The number of snowdays of District 241 are 4.
Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data.
We have to find the number of snowdays of District 241
School District District201 District211 District221 District231 District 241
Number of 4 8 3 6 ?
snow days
Mean of snow days is 5.
Mean =Sum of observations/Number of observations
5=4+8+3+6+x/5
25=21+x
Subtract 21 from both sides
x=4
Hence, the number of snow days of District 241 are 4.
To learn more on Statistics click:
#SPJ3
Answer:
DIstrict 241 had 4 snow days.
Step-by-step explanation:
5 * 5 = 25
Add the ones you know
4 + 8 + 3 + 6 = 21
Then
25 - 21 = 4
So District 241 had 4 snow days.
I know this answer is 100% correct. I answered it correctly. This problem wasn't that hard. Let me know if you need help with anything else.
Answer:
20/9
Step-by-step explanation:
Alright, so by using the "keep me, change me, turn me over" method - we can easily solve this:
I hope I was of assistance!#SpreadTheLove <3
Answer:
x = 5°
Step-by-step explanation:
We know that in a triangle, the measure of an exterior angle is equal to the sum of its two remote interior angles, therefore:
7x + 4 + 61 = 20x
7x + 65 = 20x
13x = 65
x = 5°
Answer:
Solution given:
61°+(7x+4)°=20x [ exterior angle is equal to the sum of two opposite interior angle]
65+7x=20x
65=20x-7x
13x=65°
x==5°
value of x=5°
slope = -5 , y - int = -1
Answer:
Y=-1/5x-1
Step-by-step explanation:
Combine the slope then the slope intercept.
2) 94, 124, 154
3) 8, 12, 16
4) -428, -528, -628
5) - 98, -118, 138
6) 13, 18, 23
Answer:
The set is not a basis. It is not linearly independent and doesn't span the given vector space
Step-by-step explanation:
Let u = (1,0,3), v = (-3,1,-7) and w=(5,-1,13). We want to check if the set {u,v,w} is a basis for . By definition, a basis is a linearly independent set that spans the vector space. So, if it is a basis, it automatically is linearly independent and spans the whole space. Since we have 3 vectors in
which is the matrix whose columns are u,v,w. To check that the set {u,v,w} is linearly independent,it is equivalent to check that the row-echelon form of A has 3 pivots.
The step by step calculation of the row-echelon form of A is ommited. However, the row-echelon form of A is
In this case, we have only 2 pivots on the first and second column. This means that the columns 1,2 of matrix A are linearly independent. Hence, the set {u,v,w} is not linearly independent, and thus, it can't be a basis for . Since it is not a basis, it can't span the space.