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andre [41]
2 years ago
7

In an election, Candidate A received 75 votes and candidate B received 25 votes, after the first 100 votes were counted. Assume

this ratio is maintained. If Candidate B received 750 votes, how many votes did Candidate A receive
Mathematics
1 answer:
Irina18 [472]2 years ago
7 0

Answer:

2250 votes

Step-by-step explanation:

Given,

Candidate A  :   Candidate B

        75            :          25

ratio:  3            :            1

If,  candidate B = 750 votes

1 --> 750 votes

3 --> 750 * 3 = 2250 votes

Candidate A receives 2250 votes.

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An angle measures 16° more than the measure of its supplementary angle. What is the measure of each angle?
user100 [1]

Answer:

the surface are of a rectangle box of 5cm long, 3cm wide and 4cm wide is 94cm²

3 0
3 years ago
The inverse of F(C)=9/5C+32
Nataliya [291]
F(C) = 9/5C + 32

I'll just change F(C) to y and 9/5C to 9/5x.
y = 9/5x + 32

inverse...

x = 9/5y + 32
9/5y = x - 32
y = 5/9x - 160/9

Sooooo
F(C) = 5/9C - 160/9
6 0
3 years ago
Mary has saved $24. Each week, she saves $12 more. How much money would she have 32 weeks later?
nikitadnepr [17]

Answer:

$408

Step-by-step explanation:

It’s very simple. Since gets 12 dollars every week, multiply the number of weeks and how much money she gets. Plus the amount she already had.

32 x 12 = 384

384 + 24 = 408

6 0
2 years ago
Help pls I'll give 25 points
RideAnS [48]

Answer:

\text{C. }60

Step-by-step explanation:

<u>Question from image</u>:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the <em>x</em> marks where it cannot be.

\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}

Let's start with the first position:

\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in 3!=6 ways. Therefore, there are 4\cdot 6=\boxed{24} possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in 3!=6 ways. Therefore, there are 3\cdot 6=\boxed{18} possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are 3!=6 ways to rearrange the remaining 3 digits for a total of 2\cdot 6=\boxed{12} possible numbers when 1 is the third digit of the number.

Lastly, when 1 is the fourth digit of the number, there is only 1 place the 2 can be. For this one place, there are still 3!=6 ways to rearrange the remaining three numbers. Therefore, there are 1\cdot 6=\boxed{6} possible numbers when 1 is the fourth digit of the number.

Thus, there are 24+18+12+6=\boxed{60} numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.

5 0
3 years ago
2 4 8 16 what is the rule and the sixth number in the sequence answers
aleksklad [387]
The sequence is incrementing by
{2}^{n}
Therefore, the sixth number is
{2}^{6}
Which is 64.
7 0
4 years ago
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