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3 years ago

Can someone help me solve these questions?

Mathematics

1 answer:

love history [14]3 years ago

8 0

Answer:

0, 5, 8, 11

Step-by-step explanation:

(a)

f(0) → x = 0 in interval - 2 ≤ x < 1 , thus x³ , that is

0³ = 0 ⇒ f(0) = 0

(b)

f(1) → x = 1 in interval 1 ≤ x ≤ 4, thus 3x + 2, that is

f(1) = 3(1) + 2 = 3 + 2 = 5

(c)

f(2) → x = 2 in interval 1 ≤ x ≤ 4, that is

f(2) = 3(2) + 2 = 6 + 2 = 8

(d)

f(3) → x = 3 in interval 1 ≤ x ≤ 4, that is

f(3) = 3(3) + 2 = 9 + 2 = 11

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What is 7 9/14 in simplest form plz need help now plz plz plz plz

PSYCHO15rus [73]

The anwser is 5/14...... hope this helps! :)

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Can anybody write an equation for the following word story:

Rudiy27

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3 years ago

the first box measures 18 inches by 8 inches by 15 inches. the second box is 10 inches tall and the area of the base measures 19

Mandarinka [93]

Answer:

Box 1 has the greater volume, by 200 in³

Step-by-step explanation:

The largest of two volumes can be found by computing each volume and comparing the results.

The volume is computed from dimensions by ...

V = LWH

V = (18 in)(8 in)(15 in) = 2160 in³

The volume is computed from area and height by ...

V = Bh . . . . . where B is the base area, and h is the height

V = (196 in²)(10 in) = 1960 in³

<h3>Comparison</h3>

The units of the two volumes are the same, so we can decide which is larger by comparing the numbers. Box 1 has a volume that is more than 2000 in³, and Box 2 has a volume that is less than 2000 in³.

Box 1 has a greater volume

The amount by which the volume is greater can be found by subtracting the smaller volume from the larger:

2160 in³ -1960 in³ = 200 in³

Box 1 is larger by 200 in³

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<em>Additional comment</em>

The percentage by which Box 1 is larger than Box 2 can be found by multiplying 100% by the ratio of the difference to the Box 1 volume:

200/1960 × 100% ≈ 10.2%

Box 1 is about 10% larger than Box 2.

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2 years ago

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nevsk [136]

Answer:

g(h(- 8)) = 119

Step-by-step explanation:

Evaluate h(- 8), then substitute the result obtained into g(x), that is

h(- 8) = (- 8)² - 2 = 64 - 2 = 62, then

g(62) = 2(62) - 5 = 124 - 5 = 119

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3 years ago

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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