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mestny [16]
3 years ago
5

A bakery owner sold her last two cakes for $36 each. She made a profit of 24% on one cake, but had an 8% loss on the other. What

was the total profit or loss on the two sales?
Mathematics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

loss=8

one cake= 4

another one=4

Step-by-step explanation:

36-24 =12

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Alexeev081 [22]
The area of the scale drawing will be:
600 square feet.
Step-by-step explanation:
The actual length of a wall in the room is 46 feet and the actual width of the room is 69 feet.
It is given that:
The scale factor of a room for a scale drawing is 2.3.
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46÷2.3=20 feet
and the width of the room according to scale drawing will be:
69÷2.3=30 feet
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This means that the area of scale drawing will be:
Length×Width
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7 0
3 years ago
A cupcake is 3 inches wide , 3 inches long , 3 inches high . How many cupcakes will fit in a box measuring 15 inches high , 7 cm
Volgvan

Answer:

23 cupcakes

Step-by-step explanation:

Volume is the key to this question;

Calculate the volume of the cupcake;

3 × 3 × 3 = 27 inches^3

Then the volume of the box;

15 × 7 × 6 = 630 inches^3

Then divide the volume of the box by the volume of the cupcake to see how many can fit in;

630 ÷ 27 = 23.3333

;Hence you can fit... 23 cupcakes

3 0
3 years ago
Which will result in a perfect square trinomial?
Andru [333]

There no awnser to give if you dont give a pictre or anything?

7 0
3 years ago
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
a coin bank holds nickels, dimes, and quarters. there are 54 coins in the bank and the value of the coins is $4.75. if there are
TEA [102]
Not possible, you know you have to have at least 5 nickels so substract those first, now your at 4.50.  you cant get to 54 coins after that
7 0
4 years ago
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