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3 years ago

15

Please help me with this

2 answers:

AfilCa [17]3 years ago

6 0

20,000 + 70 + 3 = 20,073

Troyanec [42]3 years ago

6 0

To solve this problem, you need to understand place value concepts. Each digit added makes a number greater, but when adding numbers you have to make sure that you add your other digits in the right places.

To solve 20,000 + 70 + 3, first we can add 70 + 3. This may seem like a simple problem, but it is a good example of place values. Because 3 is in the ones place, you add it to the ones place of the other number (currently a 0). 3 + 0 = 3, so you replace the digit 0 with a 3, giving you the answer 73.

Now, we must solve 20,000 + 73 in a similar manner. Because 73 takes up the tens and the ones places, we should add these numbers to the preexisting digits in those values. Because they are zeros, your answer will still be 73, making your final answer 20,073.

Hope this helps!

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Tickets to a local movie were sold at $12.00 for general admission and $9.00 for seniors. If 155 tickets were sold for a total o

kondaur [170]

We have a problem of a sysmtem of equation

x is the number tickets sold of general admission

y is the number o tickets sold for seniors

The first equation is about the number of ticktes sold

x+y=155

the second equation is about the amount of money

12x+9y=1680

we isolate x of the first equation

x=155-y

we substitute the equatio above in the second equation

12(155-y)+9y=1680

1860-12y+9y=1680

we isolate the y

-3y=1680-1860

-3y=-180

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y=60

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x=155-y

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7 0

1 year ago

2. Four pieces of candy are drawn at random from a bag containing five orange pieces and seven brown pieces.

alexandr402 [8]

Answer:

a) $12C4 = \frac{12!}{4! (12-4)!}= \frac{12!}{4! 8!}=\frac{12*11*10*9*8!}{4! 8!}= 495$ ways

b) $5C2 = \frac{5!}{2! (5-2)!}= \frac{5!}{2! 3!}=\frac{5*4*3!}{2! 3!}= 10$ ways

c) $7C2 = \frac{7!}{2! (7-2)!}= \frac{7!}{2! 5!}=\frac{7*6*5!}{2! 5!}= 21$ ways

Step-by-step explanation:

Combinatory means combination or arrangement of different elements.

If we have n total elements and we want to find in how many ways we can select x we can use this general formula:

$nCx= \frac{n!}{x! (n-x)!}$

Where $n! = n *(n-1)!$

a. How many different ways can four pieces be selected from the 12 colored pieces?

For this case we have:

$12C4 = \frac{12!}{4! (12-4)!}= \frac{12!}{4! 8!}=\frac{12*11*10*9*8!}{4! 8!}= 495$ ways

b. How many different ways can two orange pieces be selected from five orange pieces?

For this case we have:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5!}{2! 3!}=\frac{5*4*3!}{2! 3!}= 10$ ways

c. How many different ways can two brown pieces be selected from seven brown pieces?

For this case we have:

$7C2 = \frac{7!}{2! (7-2)!}= \frac{7!}{2! 5!}=\frac{7*6*5!}{2! 5!}= 21$ ways

6 0

4 years ago

Please help i will give a brainlist if you get it right

Dafna1 [17]

C. 69.08 inches thanks and have a great day !

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4 years ago