HI! I really need help (no links!) ignore my answer.

How to prove all pairs of corresponding sides and angles are congruent

Lera25 [3.4K]

The lines on the triangles

5 0

3 years ago

Use the least common denominator to write an equivalent fraction for each fraction for 5/9 and 2/15

Masteriza [31]

75/135 and 18/135
just times 9 and 15, 9 and 2, and then 15 and 5

8 0

3 years ago

Christine Wong has asked Dave and Mike to help her move into a new apartment on Sunday morning. She has asked them both in case

olga nikolaevna [1]

Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> = Mike will show up.

Given:

$P(D^{c})=0.55\\P(M^{c})=0.45$

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

$P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25$

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

$=1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75$

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

$P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25$

Thus, the probability that neither Dave nor Mike will show up is 0.25.

6 0

3 years ago

Read 2 more answers

A bag has six balls labeled a,b,c,d.e and f . One ball will be randomly picked, and its letter will be recorded as the outcome.

Dvinal [7]

<h3>Sample space = {a,b,c,d,e,f}</h3><h3>Event space = {a,c}</h3>

We simply list all of the letters mentioned as they are the possible outcomes. We can only pick one item from the sample space. The event space is the set of outcomes where we want to happen (picking either an 'a' or 'c').

8 0

3 years ago

Read 2 more answers

Natalie has 300 cookies for a bake sale. She makes 8 more batches with 12 cookies in each batch. She sells 6 bags of cookies wit

vitfil [10]

Answer:

I would say it's not reasonable. the answer would be the first one

Step-by-step explanation:

6 0

2 years ago