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3 years ago

A class of 33 students elected a class treasurer. there were 111votes for candidate a and 15 votes for candidate

1 answer:

3 0

Since 111 is greater than 33, I will assume that you meant 11 votes for a candidate. Also, assuming that each student can vote only once, the answer would be

11/33 x 100=33.333333...

About 33.333 of the voters voted for candidate a.

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Which of the following statements is true?

Vitek1552 [10]

Answer:

Step-by-step explanation:

1/9 can be simplified to 1/3 as 1/3 x 1/3 = 1/9 and 49/25 can be simplified to 7/5 as 7/5 x 7/5 = 49/25.

Now looking at both the fraction 1/3 and 7/5

1/3 simplifies to 0.34

It is an irrational value as the 3 keeps repeating.

7/5 simplifies to 1.4

It is a rational value as the decimal value is not repeating.

Thus making the second option the answer.

6 0

2 years ago

Susan completed a 26-mile rollerblade marathon in 2.5 hours. If Susan skated at a constant rate, how fast was she skating? A. 10

Bond [772]

Answer:

A. 10.4 mph

Step-by-step explanation:

26÷2.5=10.4

Double checking your answer:

If she does 10.4 miles in an hour, she can get 20.8 in two hours

then you add half of how much she can go in an hour which is 5.2 (10.4÷2)

you add it together, and you will get 26 miles in total

(20.8+5.2=26)

6 0

3 years ago

Alex spent $23 on a bus ride. The bus charges $0.35 a mile plus $2 a trip. How

Svetlanka [38]

67.7 cus 23 divided by .35 is 65.7+2 =67.7

7 0

3 years ago

B(n)=2^n A binary code word of length n is a string of 0's and 1's with n digits. For example, 1001 is a binary code word of len

MaRussiya [10]

Answer:

More number of words that can be made: $\bold{2^n}$

Please refer to below proof.

Step-by-step explanation:

Given that:

The number of binary code words that can be made:

$B(n) =2^n$

where n is the length of binary numbers.

Binary numbers means 2 possibilities either 0 or 1.

Here, suppose if we have 5 as the length of binary number.

And there are 2 possibilities for each digit.

So, total number of possibilities will be $2\times 2\times 2\times 2\times 2 = 2^5$

If the length of binary number is 2.

The total words possible are $2^2$ .

These numbers are:

{00, 01, 10, 11}

If the length of binary number is 3. (increasing the 'n' by 1)

The total words possible are $2^3$ .

These words are:

{000, 001, 010, 100, 011, 101, 110, 111}

So, number of More binary words = 8 - 4 = 4 or $2^2$ or $2^n$ .

So, the answer is $2^n$ .

Let us try to prove in generic terms:

$B(n) = 2^n$

Increasing the n by 1:

$B(n+1) = 2^{n+1}$

Number of more words made by increasing n by 1:

$B(n+1) -B(n)= 2^{n+1} -2^n\\\Rightarrow 2\times 2^{n} -2^n\\\Rightarrow 2^n(2-1)\\\Rightarrow \bold{2^n}$

Hence, proved that:

More number of words that can be made: $\bold{2^n}$

6 0

3 years ago

How do you find the slope? I know y=mx=b but I still dont understand

Novosadov [1.4K]

If you have an equation, then the number before the x is your slope, if you have points, use the equation (y2-y1)/(x2-x1)

6 0

3 years ago

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