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Mars2501 [29]
3 years ago
7

What is the least common multiple of 12 and 10?​

Mathematics
1 answer:
attashe74 [19]3 years ago
4 0
The Least Common Multiple is 60
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Plase dont awnser just for points
egoroff_w [7]
So 3 2/3 is about 3.67 so it would go in between 3.65 and 3.7. So it would be the second answer.
5 0
3 years ago
Read 2 more answers
A large manufacturing plant uses lightbulbs with lifetimes that are normally distributed with a mean of 1200 hours and a standar
a_sh-v [17]

Answer:

The bulbs should be replaced each 1060.5 days.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 1200, \sigma = 60

How often should the bulbs be replaced so that no more than 1% burn out between replacement periods?

This is the first percentile, that is, the value of X when Z has a pvalue of 0.01. So X when Z = -2.325.

Z = \frac{X - \mu}{\sigma}

-2.325 = \frac{X - 1200}{60}

X - 1200 = -2.325*60

X = 1060.5

The bulbs should be replaced each 1060.5 days.

4 0
3 years ago
Add the product of 2 and 5 to the product of 9 and 8
Lerok [7]

Answer:

82

Step-by-step explanation:

2*5 + 9*8 = 10 + 72 = 82

4 0
3 years ago
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Solve on the interval [0,2pi):<br>3 sec x-1 = 2​
Lelechka [254]

Answer:

0

Step-by-step explanation:

3 0
3 years ago
Find the area of the rectangle.
aev [14]

Answer:

69y+23

Step-by-step explanation:

The area of a rectangle will always be length times width or width times length, doesn't matter which way.

In this case, the width is 3y+1 miles and the length is 23 miles. So, the area is 23(3y+1)=23*3*y+23*1=69*y+23.

So, the area is \boxed{69y+23} and we're done.

3 0
2 years ago
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