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3 years ago

12

Find the distance between the two points. Round to the nearest tenth, if necessary. (Point 1 is at (-5,2) and point 2 is at (3,-

3))

1 answer:

aivan3 [116]3 years ago

3 0

Answer:

d = 9.4

Step-by-step explanation:

d= $\sqrt{(3-(-5))^2 +(-3-2)^2\\$

d= $\sqrt{(8)^2+(-5)^2}$

d= $\sqrt{64+25}$

d= $\sqrt{89$

d=9.4

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A ship leaves port on a bearing of 34.0 and travels 10.4 mi. the ship then turns due east and travels 4.6mi. how far is the ship

Natasha2012 [34]

A bearing of 34° corresponds to corresponding angle of θ=90-34=56°
The (x,y) values for the position of the ship after completing its first heading are:
x=(10.4cos 56)
y=(10.4sin56)

The trigonometric angle for θ=90-90=0
The (x,y) values for the postion of the ship after completing the second bearing is:
x=(10.4 cos56)+(4.6cos0)≈10.4 mi
y=(10.4sin56)+(4.6sin0)≈8.6mi

the distance from the port will therefore be:
d=√(10.4²+8.6²)≈13.5 miles

It trigonometric angles is:
θ=arctan(y/x)
θ=arctan(8.6/10.4)
θ≈39.6°
Thus the bearing angle is:
90-39.6=50.4°

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3 years ago

What is the GCF of 24,48m

Phoenix [80]

24
It goes into 24 once and into 48 twice

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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

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padilas [110]

I don't understand what you mean but this might be the answer
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NNADVOKAT [17]

Answer:

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Step-by-step explanation:

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