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photoshop1234 [79]
3 years ago
14

HELP ASAP!!! have a great day!!!

Mathematics
2 answers:
leonid [27]3 years ago
3 0

Answer:

D

Step-by-step explanation:

First, we get the mean and median of the first data set WITH 5 years : mean = 15 median = 14, then remove the 5 years mean = 15.222 median = 14

ohaa [14]3 years ago
3 0

Answer:

The mean will increase but the median will remain the same.

Step-by-step explanation:

With the 5 years:

Mean = 15

Median = 14

Without the 5 years:

Mean = 15.222222222222

Median = 14

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Add the following fractions. 3/8 + 5/12 = <br> a 8/20 <br> b 19/24 <br> c 8/24 <br> d 1/3
loris [4]
3/8 + 5/12 = 

B. 19/24
8 0
3 years ago
Read 2 more answers
Rolling two die, how many ways can a five show up on at least one die
IRINA_888 [86]
Total possibilities would be: 

(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

You want 5 on at-least one, so mark 5th column & 5th raw.
There are 6 in both lines with 1 common = (5,5)

In short, Your Answer would be 11

Hope this helps!
3 0
3 years ago
Solve: -4.878-(-8.96)=
Pepsi [2]
-4.878-(-8.96) can be changed to -4.878 + 8.96 because when you subtract a negative it becomes a positive so the answer would be 4.082 or B
7 0
3 years ago
Read 2 more answers
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

3 0
3 years ago
There is a line that includes the point (2, 10) and has a slope of 1. What is its equation in
rjkz [21]

Answer:

The line is y=1x+8

Step-by-step explanation:

We can use point-slope form of a line to make an equation. Point-slope form is as follows:

y-y₁=m(x-x₁)

**The variables y₁ and x₁ are where we will plug in our given coordinates and the variable m is your slope.

1. Plug in the given info:

y-10=1(x-2)

2. Distribute the 1:

y-10=1x-2

3. Add 10 to both sides:

y=1x+8

5 0
3 years ago
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