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Vikentia [17]
2 years ago
14

PLEASE HELPPPPPPPPP!!!!?????

Mathematics
1 answer:
natta225 [31]2 years ago
8 0

Answers:

  • Mean = 69
  • Median = 69
  • Mode = {68, 70}

===============================================

Explanations:

To get the mean, you add up all of the values and then you divide by N, where N is the number of values in the list. In this case, you'll divide by N = 8.

Adding up all the values gets you 70+68+70+72+68+66+65+73 = 552. Dividing that over 8 then leads to the arithmetic mean of 552/8 = 69

-------

To find the median, you'll first sort the values from smallest to largest. You should have this set {65, 66, 68, 68, 70, 70, 72, 73} after sorting. Next, cross off the smallest and largest values to end up with this smaller set {66, 68, 68, 70, 70, 72}. We do this so we can locate the middle-most item.

Repeat the process of crossing off the smallest and largest items to get {68, 68, 70, 70}. Do so one more time and we get {68,70}.

The midpoint of those two values is (68+70)/2 = 138/2 = 69. The median is 69. The mean and median may not always be the same value.

--------

The mode is the most frequent value. In this case, we have 68 and 70 show up exactly twice for each unique value. This is the most compared to the other unique values that only show up exactly once.

It is possible to have more than one mode, as long those values are more frequent than the rest. We denote the mode as a set, so the mode is {68, 70}

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AlexFokin [52]

Answer:

The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

P=\frac{3}{36}

P=\frac{1}{12}

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is \frac{1}{12}

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For this case we propose a system of equations:

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y: Variable that represents the amount of hours that Olivia's father drove

So, we have:

x + y = 14\\65x + 60y = 895

From the first equation we have:

x = 14-y

Substituting in the second equation:

65 (14-y) + 60y = 895\\910-65y + 60y = 895\\-5y = 895-910\\-5y = -15\\y = \frac {15} {5}\\y = 3

Thus, Olivia's father drove 3 hours.

x = 14-3 = 11

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Answer:

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we know that

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