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torisob [31]
3 years ago
14

If the range of a distribution were 89 and the data were reported as whole numbers, what would the width of the class interval b

e if one chose to group the distribution into approximately 14 class intervals
Mathematics
1 answer:
faltersainse [42]3 years ago
8 0

Answer: The width of the class interval would be 6.

Step-by-step explanation:

Formula to find the class width :

\text{Class width }=\dfrac{\text{Range} }{\text{Number of classes}}

Given : The range of a distribution were 89 .i.e. Range = 89

The approximate number of class intervals = 14

Then, the width of the class interval would be

 \dfrac{89}{14}=6.35714285714\approx6

[Round to the nearest integer.]

Hence, the width of the class interval would be 6.

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4 2/3. add the twos and keep the 2/3
4 0
3 years ago
NEED ANSWERED ASAP WILL REWARD BRAINLIEST
Luda [366]

Answer:

The sum of the first 100 terms is 60400

Step-by-step explanation:

* Lets revise the arithmetic sequence

- There is a constant difference between each two consecutive

  numbers

- Ex:

# 2  ,  5  ,  8  ,  11  ,  ……………………….

# 5  ,  10  ,  15  ,  20  ,  …………………………

# 12  ,  10  ,  8  ,  6  ,  ……………………………

* General term (nth term) of an Arithmetic sequence:

- U1 = a  ,  U2  = a + d  ,  U3  = a + 2d  ,  U4 = a + 3d  ,  U5 = a + 4d

- Un = a + (n – 1)d, where a is the first term , d is the difference

 between each two consecutive terms n is the position of the

 number

- The sum of first n terms of an Arithmetic sequence is calculate from

 Sn = n/2[a + l], where a is the first term and l is the last term

* Now lets solve the problem

- We will use method (1)

- From the table the terms of the sequence are:

 10 , 22 , 34 , 46 , 58 , 82 , 94 , ............., where 10 is the first term

∵ an = a1 + (n - 1) d ⇒ explicit formula

∵ a1 = 10 and a2 = 22

∵ d = a2 - a1

∴ d = 22 - 10 = 12

- The 100th term means the term of n = 100

∴ a100 = 10 + (100 - 1) 12

∴ a100 = 10 + 99 × 12 = 10 + 1188 = 1198

∴ The 100th term is 1198

- Lets find the sum of the first 100 terms of the sequence

∵ Sn = n/2[a1 + an]

∵ n = 100 , a = 10 , a100 = 1198

∴ S100 = 100/2[10 + 1198] = 50[1208] = 60400

* The sum of the first 100 terms is 60400

7 0
3 years ago
Solve each inequality -3x&lt;12
Vlad1618 [11]

Answer:

-3x<12

/-3  /-3

x<-4

Step-by-step explanation:

5 0
4 years ago
aren wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 51
torisob [31]

Answer:

The 95% confidence interval for the number of chocolate chips per cookie for Big Chip cookies is between 4.68 and 13.52.

Step-by-step explanation:

We have the standard deviation of the sample, so we use the students' t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 51 - 1 = 50

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 50 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975([tex]t_{975}). So we have T = 2.01

The margin of error is:

M = T*s = 2.01*2.2 = 4.42

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 9.1 - 4.42 = 4.68

The upper end of the interval is the sample mean added to M. So it is 9.1 + 4.42 = 13.52

The 95% confidence interval for the number of chocolate chips per cookie for Big Chip cookies is between 4.68 and 13.52.

8 0
3 years ago
Calcule as expressões:
Nata [24]

Step-by-step explanation:

a, 1/4+2/5=13/20

b,-1/8+1=7/8

c, 4/25×-1/8=-1/50

d, 2×1/9-1/2=2/9-1/2=-5/18

e, 1+2/5-1/4=7/5-1/4=23/20

8 0
3 years ago
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