Susan spent a total of 45$ at the grocery store I’d this amount she spent $18 on fruit what percentage of the total did she spen
1 answer:
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First, we need to work out the total number of students who were being surveyed.
We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population
That means total number of students being surveyed is 13+13=26 students
Then we work out the probability
P(One pet) = 8/26 = 4/13
P(Two pets) = 1/2
P(Three pets) = 3/26
P( Four pets) = 2/26 = 1/13
The probability distribution is shown in the table below. Let
be the number of pets and 
is the probability of owning the number of pets
We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population
That means total number of students being surveyed is 13+13=26 students
Then we work out the probability
P(One pet) = 8/26 = 4/13
P(Two pets) = 1/2
P(Three pets) = 3/26
P( Four pets) = 2/26 = 1/13
The probability distribution is shown in the table below. Let
Answer:
A(t) = 200+15t(1+0.02)^{t}
Step-by-step explanation:
Since the interest is calculated on the new balance every year.
Hence the formula used for compound interest is:
A = P(1+^{nt}
where, A =Amount after t years
P =Principal amount
200 is the initial balance and Since, here the $15 is added to the balance each year. Therefore, P = 200+15t
r = rate each year (0.02)
t = time (in years) (t)
n = no. of times the interest is compounded in a year (n=1)
Therefore, the recursive formula is:
A(t) = 200+15t(1+0.02)^{t}
Remark
This question likely should be done before the other one. What you are trying to do is give C a value. So you need to remember that C is always part of an indefinite integral.
y =
y = sin(x) - cos(x) + C
y(π) = sin(π) - cos(π) + C = 0
y(π) = 0 -(-1) + C = 0
y(π) = 1 + C = 0
C = - 1
y = sin(x) - cos(x) - 1 <<<<< Answer
Problem Two
Remember that
y( - e^3 ) = ln(|x|) + C = 0
y(-e^3) = ln(|-e^3|) + C = 0
y(-e^3) = 3 + C = 0
3 + C = 0
C = - 3
y = ln(|x|) - 3 <<<< Answer
This question likely should be done before the other one. What you are trying to do is give C a value. So you need to remember that C is always part of an indefinite integral.
y =
y = sin(x) - cos(x) + C
y(π) = sin(π) - cos(π) + C = 0
y(π) = 0 -(-1) + C = 0
y(π) = 1 + C = 0
C = - 1
y = sin(x) - cos(x) - 1 <<<<< Answer
Problem Two
Remember that
y( - e^3 ) = ln(|x|) + C = 0
y(-e^3) = ln(|-e^3|) + C = 0
y(-e^3) = 3 + C = 0
3 + C = 0
C = - 3
y = ln(|x|) - 3 <<<< Answer