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Ghella [55]
3 years ago
14

The number of patients treated at Dr. Artin’s dentist office each day was recorded for nine days. These are the data: 6, 6, 6, 5

, 5, 6, 5, 6, 5. Find the mean, median, mode, and range of the data. If necessary, round to the nearest tenth. A. 6, 4.6, 6, 1 B. 5.6, 6, 6, 11 C. 4.6, 6, 6, 11 D. 5.6, 6, 6, 1
Mathematics
1 answer:
klemol [59]3 years ago
6 0

Answer:

D) 5.6, 6, 6, 1

Step-by-step explanation:

Before we start solving this problem, we need to first rearrange the provided data from smallest to greatest, so we get:

5, 5, 5, 5, 6, 6, 6, 6, 6

now we can find the mean. The mean is nothing but the average of the provided values, so we need to add them and then divide them into the total amount of data:

mean=\frac{5+5+5+5+6+6+6+6+6}{9}=5.6

next, in an odd number of data, the median is the middle number when written in order. In this case the middle number (the one in the position 5) is 6, so:

median=6

the mode will tell you what is the value that has the greatest number of occurrencies. This is the number that appears the most on our list.

Mode=6

and the range is the difference between the greatest value and the smallest value:

Range=6-5=1

so the answer is D.

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3 years ago
Given the Arithmetic series A1+A2+A3+A4 7 + 11 + 15 + 19 + . . . + 91 What is the value of sum?
aivan3 [116]

Answer:

The value of the sum is 1078

Step-by-step explanation:

* Lets revise how to find the sum of the arithmetic series

- In the arithmetic series there is a constant difference between each

 two consecutive terms

- Ex:

# 4 , 9 , 14 , 19 , 24 , .......... (+ 5)

# 25 , 15 , 5 , -5 , -15 , .......... (-10)

- So if the first term is a and the constant difference between each two

  consecutive terms is d, then

  U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , ..........

- Then the nth term is Un = a + (n - 1)d

- The sum of n terms is Sn = n/2[a + L] , where L is the last term in

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* Lets solve the problem

∵ The arithmetic series is 7 + 11 + 15 + 19 + ......................... + 91

∴ The first term (a) is 7

∴ The last term is (L) 91

- Lets find the constant difference

∵ 11 - 7 = 4 and 15 - 11 = 4

∴ The constant difference (d) is 4

- Lets find the sum of the series from 7 to 91

∵ Sn = n/2[a + L]

∵ a = 7 and L = 91

∴ Sn = n/2[7 + 91]

- We need to know how many terms in the series

∵ L is the last term and equals 91 lets find its position in the series

∵ Un = a + (n - 1)d

∵ a = 7 , d = 4 and Un = 91

∴ 91 = 7 + (n - 1)(4) ⇒ subtract 7 from both sides

∴ 84 = (n - 1)(4) ⇒ divide both sides by 4

∴ 21 = n - 1 ⇒ add 1 to both sides

∴ n = 22

∴ The number of the terms in the series is 22

- Lets find the sum of the 22 terms (S22)

∴ S22 = 22/2[7 + 91]

∴S22 = 11[98] = 1078

* The value of the sum is 1078

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