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3 years ago

8

kirk has 42 pieces of candy to divide evenly between his three children. if he puts the pieces into three boxes how many pieces

of candy are there per box

1 answer:

Gekata [30.6K]3 years ago

8 0

14 pieces of candy in the box

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$4,000 invested in Fund A returned 5% profit. Amount invested in Fund B returned a 2% profit. How much was invested in Fund B if

kobusy [5.1K]

as I read it, what I get is that

x = returned profits or yielded interest from investment in A

y = returned profits or yielded interest from investment in B

T = total amount invested or namely A + B.

4000 were invested in A, and it yielded 5%, what's 5% of 4000? (5/100)(4000) = 200 = x.

we know the total amount is T, since A get 4000, B must have gotten T - 4000, or the slack. We also know that B yielded a 2% profit, well, what's 2% of T - 4000? (2/100)(T-4000) = y.

we also know that, whatever "x" and "y" are, their sum total yielded a 4% returns from T, or the total principal, what's 4% of T? (4/100)T = 0.04T.

$\bf \begin{cases} T=\textit{total principal}\\[-0.5em] \hrulefill\\ A=4000\\ x = \stackrel{\textit{5\% of A}}{200}\\[-0.5em] \hrulefill\\ B=T-4000\\ y=\stackrel{\textit{2\% of B}}{0.02(T-4000)} \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{5\% of A}}{200}+\stackrel{\textit{2\% of B}}{0.02(T-4000)}~~=~~\stackrel{\textit{4\% of T}}{0.04T} \\\\\\ 200+0.02T-80=0.04T\implies 120+0.02T=0.04T\implies 120=0.02T \\\\\\ \cfrac{120}{0.02}=T\implies 6000=T~\hfill \stackrel{~\hfill \textit{invested in B}}{6000-4000\implies 2000}$

8 0

3 years ago

Variable y varies directly with variable x, and y = 6 when x = 4.

worty [1.4K]

You need to plug K in and solve for K.
6=K4
6/4=K
you can reduce to 3/2=K

5 0

3 years ago

The distributions of X and of Y are described here. If X and Y are independent, determine the joint probability distribution of

max2010maxim [7]

Answer:

The joint probability distribution of X and Y is shown below.

Step-by-step explanation:

The distributions of X and of Y are described as follows:

X : 0 1

P (X) : 0.23 0.77

Y : 1 2 3

P (Y) : 0.40 0.22 0.38

It is provided that X and Y are independent.

That is:

P (X ∩ Y) = P (X) × P (Y)

Compute the joint probability distribution of X and Y as follows:

$P(X=0,Y=1)=P(X=0)\times P(Y=1)=0.23\times 0.40=0.92\\\\P(X=0,Y=2)=P(X=0)\times P(Y=2)=0.23\times 0.22=0.0506\\\\P(X=0,Y=3)=P(X=0)\times P(Y=3)=0.23\times 0.38=0.0874\\\\P(X=1,Y=1)=P(X=1)\times P(Y=1)=0.77\times 0.40=0.308\\\\P(X=1,Y=2)=P(X=1)\times P(Y=2)=0.77\times 0.22=0.1694\\\\P(X=1,Y=3)=P(X=1)\times P(Y=3)=0.77\times 0.38=0.2926$

X 0 1

<u>Y </u>

1 0.9200 0.3080

2 0.0506 0.1694

3 0.0874 0.2926

6 0

3 years ago

What is the derivative of e to the poker minus x

WARRIOR [948]

$(e^{-x})'=e^{-x}\cdot(-x)'=e^{-x}\cdot(-1)=-e^{-x}$

7 0

3 years ago

Read 2 more answers

Rectangle ABCD has vertices A(–6, –2), B(–3, –2),

Tatiana [17]

Answer:

A. T–4, 3(x, y)

Step-by-step explanation:

We are given that the transformation changes the rectangle ABCD to the rectangle A'B'C'D'.

The changes in the co-ordinates are given by,

A = (-6,-2) = (-6,-2) + (-4,3) = (-10,1)

B = (-3,-2) = (-3,-2) + (-4,3) = (-7,-1)

C = (-3,-6) = (-3,-6) + (-4,3) = (-7,-3)

D = (-6,-6) = (-6,-6) + (-4,3) = (-10,-3)

So, we see that the rectangle ABCD is transformed by the co-ordinates (-4,3).

That is, it is translated 4 units to the left and 3 units upwards.

Thus, option A is correct.

3 0

3 years ago