Which inequality explains why these three segments cannot be used to construct a triangle?

Mathematics

1 answer:

Anna11 [10]2 years ago

5 0

Answer:

Iliana cannot construct a triangle because the sum of the two smaller sides will not be greater than the third side. Iliana cannot construct a triangle because the sum of the two smaller sides will be greater than the third side.

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9,13,43,55 what is the mean absolute deviate (MAD) of their ages

lisov135 [29]

Answer:

The mean absolute deviation of this data $\{9,13,43,55\}$ is $MAD =19$ .

Step-by-step explanation:

The mean absolute deviation (MAD) of a dataset is the average distance between each data point and the mean. It gives us an idea about the variability in a dataset.

The steps to find the MAD include:

find the mean (average)
find the difference between each data value and the mean
take the absolute value of each difference
find the mean (average) of these differences

To find the mean absolute deviation of this data $\{9,13,43,55\}$ you must

Step 1: Calculate the mean.

$\:mean=\bar{x}= \frac{9+13+43+55}{4} =\frac{120}{4}=30$

Step 2: Calculate the distance between each data point and the mean and take the absolute value of each difference .

$|9-30|=21\\|13-30|=17\\|43-30|=13\\|55-30|=25$

Step 3: Add the distances together.

$21+17+13+25=76$

Step 4: Divide the sum by the number of data points.

$MAD = \frac{76}{4} =19$

7 0

3 years ago

If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei

MAXImum [283]

Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is ${11 \choose 3} = 165 .$ As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is ${7 \choose 3} = 35$ . Thus, there are only 35 ways to distribute the blackboards in this case.