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ryzh [129]
3 years ago
9

5 divided by a =60. Solve a

Mathematics
1 answer:
Mandarinka [93]3 years ago
4 0

Answer:

a = 300

Step-by-step explanation:

so 5/a=60

multiply 60 by 5

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3 years ago
I am in fourth grade and I have no idea how to find whole numbers between a fraction can you show me?
allsm [11]

Answer:

To find a a fraction of a whole number, you multiply the numerator of the fraction by the given number and then divide the product by the denominator of the fraction

Solved example for finding a fraction of a whole number:

1)

Find 1/3 of 21.

To find 1/3 of 21, we multiply the numerator 1 by the given whole number 21 and then divide the product 21 by the denominator 3.

1/3 × 21 = 1 × 21/3 = 21/3 = 7

So, 1/3 of 21 = 7.

1/3 of 21 is 7

5 0
3 years ago
Read 2 more answers
Find the cost of this trip: Find the airfare for flying round trip from Denver to Detroit if a one-way ticket is $235.00, there
Rzqust [24]

5% of 235 is: 235*(5/100)= 235/20 = 11,75 or 5% of 470 is: 470/20 = 23,5

Total cost = 2*(235 + 11,75) = 2*246,75 = 493,5 or Total cost = 470 + 23,5 = 493,5

6 0
3 years ago
Read 2 more answers
Amy has a box containing 6 white, 4 red, and 8 black marbles. She picks a marble randomly. It is red. The second time, she picks
Sladkaya [172]
After the 3rd pick we have: 4 white, 3 red and 8 black marbles.
The probability that the 4th marble is black is:
P ( Black ) = 8 / 15 = 0.5333
After the 4th pick we have: 4 white, 3 red and 7 black marbles.
P ( Red or Black ) = 10 / 14 = 5 / 7 = 0.7143
6 0
3 years ago
Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-ma
pickupchik [31]

Answer:

Answer explained below

Step-by-step explanation:

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds

1.START T=0.....

1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.

T=1.....

EACH OF THE M/C SENDS 10 NEW MAILS ....

.................................

LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.

SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.

HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=

M[N+1]=10*M[N].......................1

THIS IS THE RECURRENCE RELATION.....

2.SOLUTION .....

M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........

M(N+1)=[10^1][M(N)]=[10^2][M(N-1)]=[10^3][M(N-2)]=..........=[10^N][M(1)]=[10^(N+1)][M(0)]

M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2

THIS IS THE NUMBER OF MAILS SENT AFTER N+1 SECONDS .....OR ....

M[N]=[10^(N+3)].............................................3

..................IS THE SOLUTION FOR NUMBER OF MAILS SENT AFTER N SECONDS.....

3.AFTER N=20 SECONDS , THE ANSWER IS ....

M[20]=10^(20+3)=10^23

8 0
3 years ago
Read 2 more answers
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