-PLEASE HELP ME-
2 answers:
Answer:
They will both be consuming 2000 calories in 6 days.
Step-by-step explanation:
First you need to make an expression for each animal. x will represent the number of days.
Simba
Cuddles
These represent the number of calories they will be consuming on any given day. Since you want to know when they will be the same, you set them equal to each other, and solve for x.
subtract 1220 from both sides
add 180x to both sides
divide both sides by 310 to get x by itself
to find out how many calories they will be eating on day 6, simply plug 6 in to either expression, since they will be the same.
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Hiii :))
Answer is in the attachment.
Identity used :-
1) cosec² α = 1 + cot² α
Trigonometric Ratio :-
1) cot 30° = √3
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Reddam is an expensive school, right?
RainbowSalt2222 ☔
First let's find the number of the elements in the sample space, that is the total number of codes that can be produced.
The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one.
and so on...
so in total there are 10*9*8*7 = 5040 codes.
a. What is the probability that the lock code will begin with 5?
Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.
Thus, there are 1*9*8*7=504 codes which start with 5.
so
P(first digit is five)=
b. <span>What is the probability that the lock code will not contain the number 0?
from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}
from which we can form in total 9*8*7*6 codes which do not contain 0.
P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6
Answer:
0.1 ; 0.6</span>
The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one.
and so on...
so in total there are 10*9*8*7 = 5040 codes.
a. What is the probability that the lock code will begin with 5?
Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.
Thus, there are 1*9*8*7=504 codes which start with 5.
so
P(first digit is five)=
b. <span>What is the probability that the lock code will not contain the number 0?
from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}
from which we can form in total 9*8*7*6 codes which do not contain 0.
P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6
Answer:
0.1 ; 0.6</span>