All categories

2 years ago

Answer ASAP

Mathematics

1 answer:

LiRa [457]2 years ago

8 0

Answer:

x = 15°

Step-by-step explanation:

Complementary angles are two angles whose sum = 90°

∠A + ∠B = 90°

3x + 45 = 90

3x = 45

x = 45/3

x = 15°

You might be interested in

Why is this app not working?

Aloiza [94]

Answer:

try restarting it or re downloading

3 0

3 years ago

Read 2 more answers

The odometer in your mother’s car reads 3440 miles less than the odometer in your uncle’s car. Your mother’s odometer reads 2958

Serhud [2]

29580 + 3440 = 33020. The is about 33,000mi. The answer is d.

6 0

3 years ago

Read 2 more answers

Which set of angle measures could be the interior angles of a triangle? A. 15°, 48°, 27° B. 56°, 45°, 46° C. 42°, 84°, 64° D. 93

TEA [102]

Answer:

D. 93°, 29°, 58°

Step-by-step explanation:

In a triangle we can only have 3 Interior angles.

The sum of the Interior angles in a triangle = 180°

Adding up the given Interior angles in the above question:

A. 15°, 48°, 27°

= 15° + 48° + 27°

= 90°

90° ≠ 180°

Option A is wrong

B. 56°, 45°, 46°

= 56 ° + 45° + 46°

= 147°

147° ≠ 180°

Option B is wrong

C. 42°, 84°, 64°

= 42° + 84° + 64°

= 190°

= 190° ≠ 180°

Option C is wrong.

D. 93°, 29°, 58°

93° + 29° + 58°

= 180°

180° = 180°

Option D is the correct Option.

Therefore, the set of angle measures that could be the interior angles of a triangle is Option D. 93°, 29°, 58°

8 0

3 years ago

What is the slope of the line? A) 1 B) 6 C) 8 D) 1/2

antiseptic1488 [7]

8 because rise/run is 80/10

8 0

3 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