Answer:
B)5/3
Step-by-step explanation:
"3x + 5y = 15."
Rewrite in slope-intercept form.
The slope-intercept form is
y=mx+b
where m is the slope and b is the y-intercept.
y=mx+b
Subtract 3x from both sides of the equation.
5y=15−3x
Divide each term by "5" and simplify.
5y/5=15/5−3x/5
"5" and "5" cancel each other out
Divide 15 by 5
y=3-3x/5
Reorder 3 and −3x/5
y=-3x/5+3
Rewrite in slope-intercept form.
y=−3/5x+3
Slope:-3/5
The equation of a perpendicular line to y=−3x/5+3 must have a slope that is the negative reciprocal of the original slope.
<em>m</em>perpendicular=−1/(−3/5)
Simplify the result.
mperpendicular=5/3
hope this helps!
Answer:let me know please
Step-by-step explanation:
Ok
the sets are
natural numbers (or counting numbers), this is like 1,2,3,4,5 etc
whole numbers, this is including 0, so 0,1,2,3,4,5,6 etc
integers, this includes the previoius set and negatives, so -3,-2,-1,0,1,2,3,4,5 etc
rational numbers, this is the set of numbers that can be written in form a/b where b≠0, so all integers can be written like this, like example -3=-3/1, so -7/9 belongs here
-7/9 goes in the rational set
So you know the compound continuously formula is
. Thus, just plug in.
Plug that into your calc to get $1832.
Answer:
1. D. 20, 30, and 50
2. A. 86
3. B. 94
Step-by-step explanation:
1. To find the outliers of the data set, we need to determine the Q1, Q3, and IQR.
The Q1 is the middle data in the lower part (first 10 data values) of the data set (while the Q3 is the middle data of the upper part (the last 10 data values) the data set.
Since it is an even data set, therefore, we would look for the average of the 2 middle values in each half of the data set.
Thus:
Q1 = (85 + 87)/2 = 86
Q3 = (93 + 95)/2 = 94
IQR = Q3 - Q1 = 94 - 86
IQR = 8
Outliers in the data set are data values below the lower limit or above the upper limit.
Let's find the lower and upper limit.
Lower limit = Q1 - 1.5(IQR) = 86 - 1.5(8) = 74
The data values below the lower limit (74) are 20, 30, and 50
Let's see if we have any data value above the upper limit.
Upper limit = Q3 + 1.5(IQR) = 94 + 1.5(8) = 106
No data value is above 106.
Therefore, the only outliers of the data set are:
D. 20, 30, and 50
2. See explanation on how to we found the Q1 of the given data set as explained earlier in question 1 above.
Thus:
Q1 = (85 + 87)/2 = 86
3. Q3 = (93 + 95)/2 = 94
