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Gwar [14]
2 years ago
9

Suppose you want to count the number of​ four-letter passwords that can be formed using the letters in the word NUMBER. The four

letters in the password must all be different. Which of the following expressions is true concerning the number of such​ passwords? (A) There are 10 x 9 x 8 x 7 possibiities (B) It is combination question(C) There are 10000 possibilities(D) The password must not follow order matter
Mathematics
1 answer:
WITCHER [35]2 years ago
6 0

Answer:

D.The password must not follow order matter

Step-by-step explanation:

We are given that suppose you want to count the number of four letters password that can be formed using the letters in the word NUMBER.

We are given that all letters in the password must be all different .There is no nothing saying about order so we can make combination without order .

We have to find the number of 4- letter password that can be formed using letters in the word NUMBER.

By using Permutation formula we find the number of 4- letters password that can be formed by using the letters N,M,B,U,E,R.

Permutation formula : The number of arrangements when taken r objects at a time out of n objects

=\frac{n!}{(n-r)!}

Total letters in word NUMBER=N,U,M,B,E,R=6

n=6,r=4

Therefore, the number of arrangements  can be made from 6 letters when 4 letters taken at a time =\frac{6!}{(6-4)!}

The number of selection can be made from 6 letters when 4 letters taken at a time =\frac{6!}{(6-4)!}

The number of selection can be made from 6 letters when 4 letters taken at a time =\frac{6\times5\times 4\times3 \times 2!}{ 2!}

n!=n(n-1)(n-2)(n-3)...........3\cdot.2\cdot.1

The number of selection can be made from 6 letters when 4 letters taken at a time =360

Option D is correct.

Answer : D.The password must not follow order matter

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For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How
natulia [17]
<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let e the cost of the exam hours

Let w be the cost of the workshop hours

Let l be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

e=2w equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

w=2l equation (2)

Finally, we also know that 3hr exams 24hr workshops  and 12hr lectures cost $528, so:

3e+24w+12l=528 equation (1)

Now, lets find the value of l:

Step 1.  Solve for l in equation (3)

3e+24w+12l=528

12l=528-3e-24w equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

12l=528-3e-24w

12l=528-3(2w)-24w

12l=528-6w-24w

12l=528-30w equation (5)

Step 3. Replace equation (2) in equation (5) and solve for l

12l=528-30w

12l=528-30(2l)

12l=528-60l

72l=528

l=\frac{528}{72}

l=\frac{22}{3}

l=7.33

Cost of lectures  = $7.33 per hour



3 0
1 year ago
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Two new drugs are to be tested using a group of 60 laboratory mice, each tagged with a number for identification purposes. Drug
lakkis [162]

Answer:

4.979044478499338 × 10²⁶

Step-by-step explanation:

Combination can be used to determine the number of ways the mice can be selected for the drugs (A, B) and the control group.

Combination factorial is define by  ⁿCr = \frac{n!}{(n-r)!r!}

21 group of mice receiving Drug A  can be selected in ⁶⁰C₂₁ = \frac{60!}{21!39!}

(60 - 21 = 39 ) mice remained for selection of 21 mice for the second drug

Drug B 21 mice can be chosen with ³⁹C₂₁ = \frac{39!}{21!18!}

( 39 - 21 = 18)  remained for control with ¹⁸C₁₈ = \frac{18!}{18!0!}

The number of ways the mice can be chosen for drug A, drug B and the control = ⁶⁰C₂₁ × ³⁹C₂₁ × ¹⁸C₁₈ = \frac{60!}{21!39!} × \frac{39!}{21!18!} × \frac{18!}{18!0!} = 4.979044478499338 × 10²⁶

8 0
2 years ago
The distance between city A and city B is 22 miles. The distance between city B and city C is 54 miles. The distance between cit
Cloud [144]
54 is the hypotenuse because it is the longest side. Square all of the sides. So 54^2=2916
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Step-by-step explanation:

L(w) = length of beard as a function of time in weeks

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the answer would be : 2,942

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