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3 years ago

Helppppppppppppp mathhhh

1 answer:

galina1969 [7]3 years ago

6 0

A. It is indeed very possible and in matter of fact very likely because the ratio of RED marbles to BLUE marbles is so high , 10:3 and even after you pick that first red marble and don't replace it to pick the second marble the ratio of red to blue marbles would still be overwhelmingly in favor of the red marbles (9:3) ! Hope I could Help! :)

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Checking account A charges a monthly service fee of $12 and a wire transfer fee of $10.50, while checking account B charges a mo

son4ous [18]

Answer:

checking account A is the better deal.

Step-by-step explanation:

A charges a monthly service fee = $12.00

Wire transfer fee = $10.50

B charges a monthly service fee = $21.00

Wire transfer fee = $8.50

If the requirement is four wire transfer per month

A charges for 4 wires = 10.50 × 4 = $42.00

and adding monthly service fees = 42.00 + 12.00 = $54.00

B charges for 4 wires = 8.50 × 4 = $34.00

and adding monthly service fees = 34.00 + 21.00 = $55.00

Therefore A charges less than B, so checking account A is the better deal.

3 0

3 years ago

PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA POINTS*.. <br> IM GIVING 40 POINTS !! DONT SKIP :((.

grandymaker [24]

Answer:

y = x + 12

Step-by-step explanation:

y -16 = 1(x-4)

y -16 = x -4

y =x +12

5 0

3 years ago

Read 2 more answers

High school geometry

san4es73 [151]

Answer:

BD = 3.75 units

Step-by-step explanation:

Given AD is an angle bisector then it divides the opposite side into segments tat are proportional to the other 2 sides, that is

$\frac{AB}{AC}$ = $\frac{BD}{CD}$ , substitute values

$\frac{6}{8}$ = $\frac{BD}{5}$ ( cross- multiply )

8 BD = 30 ( divide both sides by 8 )

BD = 3.75 units

8 0

3 years ago

Compute the second partial derivatives ∂2f ∂x2 , ∂2f ∂x ∂y , ∂2f ∂y ∂x , ∂2f ∂y2 for the following function. f(x, y) = 2xy (x2 +

blsea [12.9K]

Answer with step-by-step explanation:

We are given that a function

$f(x,y)=2xy(x^2+y^2)^2$

Differentiate partially w.r.t x

Then, we get

$\frac{\delta f}{\delta x}=2y(x^2+y^2)^2+8x^2y(x^2+y^2)=(x^2+y^2)(2x^2y+2y^3+8x^2y)=2(5x^2y+y^3)(x^2+y^2)$

Differentiate again w.r.t x

$\frac{\delta^2f}{\delta x^2}=2(10xy)(x^2+y^2)+4x(5x^2y+y^3)=20x^3y+20xy^3+20x^3y+4xy^3=40x^3y+24xy^3$

Differentiate function w.r.t y

$\frac{\delta f}{\delta y}=2x(x^2+y^2)^2+2xy\times 2(x^2+y^2)\times 2y$

$\frac{\delta f}{\delta y}=(x^2+y^2)(2x^3+2xy^2+8xy^2)=2(x^2+y^2)(x^3+5xy^2)$

Again differentiate w.r.t y

$\frac{\delta^2f}{\delta x^2}=2(2y)(x^3+5xy^2)+20xy(x^2+y^2)=4x^3y+20xy^3+20x^3y+20xy^3=24x^3y+40xy^3$

Differentiate partially w.r.t y

$\frac{\delta^2f}{\delta y\delta x}=2(2y(5x^2y+y^3)+(x^2+y^2)(5x^2+3y^2))=10x^4+36x^2y^2+10y^4$

$\frac{\delta^2f}{\delta y\delta x}=10x^4+36x^2y^2+10y^4$ $\frac{\delta^2f}{\delta x\delat y}=2(2x(x^3+5xy^2)+(3x^2+5y^2)(x^2+y^2))=10x^4+36x^2y^2+10y^4$