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4 years ago

Please help with this question!!!

Mathematics

1 answer:

blondinia [14]4 years ago

5 0

Answer:

A = –3

B = 7

Step-by-step explanation:

Please see attached photo for explanation.

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Someone to help me please is for a maths duty

denpristay [2]

Exercise 4.

a) Since Profit (P) is (equal to) Revenue (R) minus Expenses (E) we can write P = R - E and we know that E = $10 million. So P = R - 10

b) appling the fomula written above, we put P = 25 - 10 = $15 million

c) Let's use the inverse formula. From P = R - E we need R, so move R to the left of equal with changed sign and move P to the right of equal with changed sign. We get - R = - P - E so, make it positive... R = P + E

R = 10 + 5 = $15 million

Exercise 5

a) substitute 15 to x... 3.5 * 15 = y

y = 52.5

b) substitute 144 to y

3.5x = 144 and divide both for 3.5 to find x

3.5x/3.5 = 144/3.5 -> x = 144/3.5 circa 41.1

Exercise 6

a) z = -12 * (-7) -> z = 84
b) 108 = -12x -> 12x = -108 -> 12x/12 = -108/12 -> x = -9

Exercise 7

a) y = 15.3 * (-13) -> y = 198.9
b) 15.3x = 351.9 -> 15.3x/15.3 = 351.9/15.3 -> x = 23

Exercise 8

a) y = -11 + 5 -> y = -6
b) 17 = x + 5 -> - x = -17 + 5 -> x = 17 - 5 -> x = 12

6 0

4 years ago

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

kompoz [17]

$f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}$

is a proper joint density function if, over its support, $f$ is non-negative and the integral of $f$ is 1. The first condition is easily met as long as $C\ge0$ . To meet the second condition, we require

$\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}$

b. Find the marginal joint density of $X$ and $Y$ by integrating the joint density with respect to $z$ :

$f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz$

$\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}$

Then

$\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

$\approx\boxed{0.12886}$

c. This probability can be found by simply integrating the joint density:

$\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

$\approx\boxed{0.012262}$

7 0

3 years ago

Jamie purchased a condo for $89,900 with a down payment of 20%; her credit terms were 5% for 15 years. what is jamie's monthly p

LUCKY_DIMON [66]

$\bf \qquad \qquad \textit{Amortized Loan Value}
\\\\
pymt=P\left[ \cfrac{\frac{r}{n}}{1-\left( 1+ \frac{r}{n}\right)^{-nt}} \right]
\\\\$

$\bf \begin{cases}
P=
\begin{array}{llll}
\textit{original amount}\\
\end{array}\to &
\begin{array}{llll}
\quad89900\\
\ \ -20\%\\
-17980\\
-----\\
\quad 71920
\end{array}\\
pymt=\textit{periodic payments}\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{payments are monthly, thus}
\end{array}\to &12\\
t=years\to &15
\end{cases}
\\\\\\
pymt=71920\left[ \cfrac{\frac{0.05}{12}}{1-\left( 1+ \frac{0.05}{12}\right)^{-12\cdot 15}} \right]$

8 0

3 years ago

Read 2 more answers

1.5x1.5 with a base of 2

nordsb [41]

Answer:

If you mean 1.5 * 1.5^2 then the answers 3.375

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP