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nikklg [1K]
3 years ago
14

Please help will choose brainliest!!!!!

Mathematics
2 answers:
Marianna [84]3 years ago
6 0

Answer:

The answer is (C)

x=y+z+7

MissTica3 years ago
6 0
I think its C. Well it gotta be it
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The perimeter of a rectangle is 66 inches.the length is 3 more than twice the width.what are the dimensions of the rectangle?wha
yan [13]
First, set it up as variables

W=x
L=2x+3

apply perimeter formula and solve
2(x+2x+3)=66
2(3x+3)=66
6x+6=66
6x=60
x=10

width=10
length=23
area= 230


6 0
4 years ago
I need help ASAP ok ...
tankabanditka [31]
One positive plus another is always positive because:

+ and + is +
+ and - is -
- and - is +
8 0
4 years ago
Cindy pics six times as many apples as Mary John pick eight times as many apples as Mary together they picked 135 apples how man
spin [16.1K]

Answer:

126 apples

Step-by-step explanation:

Let the number of apples picked by Mary=m

Cindy picks 6 times as many as Mary= 6Xm=6m

John picks 8 times as many as Mary= 8Xm=8m

Altogether they picked a total of 135 apples.

m+6m+8m=135

15m=135

m=135/15

m=9

We want to determine how many apples John and Cindy Picked.

John Picked 6m apples

Cindy Picked 8m apples

Their combined total= 6m+8m =14m

Since m=9

14m =14 X9 =126 apples

John and Cindy picked a total of 126 apples

3 0
3 years ago
If you bet $1 that a sum of 5 will turn up, what should the house pay (plus returning your $1 bet) if a sum of 5 turns up in ord
erma4kov [3.2K]

Here is the complete question.

1). What are the odds for rolling a sum of 5 in a single roll of two fair dice?

2). If you bet $1 that a sum of 5 will turn up, what should the house pay (plus returning your $1 bet) if a sum of 5 turns up in order for the game to be fair?

Answer:

1).  \frac{1}{8}

2).  $8.

Step-by-step explanation:

1).

The sample space (S) for rolling two fair dice is given as the following parameters illustrated below:

\left[\begin{array}{cccccc}(1,2)&(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)\end{array}\right] \left[\begin{array}{cccccc}(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)\end{array}\right]

Let F represent the event that the sum of both dice turns up is 5.

F = [(1,4),(2,3),(3,2),(4,1)}

∴ The probability for an event F  is illustrated below as:

\frac{P(F)}{P(F')}  =\frac{\frac{4}{36} }{1-\frac{4}{36} }

=\frac{\frac{4}{36} }{\frac{32}{36} }

={\frac{4}{36}}*\frac{36}{32}

= \frac{1}{8}

2). From above, we can see that the probability for rolling a sum of 5 are 1 to 8. Therefore, if you roll a sum of 5, in order for the game to be fair, the house is required to pay $8.

5 0
3 years ago
Is the set of positive intergers the same as nonnegative integers?
algol13
No there totally differnt 
8 0
3 years ago
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