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3 years ago

How can a mixed number be written as a sum

1 answer:

6 0

A mixed number can be a sum, but you can change the mixed number into an improper fraction by multiplying the whole by the denominator and adding the numerator

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I need help solving this can you guys please help me?

Kisachek [45]

Answer: # 2 hope this helps

Step-by-step explanation:

6 0

3 years ago

in the adjoining graph the point a is the position of a policeman in the point p is the position of a Thief how far is the thief

faltersainse [42]

Answer: 5 up and 7 side

Step-by-step explanation:

5 0

3 years ago

A multiple choice test has two parts. There are 4^12 ways to answer the 12 questions in Part A. There are 4^5 ways to answer the

LiRa [457]

Using the Fundamental Counting Theorem and the probability concept, it is found that:

There are $4^{17}$ ways to answer all 17 questions.

$4^{-17}$ probability you will get them all right.

<h3>What is the Fundamental Counting Theorem?</h3>

It is a theorem that states that if there are n things, each with $n_1, n_2, \cdots, n_n$ ways to be done, each thing independent of the other, the number of ways they can be done is:

$N = n_1 \times n_2 \times \cdots \times n_n$

<h3>What is a probability?</h3>

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem:

The questions in part A and in part B are independent.

For the 12 questions in part A, there are $4^{12}$ ways to answer, hence $n_1 = 4^{12}$ .

For the 5 questions in part B, there are $4^{5}$ ways to answer, hence $n_2 = 4^{5}$ .

Then:

$N = n_1 \times n_2 = 4^{12} \times 4^5 = 4^{17}$

There are $4^{17}$ ways to answer all 17 questions.

Only one outcome in which all the guesses are correct, hence:

$p = \frac{1}{4^{17}} = 4^{-17}$ probability you will get them all right.

You can learn more about the Fundamental Counting Theorem at brainly.com/question/24314866

8 0

3 years ago

If there are x teams in a sports league, and all the teams play each other twice, a total of n(x) games are played, where n

EleoNora [17]

1.
Consider a group of n objects. Assume we want to form groups of r, from these n objects.

There are in total $C(n, r)= \frac{n!}{r!(n-r)!}$ many ways of doing so.

where, r! is "r factorial", calculated as 1*2*3*...*(r-1)*r

2.

C(x, 2) is the total number of pairs out of x objects, that we can form.

so let the x objects represent the x teams, and 2, represent a group of 2, which means a game.

The total number of games is:

n(x)=2*C(x, 2) = $2* \frac{x!}{2!(x-2)!}= \frac{x!}{(x-2)!}= \frac{x(x-1)(x-2)!}{(x-2)!}=x(x-1)= x^{2} -x$

Remark, we multiplied by 2, since there will be 2 matches for each pair of teams.

Answer: $x^{2} -x$

6 0

3 years ago

Parallel Lines

Liono4ka [1.6K]

Here is the answer, the slope for both lines are the same B(or C) is just different.

7 0

4 years ago