4m+9+5m-12=43 simplify and solve

If p is inversely proportional to the square of q, and pis 24 when q is 11,

alexandr402 [8]

Answer:

726

Step-by-step explanation:

Here p = k / q², and 24 = k / 121, or k = 2904

Then p = 2904 / q²

If q = 2, p = 2904 /4 = 726

4 0

3 years ago

Read 2 more answers

Factor 16x^4 - 64y^8 completely

iris [78.8K]

16 • (x^2 + 2y^4) • (x^2 - 2y^4)

8 0

3 years ago

Wendy is creating a large soup based on a recipe. Her recipe calls for of a pint of milk, but she only has of a pint of milk, so

xeze [42]

Answer:

She can only make 5/6 of her recipe with the amount of milk that she has.

Step-by-step explanation:

Some data of this problem is missing, the data missing is:

Her recipe calls for 3/4 of a pint of milk and she only has 5/8 of a pint of milk.

Now, to know what's the portion she can do with that amount we need to divide the amount of milk she has by the total amount she needs:

$\frac{5}{8}$ ÷ $\frac{3}{4}$

When we divide we turn the second fraction upside down (the numerator becomes the denominator and viceversa) and multiply, thus:

$\frac{5}{8}$ ÷ $\frac{3}{4}=\frac{5}{8}$ × $\frac{4}{3} =\frac{20}{24}$

If we simplify this last expression we have:

$\frac{20}{24}=\frac{10}{12} =\frac{5}{6}$

Thus, she can only make 5/6 of her recipe with the amount of milk that she has.

<em>Note: In case the data missing is different, you can apply this same procedure with the fractions you have. </em>

6 0

4 years ago

Use the Venn diagram to list the set of elements of A ∩ B in roster form.

artcher [175]

Answer:

A. {e, h}

Step-by-step explanation:

In a Venn diagram, the set of elements in any intersection can simply be visualised. The elements contained in the region where the circles representing different sets overlap, are the set of elements of intersection.

In the Venn diagram given, the set of elements contained in the region where the circles representing A and B overlap are {e, h}.

{e, h} is common to both set A and set B.

7 0

3 years ago

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