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vodomira [7]
3 years ago
15

If Olivia traveled 70 miles in 2.5 hours, how many miles will Olivia travel in 9.5 hours at the same constant speed?

Mathematics
2 answers:
maw [93]3 years ago
6 0

Answer:

266

Step-by-step:

70 divided by 2.5= 28

28 times 9.5= 266

Lapatulllka [165]3 years ago
5 0

Answer:

266 miles

Step-by-step explanation:

You need to divide 70 by 2.5 equals 28. Than you need to times that by 9.5.

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The height, h, of a falling object t seconds after it is dropped from a platform 300 feet above the ground is modeled by the fun
notka56 [123]
Average rate = [h(3) - h(0)]/(3 - 0)
h(3) = 300 - 16(3)^2 = 300 - 16(9) = 300 - 144 = 156
h(0) = 300 - 16(0)^2 = 300

average rate = (156 - 300)/3 = -144/3 = -48

Therefore, the object falls with an average rate of 48 ft/s during the first 3 seconds.
4 0
3 years ago
HELP DUE AT 9:00 PM!
alina1380 [7]
A is 5,6 that is the answer for A gotta go but maybe someone else can help with the other ones
4 0
3 years ago
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A linear function has a slope of 7/3 and crosses the y-axis at 10. What is the equation of the line?
Bas_tet [7]
Y=Mx+b (m is slope, b is y-intercept)

7 0
3 years ago
3y-6 divided by 9 = 4-2 divided by -3
sweet [91]

Answer:

y = 16/9

Step-by-step explanation:

Assuming "y" is a variable

3y - 6/9 = 4 - 2/-3

3y - 2/3 =14/3

3y - 16/3 = 0

1/3 (9y - 2) = 14/3

ANSWER = y : 16/9

4 0
3 years ago
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If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
Oksana_A [137]

Answer:

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

 In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

{n \choose 5 } = \frac{n!}{5!(n-5)!}

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 {n \choose 4} .

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is 4 * {n \choose 4} .

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is {n \choose 3} .

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have {3 \choose 2 } = 3  possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of 6 * {n \choose 3}

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of {n \choose 2}  possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

4 * {n \choose 2}

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

I hope that works for you!

4 0
3 years ago
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