True answers for data in the plot are
The center is 13
The center is not 14
The Peak is 14.
It has three clusters
It is not symmetric to left or right it is bi-modal.
It is has a range from 10 to 15 most number of data points are 13 to 15.
Total number times Shelly waited is 16 times.
Step-by-step explanation:
- While taking cumulative frequency 56.75 percentage comes in 13%.
- It is the center point of the data.
- The 14 is not the center as it shows 93rd percent.
- It has three clusters 0-2 has one cluster,2-4 has 2nd cluster,5-8 third.
- It is not skewed on the left is has bi modal frequency has two heights.
- The person indeed waited for 16 times adding total dots.
- There was a zero 12 which created bi-modal distribution
If one is 50 cents, then to get to one dollar we need 2 of those. also, one dollar equals 100 cents, and 50+50=100, so then the answer would be that you need to have 2 coins of the half-dollar to get to 1 dollar. i hope that this helps you, have a great day! =)
Answer: Option 3
Step-by-step explanation:
You can make equivalent fractions by multiplying or dividing both top and bottom by the same amount. You only multiply or divide, never add or subtract, to get an equivalent fraction.
I hope this helps.
Answer:
Step-by-step explanation:
A system of linear equations is one which may be written in the form
a11x1 + a12x2 + · · · + a1nxn = b1 (1)
a21x1 + a22x2 + · · · + a2nxn = b2 (2)
.
am1x1 + am2x2 + · · · + amnxn = bm (m)
Here, all of the coefficients aij and all of the right hand sides bi are assumed to be known constants. All of the
xi
’s are assumed to be unknowns, that we are to solve for. Note that every left hand side is a sum of terms of
the form constant × x
Solving Linear Systems of Equations
We now introduce, by way of several examples, the systematic procedure for solving systems of linear
equations.
Here is a system of three equations in three unknowns.
x1+ x2 + x3 = 4 (1)
x1+ 2x2 + 3x3 = 9 (2)
2x1+ 3x2 + x3 = 7 (3)
We can reduce the system down to two equations in two unknowns by using the first equation to solve for x1
in terms of x2 and x3
x1 = 4 − x2 − x3 (1’)
1
and substituting this solution into the remaining two equations
(2) (4 − x2 − x3) + 2x2+3x3 = 9 =⇒ x2+2x3 = 5
(3) 2(4 − x2 − x3) + 3x2+ x3 = 7 =⇒ x2− x3 = −1
