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3 years ago

Help pls I will mark brainliest!!!

Mathematics

1 answer:

Sonbull [250]3 years ago

5 0

Answer:

7.5

Step-by-step explanation:

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What is the area of the figure?

Sonja [21]

Answer:

33 $\frac{1}{2}$

Step-by-step explanation:

Calculate the area of the large figure by multiplying the length and the width (7 $\frac{1}{4}$ *6). Then, calculate the area of the smaller figure (5*3 $\frac{2}{3}$ ). Finally subtract the area of the smaller figure from the area of the larger figure. ((7 $\frac{1}{4}$ *6)-(5*3 $\frac{2}{3}$ )

Hope this helps!

5 0

4 years ago

If ΔP′QR′ is a dilated image of ΔPQR, find the scale factor (k) and then find the value of x. answers: A) 7 B) 10.7 C) 6 D) 73.5

Black_prince [1.1K]

Answer:

see explanation

Step-by-step explanation:

To find k, calculate the ratio of corresponding sides, image to original, that is

k = $\frac{P'R'}{PR}$ = $\frac{8}{28}$ = $\frac{2}{7}$

Thus

QR' = $\frac{2}{7}$ × QR = $\frac{2}{7}$ × 21 = 2 × 3 = 6 , that is

x = 6 → C

5 0

3 years ago

How would you explain 2-3? (Site 1)

balandron [24]

1-2 is -1

I think that is the answer

4 0

4 years ago

7x²x + 3x3 – 2x²(x² + 2x - 1)

Bad White [126]

Answer:

− 2 x^ 4 + 6 x^ 3 + 2 x^ 2

Step-by-step explanation:

3 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