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Mumz [18]
3 years ago
13

Your round trip drive to school is 3 7/10 miles.how many miles do you drive to and from school in 5 days?

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
5 0

Answer:

18 1/2 miles

Step-by-step explanation:

3 7/10 * 5

Change to an improper fraction

(10* 3 + 7)/10    * 5

37/10 *5

Rewriting

37 * 5/10

37/2

Changing back to an mixed number

2 goes into 37 18 times with 1 left over

18 1/2

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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 length of AB please don’t do it for fun just try to do it for me hurry?
Lerok [7]

Answer:

y6%cf4yergby65thrbgtrvbh

Step-by-step explanation:

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2 years ago
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The number sentence 4x7=(4x3)+(4x4) is an example of what property
Otrada [13]
<span>The number sentence 4x7=(4x3)+(4x4) is an example of distributive property. 
The distributive property states the following:
</span><span>a × (b + c) = a × b + a × c
</span>Multiplying <span>a sum by a number is the same as </span>multiplying<span> each addend by the number and then adding the products. </span>
In our case, a=4 b+c=7, b=3, c=4

8 0
3 years ago
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PLS HELP ILL GIVE BRAINLEST <br> 1/4 ÷ 3/8<br> divide. i need it in fraction form for an answer
Advocard [28]

Answer:

answere=2/3

Step-by-step explanation:

1/4 divide 3/8

First we will reciprocal

1/4 X 8/3

Now will simplify to lowest term

1/1*2/3

Now multiply

2/3=answere

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