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3 years ago

A yearbook staff has 30 members. In how many different ways can an editor, photographer, designer, and advisor be selected?

Mathematics

1 answer:

vichka [17]3 years ago

4 0

Answer:

the answer is 30 different ways

Step-by-step explanation:

You might be interested in

Which expression is equivalent to 2(3x + 4y + x)?

Elena L [17]

$\huge\textsf{Hey there!}$

$\large\text{2(3x + 4y + x)}$

$\large\textsf{DISTRIBUTE 2 WITHIN the PARENTHESES}$

$\large\text{2(3x) + 2(4y) + 2(x)}$

$\large\text{2(3x) = \boxed{\bf 6x }}$

$\large\text{6x + 2(4y) + 2(x)}$

$\large\text{2(4y) = \boxed{\bf 8y}}$

$\large\text{6x + 8y + 2(x)}$

$\large\text{2(x) = \boxed{\bf 2x}}$

$\large\text{6x + 8y + 2x}$

$\large\textsf{COMBINE the LIKE TERMS}$

$\large\text{(6x + 2x) + (8y)}$

$\large\text{6x + 2x = \boxed{\bf 8x}}$

$\large\text{8x + 8y}$

$\boxed{\boxed{\large\text{Answer: \huge \bf 8x + 8y}}}\huge\checkmark$

$\large\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

6 0

3 years ago

Find the surface area of the cylinder to the nearest whole number. The figure is not drawn to

Anettt [7]

Answer:

correct option is 2187

Total surface area is 2,185.44 square inches.

lateral surface area is 640.56 square inches

Step-by-step explanation:

Given

radius of cylinder = 12 inches

height of cylinder = 17 inches

we will us value of pi as 3.14

surface area for cylinder consist of two part

1. lateral surface area which is gven by $2\pi rl$

where r is radius of cylinder and l is height of cylinder

Thus, lateral surface area for given cylinder is

$2\pi rl\\ =>2* 3.14*12*17\\=> 1281.12$

Thus, lateral surface area for given cylinder is 1281.12 square inches.

_________________________________

second part of surface area is area of two base (upper and lower) of cylinder.

area of base is given by area of circle $\pi r^2$

$area \ of \base = \pi r^2 = 3.14*12^2\\=>area \ of \base = 3.14*144 = 452.16$

This, area of base is for one base of cylinder

since there are 2 base

area of both base will be = 2* area of one base

= 2*452.16 = 904.32.

Thus, area of base is 904.32 sq. in.

_______________________________________

Total surface area of cylinder = area of two base+ lateral surface area of cylinder = 1281.12 square inches + 904.32 square inches

= 2,185.44 square inches.

Thus, total surface area is 2,185.44 square inches.

lateral surface area is 640.56 square inches

but option which is closest to calculated value is 2187.

Thus, correct option is 2187

___________________________________________

Note: this problem can be directly solved using formula

total surface area for cylinder = $2\pi r^2 + 2\pi rl$

5 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

Help is very much appreciated.

Alecsey [184]

Answer:

5 + 5 =10

Step-by-step explanation:

it should have been 4 +6 = 10

3 0

3 years ago

The numerator of a given fraction is 4 less than the denominator. if 3 is subtracted from the numerator and 5 is added to the de

Sergeu [11.5K]

Answer:

Step-by-step explanation:Given that the numerator of a given fraction is 4 less than its denominator.

Also given that 3 is subtracted from the numerator and 5 is added to the denominator, the fraction becomes one by fourth .

Let the fraction be

Since the numerator of a given fraction is 4 less than its denominator we have,

Numerator=Denominator-4

⇒ a=b-4

Since 3 is subtracted from the numerator and 5 is added to the denominator, the fraction becomes one by fourth we have

4(a-3)=1(b+5)

4a-12=b+5

4a-b=17

4(b-4)-b=17 ( ∵ a=b-4)

4b-16-b=17

3b=17+16

3b=33

⇒ b=11

Now put b=11 in a=b-4 we get

a=11-4

⇒ the fraction is a/b=7/11

3 0

3 years ago