3 years ago

More and more and more

Mathematics

2 answers:

Lemur [1.5K]3 years ago

8 0

Since a and 5 aren't like terms, you would write the expression as $\frac{a+5}{b}$

andreyandreev [35.5K]3 years ago

4 0

note that the denominators are the same. Add the numerators.

a + 5 = a + 5 (as you cannot combine constants and variables).

? = a + 5 is your answer

hope this helps

You might be interested in

Suppose you deposit $2,000 in a savings account that pays interest at an annual rate of %5. If no money is added or withdrawn

Inessa [10]

Answer:

For the answer, go to:

brainly.com/question/3965798

7 0

2 years ago

What is Divide 2/7 by 1.5?

aleksklad [387]

Answer:

1 ans is ur ans so pls slove this and go ahead

5 0

4 years ago

What is the area of the figure . Round to the nearest tenth.

skelet666 [1.2K]

Answer: its 40.0 m²

Step-by-step explanation:

A=wl=4·10=40.0 m²

your welcome

3 0

3 years ago

Read 2 more answers

Expand & simplify<br> 3(t+5)+5(t−6)

Reptile [31]

Answer:

8t - 15

Explanation:

⇒ 3(t+5)+5(t−6)

distribute inside parenthesis

⇒ 3(t) + 3(5) + 5(t) + 5(-6)

multiply the variables

⇒ 3t + 15 + 5t - 30

collect like terms

⇒ 3t + 5t + 15 - 30

add/ subtract like terms

⇒ 8t - 15

6 0

2 years ago

Read 2 more answers

Assume that we have m coins. We toss each one of them n times. The probability of heads showing up for each coin isp. What’s the

olga nikolaevna [1]

Answer:

$1-(1-p^n)^m$

Step-by-step explanation:

For a coin, the probability of head showing in a single toss is $p$ .

$P(H)=p$

Its complement, the probability of not head is

$P(\Sim H)=1-p$

This is a binomial distribution. In $n$ tosses, the probability of having all heads (i.e. $n$ heads) is

$P(\text{all heads})=\binom{n}{n}p^n(1-p)^0=p^n$

Let's call this value $a$ .

For $m$ coins, we determine the probability of at least 1 coin showing all heads by first finding its complement i.e. the probability of no coin showing all heads. This is also a binomial distribution.

$P(\text{no coin showing all heads})=\binom{m}{0}a^0(1-a)^m=(1-a)^m$

$P(\text{at least 1 coin showing all heads})=1-P(\text{no coin showing all heads})$

$P(\text{no coin showing all heads})=1-(1-a)^m=1-(1-p^n)^m$

8 0

3 years ago