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pogonyaev
3 years ago
10

What is the estimate of 1/7 x 20 and 1/9 of 82?

Mathematics
1 answer:
devlian [24]3 years ago
3 0
I believe the answer to 1/7 times 20 would be 2.9 that is rounded to the nearest tenth but if your rounding to the nearest whole number it would be 3
As for 1/9 of 82 it would be 9.1 rounded to the nearest tenth and it would be 9 rounded to the nearest whole number.
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A surveyor standing at A notices two posts B and C on the opposite side of a canal. The posts are 120m apart. If the angle of si
AveGali [126]

Unfortunately there isn't enough information.

Check out the diagram below. We have segment BC equal to 120 meters long. Points B, C, D and E are all on the edge of the same circle. According to the inscribed angle theorem, angles BDC and BEC are congruent. This shows that the surveyor could be at points D or E, or the surveyor could be anywhere on the circle. There are infinitely many locations for the surveyor to be at, which leads to infinitely many possible widths of this canal.

5 0
2 years ago
-7/2 + 1/4 the answer and how to get the answer, how to solve the problem
ale4655 [162]

Answer:

-13/4

Step-by-step explanation:

common denominator -

2*-7/2*2)+1/4=-14/4+1/4

= -13/4

3 0
3 years ago
A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to
melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}

Mean of X:

E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42

Variance of X:

E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}

var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58

5 0
3 years ago
The size of the overall population is __________ in determining the reliability of a poll.
lidiya [134]

The size of the overall population is <u>not important</u> in determining the reliability of a poll.

<h3>What is the purpose of the overall population in a poll?</h3>

The overall population that a poll is based on does not have anything significant to do with the poll itself which means that it is not important in determining reliability.

This is because the poll uses a sample of the population and not the population itself.

Find out more on the reliability of a poll at brainly.com/question/1347272

#SPJ4

8 0
2 years ago
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Katen [24]

Answer: You have the following information given in the problem above:

- Lou filled the tank with gas that cost ​$3.80 per gallon.

-   Lou​'s car had 5.9 gallons in the 13​-gallon tank.

2. Therefore, if he filled the tank, you can calculate the total amount of money Lou spent, by subtracting 13 gallons and 5.9 gallons and multiplying the result by ​$3.80.

3. Then, the result is:

$26.98

4 0
3 years ago
Read 2 more answers
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