get y=4.

Next I plug y=4 into the equation, the new equation should be x+4=12. I then subtract 4 from both sides to get x=8.

The two numbers are :
x=8 y=4

3 0

3 years ago

David earns $8.59 an hour. His benefits package is equal to 18 percent of his hourly wages. When you include the value of his be

meriva

A]
Amount of earning per hour=$8.59
Amount of David's benefits=18/100×8.59=:1.5462
Amount that David earn per hour including benefits is given by:
8.59+1.5462=$10.1362

b]Amount that David earns in 35 hour a week will be:
(amount per hour)*(number of hours)
=8.59*35
=$$300.65

C] amount earned by David including benefits will be:
(amount earned in 35 hours)+(total benefits in 35 hours)
total benefits=1.5462×35=$54.117
thus total amount will be:
300.65+54.117
=$354.767

4 0

3 years ago

ILL GIVE YOU BRAINLIST !!!

Alecsey [184]

Answer:

$-\frac{15}{16}$

Step-by-step explanation:

$\frac{1}{4} ^{2}$ = $\frac{1}{16}$

2*(pq)= 2( $\frac{1}{6}$ )= $\frac{1}{3}$

-3 $q^{2}$ = -3 $(\frac{4}{9} )$ = $-\frac{4}{3}$

$\frac{1}{16} +\frac{1}{3}-\frac{4}{3}$ =

$\frac{1}{16} -1$ =

$-\frac{15}{16}$

4 0

3 years ago

I’ve been having trouble with these all week!

Kobotan [32]

Answer:

Step-by-step explanation:

so you will move constant to the right side and change the sign. 72x=223-7

then you will subtract the numbers 72x=223-7 and that will give you 72=216

then you will divide both sides by 72 and that will give you

x=3 .

7 0

3 years ago

Read 2 more answers

The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh

mr_godi [17]

Answer:

a) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

b) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(10,0.15)$

Where $\mu=10$ and $\sigma=0.15$

We can calculate the probability of being defective like this:

$P(X$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

And if we replace we got:

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects.

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

Part c What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

5 0

3 years ago