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Brums [2.3K]
3 years ago
5

Forty tickets were Sold for a lottery. If Lin bought two tickets, what are the chances he will win?

Mathematics
1 answer:
ollegr [7]3 years ago
4 0

Lin has a 2/40 chance of winning,

or in percentage, a 2.5% chance.

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Help and show all work please
Anarel [89]

Answer:

1-3

2-8.9

3-6.3

4-26

5-7.8

6-17

7-10

8-32

9-7.9

10-13.2

11- is a right triangle

12- is not a right triangle

13- is a right triangle

14- is not a right triangle

15- is a right triangle

16- is not a right triangle

Step-by-step explanation:

I will do the rest it will just take me some time. I will also give you

step by step at the end.

7 0
2 years ago
Read 2 more answers
!!Help!!<br>solve for x​
Murrr4er [49]

WXY = XYW ( = 41 ) ⇒  XYW is a isosceles triangle

⇒ XW = YW

⇒ 54 = 6x + 6

⇒ 6( x + 1 ) = 54

⇒ x + 1 = 9

⇒ x = 8

ok done. Thank to me :>

4 0
2 years ago
1. What does it mean that Nikoleta is in the 10th percentile for length and weight? Is that a desirable statistic if she was in
VladimirAG [237]

Explanation:

1. It means that Nikoleta is longer and heavier than 10% of children her age.

2. We generally expect that a change from being in the 50th percentile to the 10th percentile signifies a condition that is interfering with normal growth. That is not usually considered desirable.

5 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
Points (1,3), (-4,7), and (-29,K) are collinear. Find K.<br><br> How do i solve this?
oee [108]
Co linear points have same slope
<span>Points (1,3), (-4,7), and (-29,K)
m1,2=7-3/-4-1=-4/5

m2,3=K-7/-29+4
m2,3=K-7/-25

-4/5=K-7/-25
4/5=K-7/25
4=K-7/5
20=K-7
K=27
 </span>
3 0
3 years ago
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