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3 years ago

What nitrogen base sequence is the partner c-a-t-c-g-a

Mathematics

1 answer:

nataly862011 [7]3 years ago

4 0

Answer: G-T-A-G-C-T

Step-by-step explanation: so A pairs with T While G pairs with C

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Mr. william phisics class lasts 7/8 of an hour, 3/10 is used on arming up . How many minutes do they take to warm up..NOT FRACTI

pychu [463]

7/8
3/10 of 7/8 means 3/10 times 7/8=21/80

1hour=60mins
60 times 21/80=1260/80=15.75

answer is 15.75 mins

5 0

3 years ago

What is 1000 divided by 10

Anuta_ua [19.1K]

Answer:

100

Step-by-step explanation:

just take one zero off the end of 1000 to get 100

now if you 100x10 you get 1000

5 0

3 years ago

Read 2 more answers

If the first and the last terms of an arithmetic series are 10 and 62, show that the sum of the series varies directly as the nu

Nitella [24]

Answer:

10+^@=72_^) + the series

Step-by-step explanation:

6 0

3 years ago

Which matrix equation represents this linear system?<br> <img src="https://tex.z-dn.net/?f=2x-7y%3D-1%5C%5Cx%2B3y%3D-5" id="TexF

Nikolay [14]

Answer:

The answer is the letter B.

The first column represents the x-values, and the second row represents the y-values.

For that reason, if we have:

2x - 7y = -1

x + 3y = -5

Then, the matrix will be given by:

[ 2 -7

1 3]

Then, the third colum will be the equality:

[ -1

-5]

So the correct option is the letter B.

3 0

3 years ago

Read 2 more answers

1)How many ways can the letters in the word BOOKKEEPER be arranged? (Must show set-up )

sergeinik [125]

Question 1

There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are $\frac{10!}{(10-6)!6!}$ ways of arranging the letters, which equal to 210 ways

Question 2

There are seven swimmers in total.
There are $\frac{7!}{(7-1)!1!}$ ways of choosing the first winner, which is 7 ways
There are $\frac{6!}{(6-1)!1!}$ ways of choosing the second winner, which is 6 ways
There are $\frac{5!}{(5-1)!1!}$ ways of choosing the third winner, which is 5 ways
There are 7×6×5=210 ways of choosing first, second, and third winner

Question 3

The probability of eating an orange and a red candy is $\frac{15}{31}$ × $\frac{9}{30}$ , which equals to $\frac{9}{62}$

The probability of eating two green candies is $\frac{7}{31}$ × $\frac{6}{30}$ which equals to $\frac{7}{155}$

3 0

3 years ago