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Andrei [34K]
3 years ago
6

In january of one winter, 3/10 of a wood pile was used in a wood stove. In February, 2/5 of the wood pile was used. What part of

the wood pile was used up at the end of February?
Mathematics
1 answer:
arsen [322]3 years ago
6 0
2/5 is equivalent to 4/10 so at the end of February 4/10 was used up
And at the end of January, 3/10 was used
so 4/10 + 3/10 = 7/10

At the end of February, 7/10 of the wood was used
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the standard normal curve shown below models the population distribution of a random variable. What proportion of the values in
LiRa [457]

Answer:

<u></u>

  • <u>0.3175  or 31.75%</u>

Explanation:

Please, find attached the diagram with the standard normal curve for this problem.

The<em> two z-scores</em> indiated are z = - 1.25 and z = 0.80

The <em>proportion of the values in the population that does note lie between the two z-scores indicated on the diagram</em> is equal to the the areas below the curve to the left of  z < -1.25 and to the right z > 0.80

The areas to the left or to the right of the z-scores are found in the tables of standard normal cummulative probabilities.

There are tables that show the cummulative probability to the left of the z-scores and tables that show the cummulative probability to the right of the z-scores.

Using a table for the cummulative probatility to the right of the z-score = 0.80 you find:

  • P(Z > 0.80) = 0.2119

Using the symmetry property of the standard normal distribution, P(Z<-1.25) = P(Z>1.25).

Thus, using the same table: P(Z>1.25) = 0.1056

Hence, P(Z<-1.25) + P(Z>0.8) = 0.1056 + 0.2119 = 0.3175.

Therefore, 0.3175 or 31.75% <em>of the values in the population does not lie between the two z-scores indicated on the diagram.</em>

5 0
3 years ago
if you draw 35 lines on a piece of paper so that no two lines are parallel to each other and no three lines are concurrent, how
Levart [38]

Answer:

The lines will intersect 595 times.

Step-by-step explanation:

We can do it in 2 methods.

<u>1st method:-</u> Let us count the number of intersections by drawing lines one by one.

1st line - 0 points=0

2nd line- new 1 point=1

3rd line - old 1 point + new 2 points with two old lines = 1+2

4th line - old (1+2) points+ new 3 points= 1+2+3

5th line - old (1+2+3) points + new 4 points=1+2+3+4

Like that from 35th line-1+2+3+4......34 = \frac{34X35}{2}=595

<u>2nd method:-</u>

We have to choose 2 lines from 35 lines.

We can do it in 35_{C} _{2} ways = 595 points

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miskamm [114]
2x+5= -7
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