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3 years ago

4-15 all this is overdue and I cant understand anything

Mathematics

1 answer:

attashe74 [19]3 years ago

5 0

I'm not 100% sure how to do this, but it would help if you had any notes/online notes/textbook that shows how to do it (also to see if what I did is the same as the notes)

4. Break-even point (It's basically the intersection point. Or the point when the two lines meet/the point that is the same for both equations)

I plugged in numbers for x into both equations, until C and R were the same number.

I had x = 1, so I plugged in 1 for x into both equations:

C = 15x + 150 C = 15(1) + 150 = 15 + 150 = 165

R = 45x R = 45(1) = 45

Then I had x = 2, then x = 3, and so on until C and R were the same number (meaning they had the same point), which was (5, 225)

Idk how you are suppose to check the solution because there are multiple ways to do this. I doubt this, but one option is plugging (5, 225) back into the equations:

C = 15x + 150 R = 45x

225 = 15(5) + 150 225 = 45(5)

225 = 75 + 150 225 = 225

225 = 225

5. (I made another sort of table? because I didn't know what kind you were suppose to use) Do the same as #4, and plug in numbers for x until C and R were the same number

6. Do the same as #4 (If you have any questions or information I should know then let me know)

7. y = mx + b

"m" is the slope, "b" is the y-intercept [the y value when x = 0 (0, y)]

The y-intercepts are -2 and 13, so the graph is B. Now you need to estimate the solution(point of intersection) by looking at the graph. It looks around (6,7).

Now you need to check the solution by plugging the point into both equations.

y = 1.5x - 2

7 = 1.5(6) - 2

7 = 9 - 2

7 = 7

y = -x + 13

7 = -6 + 13

7 = 7 The point is the solution

8. The graph is A. Do the same as #7 and estimate a point. I will say around (2.5,6.5) And check this by plugging the point into both equations

y = x + 4

6.5 = 2.5 + 4

6.5 = 6.5

y = 3x - 1

6.5 = 3(2.5) - 1

6.5 = 7.5 - 1

6.5 = 6.5 The point is the solution

9. Graph C, do the same as #7 and #8

10. y = mx + b

"b" is (0,b) or (0,y)

"m" is the slope

( $slope=\frac{rise}{run}$

Rise is the number of units you go up(+) or down(-)

Run is the number of units you go to the right)

You need to graph the lines of each equation, and find the point of intersection.

11-15. Do the same as #10

If "y" is not by itself, get it by itself.

#13:

x + y = 27 Subtract x on both sides to get "y" by itself

y = 27 - x

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In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

Find the area of r = 4.5

Gnom [1K]

here ya go 14.13m2 A≈14.13⁢m2

6 0

3 years ago

Randy and Sandy have a total of 20 books between them. After Sandy loses three by leaving them on the bus, and some birthday gif

nasty-shy [4]

Answer:

Randy had 13 books an Sandy had 7 books.

Step-by-step explanation:

20 - 3 = 17

30 - 17 = 13

13 x 2 = 26

26 + 4 = 30

13 + 7 = 20

Hope this helps! Please mark as brainliest.

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3 years ago

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Kamila [148]

1.
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C

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B

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3 years ago

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Nonamiya [84]

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3 years ago

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