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Amiraneli [1.4K]
3 years ago
7

Eating 2 tablespoons of peanut butter gives you 200 calories. Suppose you should get 2300 calories in one day. How many calories

will come from other foods?
Mathematics
1 answer:
Masja [62]3 years ago
6 0

Answer:

2100

Step-by-step explanation:

because you subtract the 200 from the total calories.

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DNA molecules consist of chemically linked sequences of the bases adenine, guanine, cytosine and thymine, denoted A, G, C and T.
Dmitry [639]

Answer:

1. See the attached tree diagram (64 different sequences); 2. 64 codons; 3. 8 codons; 4. 24 codons consist of three different bases.

Step-by-step explanation:

The main thing to solve this kind of problem, it is to know if the pool of elements admits <em>repetition</em> and if the <em>order matters</em> in the sequences or collections of objects that we can form.

In this problem, we have the bases of the DNA molecule, namely, adenine (A), thymine (T), guanine (G) and cytosine (C) and they may appear in a sequence of three bases (codon) more than once. In other words, <em>repetition is allowed</em>.

We can also notice that <em>order matters</em> in this problem since the position of the base in the sequence makes a difference in it, i.e. a codon (ATA) is different from codon (TAA) or (AAT).

Then, we are in front of sequences that admit repetitions and the order they may appear makes a difference on them, and the formula for this is as follows:

\\ Sequences\;with\;repetition = n^{k} (1)

They are sequences of <em>k</em> objects from a pool of <em>n</em> objects where the order they may appear matters and can appeared more than once (repetition allowed).

<h3>1 and 2. Possible base sequences using tree diagram and number of possible codons</h3>

Having all the previous information, we can solve this question as follows:

All possible base sequences are represented in the first graph below (left graph) and are 64 since <em>n</em> = 4 and <em>k</em> = 3.

\\ Sequences\;with\;repetition = 4^{3} = 4*4*4 = 64

Looking at the graph there are 4 bases * 4 bases * 4 bases and they form 64 possible sequences of three bases or codons. So <em>there are 64 different codons</em>. Graphically, AAA is the first case, then AAT, the second case, and so on until complete all possible sequences. The second graph shows another method using a kind of matrices with the same results.

<h3>3. Cases for codons whose first and third bases are purines and whose second base is a pyrimidine</h3>

In this case, we also have sequences with <em>repetitions</em> and the <em>order matters</em>.

So we can use the same formula (1) as before, taking into account that we need to form sequences of one object for each place (we admit only a Purine) from a pool of two objects (we have two Purines: A and G) for the <em>first place</em> of the codon. The <em>third place</em> of the codon follows the same rules to be formed.

For the <em>second place</em> of the codon, we have a similar case: we have two Pyrimidines (C and T) and we need to form sequences of one object for this second place in the codon.

Thus, mathematically:

\\ Sequences\;purine\;pyrimidine\;purine = n^{k}*n^{k}*n^{k} = 2^{1}*2^{1}*2^{1} = 8

All these sequences can be seen in the first graph (left graph) representing dots. They are:

\\ \{ATA, ATG, ACA, ACG, GTA, GTG, GCA, GCG\}

The second graph also shows these sequences (right graph).

<h3>4. Possible codons that consist of three different bases</h3>

In this case, we have different conditions: still, order matters but no repetition is allowed since the codons must consist of three different bases.

This is a case of <em>permutation</em>, and the formula for this is as follows:

\\ nP_{k} = \frac{n!}{n-k}! (2)

Where n! is the symbol for factorial of number <em>n</em>.

In words, we need to form different sequences (order matters with no repetition) of three objects (a codon) (k = 3) from a pool of four objects (n = 4) (four bases: A, T, G, and C).

Then, the possible number of codons that consist of three different bases--using formula (2)--is:

\\ 4P_{3} = \frac{4!}{4-3}! = \frac{4!}{1!} = \frac{4!}{1} = 4! = 4*3*2*1 = 24

Thus, there are <em>24 possible cases for codons that consist of three different bases</em> and are graphically displayed in both graphs (as an asterisk symbol for left graph and closed in circles in right graph).

These sequences are:

{ATG, ATC, AGT, AGC, ACT, ACG, TAG, TAC, TGA, TGC, TCA, TCG, GAT, GAC, GTA, GTC, GCA, GCT, CAT, CAG, CTA, CTG, CGA, CGT}

<h3 />

6 0
3 years ago
What is the equation of the function graphed below?
Andreas93 [3]

Answer:

D

Step-by-step explanation:

sorry if the answer is wrong

5 0
3 years ago
One senior from the high schools will be selected at random. What is the probability that the senior selected will not be from h
DochEvi [55]

Using the probability concept, it is found that there is a 0.7361 = 73.61% probability that the senior selected will not be from high school b given that the senior responded with a choice other than college.

<h3>What is a probability? </h3>
  • A <em>probability </em>is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

Researching the problem on the internet, it is found that:

  • 538 seniors responded with a choice other than college.
  • Of those students, 99 + 83 + 49 + 31 + 63 + 71 = 396 are not from high school b.

Hence:

p = \frac{396}{538} = 0.7361

0.7361 = 73.61% probability that the senior selected will not be from high school b given that the senior responded with a choice other than college.

You can learn more about the probability concept at brainly.com/question/26148436

8 0
3 years ago
Qustion to give kagimatoby a brainly
Colt1911 [192]

Answer:

wdym

Step-by-step explanation:

7 0
3 years ago
Please help i would really appreciate it.
vredina [299]
(1,4) this is because the flip over y axis does not touch the y value but gives the opposite of the x value
8 0
3 years ago
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