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bekas [8.4K]
3 years ago
7

52.75 with a 25% discount. How much will he pay for it?

Mathematics
2 answers:
Aliun [14]3 years ago
8 0

Answer:

39.56

Step-by-step explanation:

Allisa [31]3 years ago
7 0

Answer:

He pays $39.56

Step-by-step explanation:

If finding the discount, always divide the percent by 100.

25% = 0.25

52.75*0.25=13.1875

Because its money, round to hundredths place, $13.19

Subtract the "percent" (13.19) from 52.75

52.75 - 13.19 = $39.56



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dusya [7]

Answer: Correct me if I’m wrong but the answer should be 6,000 kg

Step-by-step explanation:

If you multiply 50 by 120 you get 6,000 or you can multiply 50 by 20 which equals to 1,000 and then do 1,000 multiplied by 6 and you get ur answer (this should be correct)

5 0
3 years ago
Graph f(x)=log 1/2 (x+1)
iVinArrow [24]

Please keep in mind that f(x)=log 1/2 (x+1) can also be written as:

f(x)=log ((x+1)/2)

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4 years ago
Pls I need help! I’ll give Brainlyist!
Phoenix [80]

Answer:

I would think it to be the last one?it’s seems the most accurate of them all?

Step-by-step explanation:

7 0
3 years ago
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
3 years ago
Please help on this khan problem i can't word it right so there is attachments for the introductions and options for answers
Gwar [14]

Answer:

c

Step-by-step explanation:

because it is a outlier we can assume these 2 things:

it influences the mean and median, with the mean being effected more

because we know this, we can now solve the problem

both the median and mean decreased, but the mean decreased more than the median

7 0
3 years ago
Read 2 more answers
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