All categories

1 year ago

How does Bolsa describe Billy Joe? smart funny crazy

Mathematics

1 answer:

Ne4ueva [31]1 year ago

3 0

Answer:

The ansswer is the last one. (Crazy)

Step-by-step explanation:

You might be interested in

I FAILED! Find the correct result. 1+4=5 2+5=12 3+6=21 5+8=_________ If your answer is wrong, you must upload it on your timelin

soldi70 [24.7K]

Answer:13

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

I don’t know how to solve this

sleet_krkn [62]

9514 1404 393

Answer:

x ∈ (-∞, 9) U (-1, ∞)

Step-by-step explanation:

Get the absolute value by itself. Do that by multiplying by 4.

|3x +15| > 12

This resolves to two cases:

x +5 > 4 . . . . divide by 3

x > -1 . . . . . . .subtract 5

x +5 < -4 . . . . divide by -3

x < -9 . . . . . . . subtract 5

Then the solution in interval notation is ...

x ∈ (-∞, 9) U (-1, ∞)

6 0

2 years ago

Sin law or cos law?!!!!!!!!!!!!!

MA_775_DIABLO [31]

Answer:

Law of Sines

Step-by-step explanation:

Law of Sines: $\frac{A}{sina} =\frac{B}{sinb} =\frac{C}{sinc}$

Step 1: Solve for m∠B

$\frac{25}{sin36} =\frac{27}{sinx}$

25sinx = 27sin36°

sinx = 27sin36°/25

$x = sin^{-1}(\frac{27sin36}{25} )$

m∠B ≈ 39.4°

Step 3: Find m∠C

180 - (36 + 39.4)

180 - 75.4

m∠C = 104.6°

5 0

2 years ago

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

2 years ago

Help I need this now

Deffense [45]

Answer:

22m

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers