Multiply 2 by -6.
y=2x-1
y=2(-6)-1
y= -12-1
y= -13
Answer:
Probability that Caroline buys fruit, a CD or both is 0.76.
Step-by-step explanation:
Let event A = Caroline buys fruit, event B = Caroline buys CD, Ac and Bc are complementary events.
Events AB, ABc, AcB and AcBc are jointly exhaustive and disjoint, hence P(AB) + P(ABc) + P(AcB) +P(AcBc) =1.
Events A and B independent, hence Ac and Bc independent too and probability P(AcBc) = P(Ac)*P(Bc) = (1 - P(A))(1-P(B)) = 0.6*0.4 = 0.24.
Required probability P(AB + ABc + AcB ) = P(AB) + P(ABc) + P(AcB) = 1- P(AcBc) = 1 - 0.24 = 0.76.
Answer:
See attachment
Step-by-step explanation:
You can obtain any two points on the graph of 
and use it to draw its graph.
When x=0, 
So you plot (0,2)
When x=1, 
You again plot (1,-71).
With a straight edge you can now draw a straight line through the two points.
See attachment
a cause their really isnt any ADEF like in letters
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
