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d1i1m1o1n [39]
4 years ago
6

How many centi are in 12 in

Mathematics
2 answers:
Deffense [45]4 years ago
5 0
30.48 centimenters are in 12 inches.
Marina CMI [18]4 years ago
3 0
30.48 centimeters in 12 inches
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Choose the arithmetic series that is associated with this summation.
Verizon [17]
Plug n = 1 into the expression 20-9n and we get
20-9n = 20-9*1 = 20-9 = 11

So 11 is the first term of the series that we add up. Therefore the answer must be choice C, which has 11 listed as the first term added
7 0
4 years ago
I need help with all except graph on first question please! Work shown or not doesn’t matter
Jobisdone [24]

Answer:

y = 7/5x + 18/5

Step-by-step explanation:

so we have y = mx + b

f(1) = 5

f(-4) = -2

so let's find b

so like if you do (5 - (-2))/(1 - (-4)) you get (5 + 2)/(1 + 4) = 7/5

y = 7/5x + b

so now we can plug in a value like f(1) = 5

5 = (7/5)1 + b

b = 18/5

5 0
3 years ago
Y = A system of equations. Y equals StartFraction 2 over 3 EndFraction x plus 1. Y equals negative StartFraction 2 over 3 EndFra
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5 0
4 years ago
Pat bounces a basketball 25 times in 30 seconds. At that rate, approxiaetely how many times will Pat bounce the ball in 150 seco
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Answer:

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6 0
3 years ago
An expirement consists of rolling two fair number cubes. What is the probability that the sum of the two numbers will be 4? Expr
ExtremeBDS [4]

Answer:

\dfrac{1}{12}

Step-by-step explanation:

Given:

Two fair number cubes i.e. two dice consisting the numbers 1, 2, 3, 4, 5, 6 on their faces and have equal probability of each number.

The dice are rolled.

To find:

Probability of getting the sum of two numbers as 4.

Solution:

First of all, let us have a look at the total possibilities when two dice are rolled:

([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])

These are total 36 possible outcomes.

For getting a sum as 4:

Possible number of favorable cases are 3 (as highlighted in BOLD in above)

Formula for probability of an event E can be observed as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

Required probability is:

\dfrac{3}{36} = \bold{\dfrac{1}{12}}

5 0
3 years ago
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