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Alisiya [41]
3 years ago
14

Xavier worked 10 hours on Monday and 15 hours on Wednesday. His total pay was $280.00. What is his rate per hour?

Mathematics
2 answers:
mario62 [17]3 years ago
6 0
The answer is B, $11.20. First you add the total hours Xavier worked which is 25 hours. Since the total pay is $280.00, you can divide $280 by 25 to find out the rate per hour. 280 divded by 25 is 11.2, equivalent to 11.20. The answer is $11.20, B.
Alex3 years ago
3 0

Answer:

280/(10+15) = 11.20. Xavier's rate per hour is $11.20

Step-by-step explanation:

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Rama09 [41]

Answer:

the equal sign

Step-by-step explanation:

7 0
2 years ago
WILLL GIVE BRAINLEST <br> WILL GUVE BRAINLEST<br> Photo is attached!! Thanks I appreciate if u help
sp2606 [1]

Answer:

1 false

2 true

3 true

4 false

5 true

Step-by-step explanation:

f(a) = (2a - 7 + a^2) and g(a) = (5 – a).

1 false f(a) is a second degree polynomial and g(a) is a first degree polynomial

When added together, they will be a second degree polynomial

2. true  When we add and subtract polynomials, we still get a polynomial, so it is closed under addition and subtraction

3.  true  f(a) + g(a) = (2a - 7 + a^2) + (5 – a)

Combining like terms = a^2 +a -2

4.  false   f(a) - g(a) = (2a - 7 + a^2) - (5 – a)

Distributing the minus sign (2a - 7 + a^2) - 5 + a

Combining like terms  a^2 +3a -12

5.  true  f(a)* g(a)  = (2a - 7 + a^2)  (5 – a).

Distribute

                      (2a - 7 + a^2)  (5)  – (2a - 7 + a^2)  (a)

                       10a -35a +5a^2 -2a^2 -7a +a^3

Combining like term

-a^3 + 3 a^2 + 17 a - 35

3 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
Given: sinθ = -3/5, θ is a third quadrant angle, and tan φ = -7/24, φ is a second-quadrant angle; find cos(θ + φ])
Natali [406]

Answer:

First option: cos(θ + φ) = -117/125

Step-by-step explanation:

Recall that cos(θ + φ) = cos(θ)cos(φ) - sin(θ)sin(φ)

If sin(θ) = -3/5 in Quadrant III, then cos(θ) = -4/5.

Since tan(φ) = sin(φ)/cos(φ), then sin(φ) = -7/25 and cos(φ) = 24/25 in Quadrant II.

Therefore:

cos(θ + φ) = cos(θ)cos(φ) - sin(θ)sin(φ)

cos(θ + φ) = (-4/5)(24/25) - (-3/5)(-7/25)

cos(θ + φ) = (-96/125) - (21/125)

cos(θ + φ) = -96/125 - 21/125

cos(θ + φ) = -117/125

8 0
2 years ago
Write the coordinate of G'after G(1, 2) was reflected over the line y=x.
Vikentia [17]

Answer:

(-1,2)

Step-by-step explanation:

4 0
3 years ago
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