Antoinette solves the linear equation 3(x-3)+2x+9 = 2X+ 2(x-1) using the steps shown below.
2 answers:
Answer:
Antoinette's error
Option A) She distributed incorrectly.
Step-by-step explanation:
We are given the following information in the question:
Antoinette solves the linear equation
Antoinette made mistake in step 1 that is she did not distributed correctly.
If she would have distributed correctly, then she would have got the following solution:
Substituting x = -1 in the given equation, we gwt:
Antoinette's error
Option A) She distributed incorrectly.
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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>