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3 years ago

How many years is tween 1844 to2015

Mathematics

2 answers:

Blizzard [7]3 years ago

6 0

2015-1844=171

171 years is between year 2015 and 1844.

STatiana [176]3 years ago

4 0

To find the number of years between 1844 to 2015 you have to subtract.

Lets say that x is the # of years in between 1844 and 2015

Then we have:
$\sf{2015 - 1844 =x}$

After doing the subtraction using a calculator or a pencil and paper, you get that:

$\huge{\boxed{\bf{x=171}}}$

So 171 years have passed from 1844 to reach to 2015.

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What is the range of the function represented by the table of values below?

Viefleur [7K]

The range is the interval of y values.
Smallest y values is -9 and the highest y values is 9.
So the range is $-9 \leq y \leq 9$

4 0

3 years ago

Solve this system of equations using substitution:

yanalaym [24]

I’m thinking your r in the first equation is actually an x.

X + 2(2x -4) =10

X +4x -8 =10

5x = 18

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2 years ago

Lizzie rolls two dice. What is the probability that the sum of the dice is:

zhenek [66]

Answer:

$A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\$

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

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

$P(A) = \dfrac{12}{36} = \dfrac{1}{3}$

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)} Count 1

Possible cases for sum 3: {(1,2), (2,1)} Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)} Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)} Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

$P(B) = \dfrac{15}{36} = \dfrac{5}{12}$

C. Divisible by 3 and less than 7:

$P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}$

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

$P(A \cap B) = \dfrac{7}{\text{36}}$

5 0

2 years ago

Plz help right now illl give brainlist 5.0 and say thanks

Bess [88]

4+4+1 is the answer

8 0

2 years ago

Read 2 more answers

HELP ME PLEASE I WILL GIVE BRAINLIEST FOR BOTH PARTS PLEASE HELP

Katen [24]

For part A: two transformations will be used. First we will translate ABCD down 3 units: or the notation version for all (x,y) → (x, y - 3) so our new coordinates of ABCD will be:
A(-4,1)
B(-2,-1)
C(-2,-4)
D(-4,-2)

The second transformation will be to reflect across the 'y' axis. Or, the specific notation would be: for all (x,y) → (-x, y) New coordinates for A'B'C'D'
A'(4,1)
B'(2,-1)
C'(2,-4)
D'(4,-2)

Part B: The two figures are congruent.. We can see this a couple of different ways.
- first after performing the two transformations above, you will see that the original figure perfectly fits on top of the image.. exactly the same shape and size.
- alternatively, you can see that the original and image are both parallelograms with the same dimensions.

5 0

3 years ago