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Anarel [89]
3 years ago
7

What is the median score? A. 85 B.87 C.89 D.91

Mathematics
2 answers:
yaroslaw [1]3 years ago
6 0
B. 87 because the median is the middle number when a group of numbers is order least-greatest hope that helped!
777dan777 [17]3 years ago
5 0
B. 87 because the median is when you rank the all the numbers smallest to biggest, and the number in the middle is the median
You might be interested in
Rock musician Donny West is paid 15% on his CD sales and tour video sales. Last year, he sold one million CDs and 550,000 videos
olganol [36]
He had $5,000,000 in CD sales; $3,300,000 in video sales; and he earned $1,245,000 in royalties.

1,000,000 CDs * $5 each = $5,000,000
550,000 videos * $6 each = $3,300,000

Total sales = 5,000,000+3,300,000 = 8,300,000
15% royalties = 0.15(8,300,000) = 1,245,000
5 0
4 years ago
The ratio of black tiles to red tiles is 3 to 7. If 63 back tiles are used for a prject,how many red tiles would be used.
Vladimir79 [104]

Answer:

63:147

Step-by-step explanation:

Because 3*21= 63 and 7*21= 147

7 0
3 years ago
Help! All is good, help if you can!
aliina [53]

Answer:

brainliest plz

Step-by-step explanation:

what do you need help with?

4 0
3 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
Hi its getting pretty late and i need a little help<br> (ill give u brainliest)
Anarel [89]

Answer:

yes

yes

no

no

yes

have a good day!

Step-by-step explanation:

ive just done this today, it was really easy!

7 0
3 years ago
Read 2 more answers
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