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monitta
3 years ago
6

On the last 4 math assignments, Andrew scored the following:

Mathematics
2 answers:
iVinArrow [24]3 years ago
7 0
The correct answer would be 90. Im very good with finding medians.
kap26 [50]3 years ago
3 0

Answer:

median is 90

Step-by-step explanation:

75, 88, 92, 100

(88+92)/2 = <u>90</u>

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Use pictures to explain how to use the sum 13 to show how to add in any order
sukhopar [10]
$$$$$$$$$+$$$$=13 $ this is one way
6 0
3 years ago
-20y + 15 = 2 - 16y + 11
shtirl [24]
In order to answer this, we will first need to combine like terms on each side. on the left, you can leave them alone. however on the right, we will need to combine 2 and 11. this is 13. the right side becomes 13-16y. after that, we can add 20y to both sides. that equals 15=13+4y now we can subtract 13 from both sides. 2=4y. then we divide by 4 on both sides to find y. y=.5
7 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
Need help ASAP Thankyou
sineoko [7]

Answer:

Volume of the tank = 636 ft³

Step-by-step explanation:

Formula for the volume of a cylinder is,

V = πr²h

Where r = radius of the cylinder

h = height of the cylinder

By substituting the values of radius and height of the water tank given in the question,

V = \pi (4.5)^{2}(10)

   = 202.5π

   = 202.5 × (3.14)

   = 635.85 ft³

   ≈ 636 ft³

Therefore, volume of the cylindrical water tank is 636 ft³.

4 0
3 years ago
Shelly and Terrence earned points in a game by completing various tasks. Shelly completed x tasks and scored 90 points on each o
jeyben [28]
B the number of tasks Terrence completed
7 0
3 years ago
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