Answer:
Is 3(b+c) equivalent expressions?
A. True
The mean is the average of a set of numbers.
To find the mean of this data, form a number set by gathering all the numbers.
We need to find the average weekly allowance. To do this, each number in the number set should be the different allowances, and their quantity is the number of students who earned that allowance.
In this case, there would be seven 0s, five 3s, seven 5s, three 6s, and two 8s.
The numbers are:
0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 8, 8
To find the mean of these numbers, add then together then divide by the total amount of numbers.
This means doing:
(0 + 0 + 0 + 0 + 0 + 0 + 0 + 3 + 3 + 3 + 3 + 3 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 8 + 8) / 7 + 5 + 7 + 3 + 2
An easier formula could be used by using multiplication.
This would be [7(0) + 5(3) + 7(5) + 3(6) + 2(8)] / 24
This is a lot easier to read!
Now to solve it.
7 • 0 = 0
5 • 3 = 15
7 • 5 = 35
3 • 6 = 18
2 • 8 = 16
0 + 15 + 35 + 18 + 16 = 84
84 / 24 = 3.5
The mean is 3.5, or $3.50
This means that the average weekly allowance amongst these students is $3.50.
Hope this helps!
Answer:Hope This Helps ☺️
Step-by-step explanation:
She is not correct because she did not substitute the same number in both expressions in Step 1
Step-by-step explanation:
CASE 1: substitute 1 for x to both sides of the equations
L.H.S
-(4x-5)+2(x-3)
-(4 (1) - 5)+ 2(1-3) = - (-1) + 2(-2) = 1 - 4 = -3
R.H.S
-2x - 5
-2(1) - 5 = -2-5 = -7
Hence for x= 1
-(4x-5)+2(x-3) ≠ -2x -5
Because -3 ≠ -7
CASE 2: substitute -1 for x to both sides of the equations
L.H.S
-(4x-5)+2(x-3)
-(4 (-1) - 5)+ 2(-1-3) = - (-9) + 2(-4) = 9 - 8 = 1
R.H.S
-2x - 5
-2(-1) - 5 = 2-5 = -3
Hence for x= -1
-(4x-5)+2(x-3) ≠ -2x -5
Because 1 ≠ -3
Answer:
She is not correct because she did not substitute the same number in both expressions in Step 1
Answer:
Consider the following calculations
Step-by-step explanation:
Since 1 Blimp uses 2 components of B and C each
=> choosing 2 components of B(remaining after using in other prototypes) for 1st model= 22C2
choosing 2 components of B(remaining after using in other prototypes) for 2nd model= 21C2
choosing 2 components of B(remaining after using in other prototypes) for 3rd model= 20C2
choosing 2 components of B(remaining after using in other prototypes) for 4th model= 19C2
choosing 2 components of B(remaining after using in other prototypes) for 5th model= 18C2
and choosing 2 components of C(remaining after using in other prototypes) = 24C2
Similarly for C
P(5 prototypes of Blimp created)=[(22C2 / 25C2 )*(24C2 / 25C2 )] + [(21C2 / 25C2 )*(23C2 / 25C2 )]+[(20C2 / 25C2 )*(22C2 / 25C2 )]+[(19C2 / 25C2 )*(21C2 / 25C2 )]+[(18C2 / 25C2 )*(20C2 / 25C2 )]
I think your answer would be G:)