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pentagon [3]
3 years ago
15

How many different combinations are possible if each lock contains the numbers 0 to 39, and each combination contains three dist

inct numbers?(Combinations like 15-15-15 and 12-12-5 are not used.)
Mathematics
1 answer:
Georgia [21]3 years ago
8 0
(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 9<span>5</span>
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21. If the area of the plane shape below is<br>36cm, find its height.​
mariarad [96]

Answer:

4 cm

Step-by-step explanation:

The figure is a parallelogram with area (A) calculated as

A = bh ( b is the base and h the perpendicular height )

Here A = 36 and b = 9 , thus

9h = 36 ( divide both sides by 9 )

h = 4 cm

4 0
3 years ago
Read 2 more answers
Can someone smart please help me?
ollegr [7]
Hello! 

For a:

How I would do this, is I would first say if all 46 animals (heads) were chickens, how many legs would there be? Each chicken has 2 legs, so 46 * 2 = 92. The total amount of legs is 96 as stated in the question, so if all of the animals were chickens, the farmer would be 4 legs short.

Now to add rabbits into the equation. Rabbits have 4 legs, and chickens have 2. You want to find the difference between the two, because as you add rabbits to the animals the farmer has, then you have to take away chickens at the same time. 4-2 = 2, so for each rabbit you replace, you add 2 legs. 

Since the farmer is 4 legs short with all chickens, then you just divide that 4 by the 2 legs you add by replacing a chicken with a rabbit.

4 / 2 = 2 rabbits

So that means there are 2 rabbits. Since there are 46 heads in total, if 2 are rabbits, that means there are 44 chickens.

So there are 44 chickens and 2 rabbits.

b)

You can follow the same steps: I'm assuming all are child tickets for now:

3.05 * 100 = $305

And now you find how much money short you are.

498.6 - 305 = 193.6

Next, you find the difference in the ticket costs.

5.25 - 3.05 = 2.20

And you divide to find the number of adult tickets.

193.6 / 2.2 = 88

Since 100 tickets were sold, and 88 adult tickets were sold, that means 12 child tickets were sold.
7 0
3 years ago
True or False 68oz &lt; 4lb 6oz
brilliants [131]

Answer:true!!!

Step-by-step explanation:

6 0
3 years ago
???????????anybody helppp
fredd [130]

Answer:

B) 24 p-35

Step-by-step explanation:

<u>Step :1</u>

<u>A</u>pply distributive property  a.(b+c) = a.b+a.c

Given data 1+4(6 p-9)

                                 = 1+4.6 p - 4.9

multiply

                                 = 1+ 24 p - 36

subtracting

                                 = 24 p - 35

5 0
3 years ago
Solving the whole thing
erik [133]
Let
x---------------> distance from people living to the city center

we Know that
Zone 1 covers people living within three miles of the city center
Zone 1 ------------> [x < 3 miles]

Zone 2 covers those between three and seven miles from the center
Zone 2 ------------> [ 3 <= x < = 7  miles]

Zone 3 covers those over seven miles from the center
Zone 3 ------------> [  x > 7  miles]

<span>calculate the distance between two points to find the value of x
</span>
case A) point (0,0)  point (3,4)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(4-0)² +(3-0)²]------> √[16+9]
x=√25-------------> x=5 miles

the answer Part A)
people living in (3,4)
x=5 miles -------------> covers Zone 2 [ 3 < =x <= 7  miles]


case B) point (0,0)  point (6,5)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(5-0)² +(6-0)²]------> √[25+36]
x=√61-------------> x=7.81 miles

the answer Part B)
people living in (6,5) 
x=7.81 miles -------------> covers Zone 3 [  x > 7  miles]

case C) point (0,0)  point (1,2)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(2-0)² +(1-0)²]------> √[4+1]
x=√5-------------> x=2.23 miles

the answer Part C)
people living in (1,2) 
x=2.23 miles -------------> covers Zone 1 [ x < 3  miles]

case D) point (0,0)  point (0,3)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(3-0)² +(0-0)²]------> √[9]
x=√9-------------> x=3 miles

the answer Part D)
people living in (0,3) 
x=3 miles -------------> covers Zone 2 [ 3 < =x <= 7  miles]

case E) point (0,0)  point (1,6)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(6-0)² +(1-0)²]------> √[36+1]
x=√37-------------> x=6.08 miles

the answer Part E)
people living in (1,6) 
x=6.08 miles -------------> covers Zone 2 [ 3 < = x <= 7  miles]

4 0
4 years ago
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