P(8 or 12) = 1/6
P(not 8 or 12 = 5/6
Let the expected points fro one roll be X.

The answer is -(1/6) points.
X + 19 + 7x + 9 = 180
8x + 28 = 180
8x = 180 - 28
8x = 152
X = 152 : 8
X = 19
Answer:
<u>C. 1 to 7.99</u>
Step-by-step explanation:
Let's recall that for calculating the range of a data set, we have to find out the lowest and the highest values of the data set.
As we can see in the graph below, the lowest value in the histogram is 1 and the highest is 7.99, therefore the correct answer is:
<u>C. 1 to 7.99</u>
Q = 3s - 9/6
q = 3(s - 3)/6
q = s - 3/2
6q + 9 = 3s
6q + 9/3 = s
3(2q + 3)/3 = s
2q + 3 = s
s = 2q + 3
To find W⊥, you can use the Gram-Schmidt process using the usual inner-product and the given 5 independent set of vectors.
<span>Define projection of v on u as </span>
<span>p(u,v)=u*(u.v)/(u.u) </span>
<span>we need to proceed and determine u1...u5 as: </span>
<span>u1=w1 </span>
<span>u2=w2-p(u1,w2) </span>
<span>u3=w3-p(u1,w3)-p(u2,w3) </span>
<span>u4=w4-p(u1,w4)-p(u2,w4)-p(u3,w4) </span>
<span>u5=w5-p(u4,w5)-p(u2,w5)-p(u3,w5)-p(u4,w5) </span>
<span>so that u1...u5 will be the new basis of an orthogonal set of inner space. </span>
<span>However, the given set of vectors is not independent, since </span>
<span>w1+w2=w3, </span>
<span>therefore an orthogonal basis cannot be found. </span>