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ryzh [129]
3 years ago
9

Please help me! Will give brainlst :)

Mathematics
2 answers:
Sati [7]3 years ago
7 0

Answer:

Step-by-step explanation:

kramer3 years ago
6 0

Answer:

this is a line with a slope of 1 and a y-intercept of 3

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6. A pitcher has 16 cups of water in it. During the day, Aditi drank 5⁄2 cups, Kavitha drank 15⁄4 cups, and Rahul drank 33⁄8 cup
Crank

Answer:

A. Aditi 2 1/2 cups, Kavitha 3 3/4 cups, and Rahul 4 1/8 cups

Step-by-step explanation:

2*2=4 5-4=1      2 1/2

3*4=12  15-12=3       3 3/4

4*8-32  33-32=1     4 1/8

4 0
3 years ago
523 x = 523 x = last one
Sphinxa [80]

Answer:

X has an infinite number of solutions

or

0 = 0

Step-by-step explanation:

<em>Hey there!</em>

Given,

523x = 523x

Simplify

523x = 523x

-523x to both sides

0 = 0

<u>So x has an infinite number of solutions.</u>

<em>Hope this helps :)</em>

5 0
3 years ago
Read 2 more answers
What is the probability that at least one of a pair of fair dice lands of 5, given that the sum of the dice is 8?
o-na [289]

Answer:

0.40

Step-by-step explanation:

to find out the probability that at least one of a pair of fair dice lands of 5, given that the sum of the dice is 8

Let A = sum of dice is 8

B = one lands in 5

P(B/A) = P(AB)/P(A) by conditional probability

P(AB) = sum is 8 and one is 5

So (5,3) or (3,5)

P(A) = sum  is 8.

i.e. (2,6) (2,6) (3,5) (5,3) (4,4)

Required probability

= n(AB)/n(A)

=\frac{2}{5} =0.40

4 0
3 years ago
Three stamps can be attached to each other in various ways.how many ways might three stamps be attached?
ivanzaharov [21]
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula. 
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C. 
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space. 
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.


8 0
3 years ago
Merle opened a new savings account she deposited 40,000 at 10% compounded semiannually at the start of the fourth year merle dep
UNO [17]

happy first question!!!!

4 0
2 years ago
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