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Maslowich
2 years ago
7

A swimmer can do 4 laps in 2 minutes and 30 seconds how long does it take him to do one lap​

Mathematics
2 answers:
Shkiper50 [21]2 years ago
3 0

Answer:

it will take 1 min 30 second to do one lap

Step-by-step explanation:

plz mark me as brainlist

xeze [42]2 years ago
3 0

Answer:

60 seconds make one minute. 2 minutes 30 seconds is 150 seconds

to find time per lap is 150 divided by 4 which will give you 37,5 second per lap

the answer is 37,5 seconds per lap.

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The length of the hypotenuse of a right triangle is 15 in. and one leg has length of 6 in. Find the length of the other leg.
liq [111]

Answer:

square root 189 i did the math and google says the same

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
El seño Alonso pidió prestado dinero a una tasa del 8% anual . Si el cargo por interéses ese año es de $360 cuánto recibirá de p
natima [27]

Answer:

$ 4500

Step-by-step explanation:

Dejar;

Principal sea P

El tiempo sea T

Tasa sea R

Interés sea yo

De la pregunta;

R = 8%, T = 1 año, I = $ 360, P =?

De;

I = PRT / 100

100I = PRT

P = 100 I / RT

P = 100 * 360/8 * 1

P = $ 4500

5 0
2 years ago
Model and solve 2÷2/3
NeX [460]

Answer: 1/3

Step-by-step explanation:

1 divided by 3 = 1/3

7 0
3 years ago
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Which expression is equivalent to
lapo4ka [179]
The last one
2j^2/3k^4
6 0
2 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
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