1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lawyer [7]
3 years ago
15

F(x)=-9x^2-7x+8, x=5

Mathematics
1 answer:
VashaNatasha [74]3 years ago
4 0
F(x) = -9x^2 - 7x + 8
f(5) = -9(5^2) - 7(5) + 8
f(5) = -9(25) - 35 + 8
f(5) = - 225 - 35 + 8
f(50 = - 252 <==
You might be interested in
If f(x)=x^2-7 and h(x)=7x-5, then f(h(2))=
zaharov [31]
We are given 3 equations. Using these equations we can find f(h(2)).

h(x)=7x-5
h(2)=7(2)-5
h(2)=14-5
h(2)=9

f(x)=x^2-7
f(h(2))=(7x-5)^2-7
we know the answer for h(2) by now, so we can substitute for that.
f(9)=9^2-7
f(9)=81-7
f(9)=74

so f(h(2))=74
5 0
3 years ago
Find the sum of (x + 5) and (2x + 3)
weeeeeb [17]

Answer:

3x+8

Step-by-step explanation:

8 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
2 years ago
I WILL GIVE 100 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT. Use the diagram to the right to complete the following mapping s
Oksi-84 [34.3K]
B is your answer! let me know if you need help!
5 0
3 years ago
1. write an algebraic expression for the phrase three times the sum of B and F. ​
Nezavi [6.7K]

The sum of b and f is b+f

Three times this sum is 3(b+f)

6 0
3 years ago
Other questions:
  • the spinner below has 12 congruent sections. sarah will spin the arrow on the spinner twice. what is the probability that the ar
    12·1 answer
  • The ratio of boys to girls in a class room is 5 to 7. If there are 30 boys then how many girls are there
    11·2 answers
  • What is the volume of a cube with side lengths of 5 ft?
    8·2 answers
  • I'm confused on how to get started on each one (A,B, &amp; C)
    11·1 answer
  • (9.2×10^5)(4×10^-3)​
    9·1 answer
  • | x - 2| = | 4 + x |
    15·1 answer
  • =. Diego and Priya are each saving money. Diego starts with $80 in his savings account and adds $5
    10·1 answer
  • Hi could can you please tell me the answer and tell me how you got it so I can understand this? Both 6 A and B
    8·1 answer
  • (c) The annual membership fee in 2013 is $198 for each adult and $75 for each child.
    12·1 answer
  • Which of the following term describes √36 ? *
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!