The measures of spread include the range, quartiles and the interquartile range, variance and standard deviation. Let's consider each one by one.
<u>Interquartile Range: </u>
Given the Data -> First Quartile = 2, Third Quartile = 5
Interquartile Range = 5 - 2 = 3
<u>Range:</u> 8 - 1 = 7
<u>Variance: </u>
We start by determining the mean,

n = number of numbers in the set
Solving for the sum of squares is a long process, so I will skip over that portion and go right into solving for the variance.

5.3
<u>Standard Deviation</u>
We take the square root of the variance,

2.3
If you are not familiar with variance and standard deviation, just leave it.
Because the 2 sides of the triangle are the same, they are both 21, the two bottom angles would also be the same.
The 3 inside angles of a triangle must equal 180 degrees.
The top angle is given as 38, so the remaining two angles must equal 180 - 38 = 142 degrees.
Since the 2 bottom angles would be the same, divide 142 by 2 to get x.
X = 142 / 2 = 71 degrees.
Answer:
4.56
×
10^
−
3
Step-by-step explanation:
Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the 10
. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.
Answer:
b. the area to the right of 2
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X, which is also the area to the left of Z. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X, which is the area to the right of Z.
In this problem:




Percentage who did better:
P(Z > 2), which is the area to the right of 2.
Answer:
Given System of equation:
x-y =6 .....,[1]
2x-3z = 16 ......[2]
2y+z = 4 .......[3]
Rewrite the equation [1] as
y = x - 6 .......[4]
Substitute the value of [4] in [3], we get

Using distributive property on LHS ( i.e,
)
then, we have
2x - 12 +z =4
Add 12 to both sides of an equation:
2x-12+z+12=4+12
Simplify:
2x +z = 16 .......[5]
On substituting equation [2] in [5] we get;
2x+z=2x -3z
or
z = -3z
Add 3z both sides of an equation:
z+3z = -3z+3z
4z = 0
Simplify:
z = 0
Substitute the value of z = 0 in [2] to solve for x;

or
2x = 16
Divide by 2 both sides of an equation:

Simplify:
x= 8
Substitute the value of x =8 in equation [4] to solve for y;
y = 8-6 = 2
or
y = 2
Therefore, the solution for the given system of equation is; x = 8 , y = 2 and z =0