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yulyashka [42]
3 years ago
7

What does z equal? Equation: 9 (z - 82) = 90

Mathematics
2 answers:
iVinArrow [24]3 years ago
6 0

Answer:

z = 92

Step-by-step explanation:

9 (z - 82) = 90

Divide both sides by 9

z - 82 = 90/9

z - 82 = 10

add 82 to both the sides,

z - 82 +82 = 10 +82

z = 92

matrenka [14]3 years ago
3 0

Answer:

Z=92

Step-by-step explanation:

Distribute, 9z-738=90

Add 738, 9z=828

Divide, z=92

You might be interested in
Find the cost per ounce of a gold alloy made from 30 oz of pure gold that costs $1290 per ounce and 60 oz of an alloy that costs
Romashka [77]
The answer would be $24.55 per oz of gold alloy so rounded up would be $25
6 0
3 years ago
Read 2 more answers
For which equation is m = 12 the solution? a. 4m=40 b. m+20=42. c. 4m=48. d. m-4=9
Gelneren [198K]
To solve this question, we just need to insert 12 into the m position of each question and see if the equation holds true.

a. 4m = 40

4(12) = 40

48 = 40

48 obviously does not equal 40, so it is not choice A.

b. m + 20 = 42

12 + 20 = 42

32 = 42

Again, 32 isn't the same as 42, so not choice B either.

c. 4m = 48

4(12) = 48

48 = 48

It looks like this one is true, but let's solve D also just to make sure.

d. m - 4 = 9

12 - 4 = 9

8 = 9

This is false, since 8 does not equal 9.

Therefore, choice C (4m = 48) is the correct answer.
Hope that helped! =)
6 0
3 years ago
Read 2 more answers
The explicit rule for the arithmetic sequence shown in the graph is a↓n=9.5+2(n-1)
Virty [35]

Answer:

True.

Step-by-step explanation:

The explicit form for an arithmetic sequence is:

a_n=a_1+d(n-1)

where a_1 is the first term and d is the common difference.

The common difference here is 2 because the y's are going up by 2 while the x's are going up by 1.

Yes, the common difference is the slope.

So we have d=2.

The first term is 9.5 because that is what happens when x=1.

x is n.

y is a_n.

So the answer is true.

----

You can also verify by plugging in numbers for n and see if you get the outputs mentioned in the pairs given:

Let n=1:

a_1=9.5+2(1-1)

a_1=9.5+2(0)

a_1=9.5+0

a_1=9.5

Let n=2:

a_2=9.5+2(2-1)

a_2=9.5+2(1)

a_2=9.5+2

a_2=11.5

Let n=3:

a_3=9.5+2(3-1)

a_3=9.5+2(2)

a_3=9.5+4

a_3=13.5

Let n=4:

a_4=9.5+2(4-1)

a_4=9.5+2(3)

a_4=9.5+6

a_4=15.5

Let n=5:

a_5=9.5+2(5-1)

a_5=9.5+2(4)

a_5=9.5+8

a_5=17.5

Let n=6:

a_6=9.5+2(6-1)

a_6=9.5+2(5)

a_6=9.5+10

a_6=19.5

We have confirmed that we get all 6 of the mentioned points using the equation they gave.

5 0
3 years ago
How would the mean change if the number 52 replaced the number 12 in the set?
m_a_m_a [10]

Answer:

increase

Step-by-step explanation:

First, we need to find the mean of the given data set without any change.

10 + 61 + 10 + 44 + 21 + 79 + 27 + 12 = 264

264 ÷ 8 = <u>33</u>

Now that we have the mean, we can find the mean of the data set with the change of 52.

10 + 61 + 10 + 44 + 21 + 79 + 27 + 52 = 304

304 ÷ 8 = <u>38</u>

<u />

From this, we can see that the mean has increased with the change of 12 to 52. Thus, that's the correct option.

hope this helps!

6 0
2 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
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