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Hitman42 [59]
3 years ago
14

Pls answer this math question. In a given class 12 students have cats, 15 have dogs, and 6have both. There are nine students wit

h no cat or dog. How many students are in the class? PLS ANSWER THE SAME QUESTION FOR 16 CAT STUDENTS, 10 DOG STUDENTS, 3 THAT HAVE BOTH, AND 12 STUDENTS HAVE NONE.
Mathematics
2 answers:
VLD [36.1K]3 years ago
7 0

Answer:

1st question is 42 students in the class.       2nd question is 41.

Do tell me if I'm wrong please

Step-by-step explanation:

15 + 12 + 9 + 6

   27 + 9 + 6

   27 + 15 = 42

16 + 10 + 3 + 12

  26 + 15

      41

     

Over [174]3 years ago
5 0
Working step by step I first see all my different groups then add them all together
so
12 - cats
15 - dogs
6 - both
9 - neither

12 students + 15 students + 6 students + 9 students = 42 students
therefore 42 students are in the class

16 + 10 + 3 + 12 = 41

I’m sorry if i got it wrong!
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6 0
3 years ago
The set of the consecutive odd numbers 1, 3, 5, 7, ... , N has a sum of 400. How many numbers are in the set?
Amanda [17]

Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.

First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.

Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:

         1, 3, 5, 7, 9, 11, 13, 15 .

Number of items in the set . . . 8
Sum of all the items in the set . . . 64

Hmmm.  That's interesting.  64 happens to be the square of 8 . 
Do you think that might be all there is to it ?

Let's check it out:

Even-numbered lists of odd numbers:

1, 3                                   Items = 2, Sum = 4
1, 3, 5, 7                           Items = 4, Sum = 16
1, 3, 5, 7, 9, 11                 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .

Amazing !  The sum is always the square of the number of items in the set !

For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.

I slogged through it on my calculator, and it's true.

I never knew this before.  It seems to be something valuable
to keep in my tool-box (and cherish always).


3 0
3 years ago
Need help!!! will mark brainliest!!!!!!
irga5000 [103]

Answer:

right 1 unit and down 5 units

Step-by-step explanation:

5 0
3 years ago
You try to explain the number of IBM shares traded in the stock market per day in 2005. As an independent variable you choose th
allochka39001 [22]

Answer:

This is an example of

C. simultaneous causality.

Step-by-step explanation:

Simultaneous causality eliminates the conclusion that is often taken for granted to the effect that one variable is a response variable while the other is an explanatory variable because the two variables, the price and the number of shares, influence each other at the same time.  When more shares are traded than demanded in the stock market in any day, the price tends to go down, and vice versa.

4 0
3 years ago
What is the inverse of the function g(x)=7x+3/x-5?
Ad libitum [116K]

Answer:

g(x) = (5x + 3)/ x - 7

Step-by-step explanation:

Let g(x) = y

y = (7x + 3)/x - 5

Make x the subject

xy - 5y = 7x + 3

xy - 7x = 5y + 3

x(y - 7) = 5y + 3

x = (5y + 3)/ y - 7

Therefore, the inverse of the function = (5x + 3)/ x - 7

8 0
3 years ago
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