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Alexxandr [17]
3 years ago
15

On every 3rd day Ivan goes to the gym to exercise. Every 5th day Gavin goes to the gym to exercise. Whtis the first day Ivan and

Gavin will go to the gym at the same time?
Mathematics
2 answers:
Serggg [28]3 years ago
8 0

Answer:

Ivan and Gavin will meet 15th day.

Step-by-step explanation:

Given:

On every 3rd Ivan goes to gym and every 5th day Gavin goes to the gym.

We need to find when Ivan and Gavin will go the gym at the same time.

Here we need to find the Least Common Multiple (LCM) of 3 and 5.

3 = 1 * 3

5 = 1 * 5

LCM of 3 and 5 is 1*3*5 = 15

So, Ivan and Gavin will meet 15th day.

Allisa [31]3 years ago
6 0
On the 15th day they will go the least number both go into is 15
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3 years ago
Need answer to 2/3+5/4
Anna11 [10]

Answer:

23/12 or 1 and 11/12

Step-by-step explanation:

Your first step is that you need to gain a common denomitor. In this scenario, this is 12. To get this, you can either go through the given factors such as 3 having the factors of 6, 9, and 12 and 4 having the factors of 8, and 12.

Another method is the multiply the two given denominators together, but there are certain instances where you shouldn't do that.

Now that you have a common denominator, you need to change the numerators to accomadate the denominators. For 3 to turn into 12, you need to multiply it by 4, thus you have to do the same to the numerator. For 4 to turn into 12, you need to multiply it by 3. This will give you these two new fractions:

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Hope this helps!

8 0
3 years ago
Read 2 more answers
he number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.02
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Answer:

a) 98.01%

b) 13.53\%

c) 27.06%

Step-by-step explanation:

Since a car has 10 square feet of plastic panel, the expected value (mean) for a car to have one flaw is 10*0.02 = 0.2  

If we call P(k) the probability that a car has k flaws then, as P follows a Poisson distribution with mean 0.2,

P(k)= \frac{0.2^ke^{-0.2}}{k!}

a)

In this case, we are looking for P(0)

P(0)= \frac{0.2^0e^{-0.2}}{0!}=e^{-0.2}=0.9801=98.01\%

So, the probability that a car has no flaws is 98.01%

b)

Ten cars have 100 square feet of plastic panel, so now the mean is 100*0.02 = 2 flaws every ten cars.

Now P(k) is the probability that 10 cars have k flaws and  

P(k)= \frac{2^ke^{-2}}{k!}

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P(0)= \frac{2^0e^{-2}}{0!}=0.1353=13.53\%

And the probability that 10 cars have no flaws is 13.53%

c)

Here, we are looking for P(1) with P defined as in b)

P(1)= \frac{2^1e^{-2}}{1!}=2e^{-2}=0.2706=27.06\%

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3 years ago
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3 0
3 years ago
A study of class attendance and grades among first-year students at a state university showed that, in general, students who mis
ratelena [41]

Answer:

the numerical value of the correlation between percent of classes attended and grade index is r = 0.4

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Now, given that class attendance explained 16% of the variation in grade index among the students.

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Therefore,  the numerical value of the correlation between percent of classes attended and grade index is r = 0.4

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