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ValentinkaMS [17]
3 years ago
7

The following is the number of minutes to commute from home to work for a group of 25 automobile executives. 35 36 39 43 37 35 3

4 30 36 34 30 39 37 40 38 33 31 28 39 35 35 36 41 24 36 How many classes would you recommend? What class interval would you suggest? (Round up your answer to the next whole number.) Organize the data and plot a frequency distribution on a piece of paper. Comment on the shape of the frequency distribution. It is not symmetric. It is fairly symmetric, with most of the values between 24 and 43. It is not very symmetric, but most of the values lie between 24 and 43.

Mathematics
1 answer:
jeyben [28]3 years ago
4 0

Answer:

It is not symmetric, but skewed left. Data appears more to be on the left side.

Step-by-step explanation:

The smallest value is 24 and the largest is 43 . The difference between these two values is 19 which can be divided into into intervals of 4.

19/4= 4.75 It will be rounded to 5.

The class interval can of 5. Starting from 20 we get class intervals and frequency distribution as

Class Intervals          Data                                        Frequency      

20-24                           24                                              1

25- 29                           28,                                            1

30-34                            34,30,34,30,33,31,                   6

35-39                         35,36,39,37,35,36,39,37,           14

                                         38,39,35,35,36,36

40-44                                43,40,41                                3

The class intervals are inclusive of both upper and lower limits. The difference between the lower limits of two consecutive classes or upper limits of two consecutive classes must be the same.

As we see the difference here is that of 5 between the two upper or lower limits of consecutive classes.

The histogram is attached which shows the class intervals along x- axis and data frequency along y- axis.

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1. In the following diagram, UW bisects XV at Y. Which of the
aev [14]

Answer:

Statement 4 is correct

Step-by-step explanation:

Here, we want to select which statement is true based on the given diagram;

The statement that must be true is that Y is the midpoint of XV

This is because, by bisection , we mean dividing into 2 equal parts

The line UW has divided the line XV into two equal parts

So this mean that Y is the midpoint of the line XV

6 0
3 years ago
Find the sum of <br> 2x^2- 8x– 6 and 9x^2-8
Rama09 [41]

Answer:

11x^2-8x-14

Step-by-step explanation:

(2x^2- 8x– 6)+(9x^2-8) = 11x^2-8x-14

5 0
3 years ago
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If x2 = 30, what is the value of x?<br> A. ±60 B. ±15 C. ±square root of 30 D.±square root of 15
anygoal [31]

Answer:

  C.  ±square root of 30

Step-by-step explanation:

Apply the square root function to both sides of the equation:

  \sqrt{x^2}=\sqrt{30}\\\\|x|=\sqrt{30}\\\\x=\pm\sqrt{30}

_____

The absolute value equation has two solutions. They match choice C.

6 0
3 years ago
A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow
Mrrafil [7]

Answer:

\displaystyle \frac{54}{5405}.

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

\displaystyle \left(50\atop 5\right) = 2,118,760.

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

\displaystyle \left( 28\atop 2\right) = 378.

Number of ways to choose 3 green candies out of a batch of 8:

\displaystyle \left(8\atop 3\right)=56.

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168.

The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:

\displaystyle \frac{21,168}{2,118,760} = \frac{54}{5405}.

3 0
3 years ago
Please help and tell me how to do this, please 24 pts
GREYUIT [131]

Step 1: Find common denominators

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2 3/4 - 1 2/4

Step 2: Subtract

2 - 1 = 1

3/4 - 2/4 = 1/4

Answer: 1 1/4

Hope this helps!! :)

8 0
3 years ago
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