15 because I do it like 12+3 because the 3 is negative
That outlier will appear as a single blip of data, but it will be far
away from the central part of the histogram. It will not be that visible
if there is a lot of data, since 1 outlying data point will just show a
very low height compared to the central data near the mean/median/mode,
where there will be higher frequencies and higher bars of data.
If these are the choices:
<span>A.The outlier will appear as a tall bar near one side of the distribution.
B.The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.
C.Since a histogram shows frequencies, not individual data values, the
outlier will not appear. Instead, the outlier increases the frequency
for its class by 1.
D.The outlier will appear as the tallest bar near the center of the distribution.
</span>
Then the most accurate answer will be B, since it's just a single value with a bar corresponding to a frequency of 1.
Answer:
=225/16cm^2
Step-by-step explanation:
Given
The ratio of the length to the width of the rectangle =3:2.
The perimeter of a rectangle = 60cm.
Solution
Let length=3x
width=2x
Perimeter of rectangle=2(length + width)
2(3x+2x)=60
5x*2=60
10X=60
X=60/10
X=6
SO LENGTH =3×6=18cm
WIDTH=2×6
=12cm
Let perimeter of the square=5x
Length of rectangle=6x
6x = 18
X=18/6
X=3
SO PERIMETER OF SQUARE=5×3
4SIDE=15
Side=15/4
Area of square=15/4×15/4
=225/16cm^2
Answer:
See Explanation below for proof.
Step-by-step explanation:
Let number of dimes be D
Then number of quarters = 2D
We then get the following VALUE equation: .1D + .25(2D) = 4.5
.1D + .5D = 4.5
.6D = 4.5
D, or number of dimes = 4.5/.6 = 7.5
There is no such thing as HALF a DIME, so it's NOT POSSIBLE
umm... sure but like where's the question?
