Sounds to me as tho you are to graph 3x+5y<10, and that after doing so you are to restrict the shaded answer area created by the "constraint" inequality x≤y+1. OR x-1 ≤ y OR y≥x-1. If this is the correct assumption, then please finish the last part of y our problem statement by typing {x-y<=1}.
First graph 3x+5y = 10, using a dashed line instead of a solid line.
x-intercept will be 10/3 and y-intercept will be 2. Now, because of the < symbol, shade the coordinate plane BELOW this dashed line.
Next, graph y=x-1. y-intercept is -1 and x intercept is 1. Shade the graph area ABOVE this solid line.
The 2 lines intersect at (1.875, 0.875). To the LEFT of this point is a wedge-shaped area bounded by the 2 lines mentioned. That wedge-shaped area is the solution set for this problem.
Answer:
Step-by-step explanation:
The formula of an area of a elipse:
We have:
Substitute:
There are 52 cards in a deck of cards so that makes 52 your denominator. There are 4 aces in a deck of cards, so 52 - 4 = 48 making 48 your numerator. so the probability the card won't be an ace is 48 out of 52.
48/52 chance that the card will not be an ace.
Hope this helps you. :-)
A.Calculate the mean,median and mode.(3 points each) 1.)1,2,3,4,5 2.)2,3,4,5,6,6 3.)6,7,5,4,5,6,2,5
zlopas [31]
Answer:
Step-by-step explanation:
1.)1,2,3,4,5
mean=sum of all values/number of values
=1+2+3+4+5/5
=15/5
mean=3
Mode :
In the given data, no observation occurs more than once.
Hence the mode of the observations does not exist, means mode=0.
Median
1,2,3,4,5
Middle value is 3 so the median is 3.
2.)2,3,4,5,6,6
mean=sum of all values/number of values
=2+3+4+5+6+6/6
=26/6
mean =4.33
Mode
is that value of the observation which occurs maximum number of times so here mode is 6.
Median
2,3,4,5,6,6
4+5/2
9/2
median=4.5
3.)6,7,5,4,5,6,2,5
mean=sum of all values/number of values
=6+7+5+4+5+6+2+5/8
=40/8
mean =5
Mode
is that value of the observation which occurs maximum number of times so here mode is 5
Median
2,4,5,5,5,6,6,7
5+5/2
10/2
median=5
The area of the triangle is
A = (xy)/2
Also,
sqrt(x^2 + y^2) = 19
We solve this for y.
x^2 + y^2 = 361
y^2 = 361 - x^2
y = sqrt(361 - x^2)
Now we substitute this expression for y in the area equation.
A = (1/2)(x)(sqrt(361 - x^2))
A = (1/2)(x)(361 - x^2)^(1/2)
We take the derivative of A with respect to x.
dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]
dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]
Now we set the derivative equal to zero.
(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0
-2x^2 + 361 = 0
-2x^2 = -361
2x^2 = 361
x^2 = 361/2
x = 19/sqrt(2)
x^2 + y^2 = 361
(19/sqrt(2))^2 + y^2 = 361
361/2 + y^2 = 361
y^2 = 361/2
y = 19/sqrt(2)
We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
