All categories

3 years ago

Help please need it adap

Mathematics

1 answer:

Rus_ich [418]3 years ago

7 0

Answer:

supplementary angle to x = 180 - 31 - 43 - 32 = 74°

x = 180° - 74° = 106°

You might be interested in

If a die is rolled one time, what is the probability of getting a 4

BaLLatris [955]

Answer:

1/6

Step-by-step explanation:

You have one die, with one 4 and six sides. The probability of getting a 4 with one roll is 1/6.

5 0

3 years ago

Read 2 more answers

In an aquiarium, there are 7 large fish and 6 small fish. Half of the small fish are red. One fish is selected at random. Find t

Nesterboy [21]

Answer:It's looking for square feet so it wasn't the area. The area is o

Step-by-step explanation:

6 0

3 years ago

What is greater than 3.43

zalisa [80]

Specify? Not sure how to answer

6 0

4 years ago

Read 2 more answers

Given a joint PDF, f subscript X Y end subscript (x comma y )equals c x y comma space 0 less than y less than x less than 4, (1)

ioda

(1) Looks like the joint density is

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0$

In order for this to be a proper density function, integrating it over its support should evaluate to 1. The support is a triangle with vertices at (0, 0), (4, 0), and (4, 4) (see attached shaded region), so the integral is

$\displaystyle\int_0^4\int_y^4 cxy\,\mathrm dx\,\mathrm dy=\int_0^4\frac{cy}2(4^2-y^2)=32c=1$

$\implies\boxed{c=\dfrac1{32}}$

(2) The region in which X > 2 and Y < 1 corresponds to a 2x1 rectangle (see second attached shaded region), so the desired probability is

$P(X>2,Y$

(3) Are you supposed to find the marginal density of X, or the conditional density of X given Y?

In the first case, you simply integrate the joint density with respect to y:

$f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^x\frac{xy}{32}\,\mathrm dy=\begin{cases}\frac{x^3}{64}&\text{for }0$

In the second case, we instead first find the marginal density of Y:

$f_Y(y)=\displaystyle\int_y^4\frac{xy}{32}\,\mathrm dx=\begin{cases}\frac{16y-y^3}{64}&\text{for }0$

Then use the marginal density to compute the conditional density of X given Y:

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}\frac{2xy}{16y-y^3}&\text{for }y$

6 0

3 years ago

An athletic field is a 60 yd60 yd-by-120 yd120 yd rectangle, with a semicircle at each of the short sides. A running track 101

gregori [183]

Distance of each track are:

D₁ = 428.5 yd

D₂ = 436.35 yd

D₃ = 444.20 yd

D₄ = 452.05 yd

D₅ = 459.91 yd

D₆ = 467.76 yd

D₇ = 475.61 yd

D₈ = 483.47 yd

Explanation:

Given:

Track is divided into 8 lanes.

The length around each track is the two lengths of the rectangle plus the two lengths of the semi-circle with varying diameters.

Thus,

$D = 2(120) + 2. \frac{1}{2} \pi d\\\\D = 240 + \pi d$

Starting from the innermost edge with a diameter of 60yd.

Each lane is 10/8 = 1.25yd

So, the diameter increases by 2(1.25) = 2.5 yd each lane going outward.

So, the distances are:

D₁ = 240 + π (60) → 428.5yd

D₂ = 240 + π(60 + 2.5) → 436.35 yd

D₃ = 240 + π(60 + 5) → 444.20 yd

D₄ = 240 + π(60 + 7.5) → 452.05 yd

D₅ = 240 + π(60 + 10) → 459.91 yd

D₆ = 24 + π(60 + 12.5) → 467.76 yd

D₇ = 240 + π(60 + 15) → 475.61 yd

D₈ = 240 + π(60 + 17.5) → 483.47 yd

4 0

4 years ago