Which events are mutually exclusive?Jon eats 2 apples; Jon eats 4 apples.
Jon eats 4 apples; Jon eats 1 apple.
Jon eats more than 1 apple; Jon eats 3 apples.
Jon eats 2 apples; Jon eats more than 2 apples.
Answers
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because 1 dime=10¢ and so 21 x .10 = 2.10 and 1 nickel=5¢ so .05 x 4 = .20
2.10 + .20 = 2.30
I’m not sure if the explaining made sense but it’s 4 nickels :)
Answers
Answer:
36 in.
Step-by-step explanation:
Step 1:
Area = Length × Width Equation
Step 2:
6 × 6 Multiply
Answer:
36 inches
Hope This Helps :)
Answers
Answer:
73
Step-by-step explanation:
Answers
Final answer:
The given problem is a type of recursive sequence in mathematics. We can find any term in the sequence by taking four times the negative of the previous term and adding 15. For example, the second term is -13, calculated from the given first term of 7 as -4*7+15.
Explanation:
The problem given states that F(1)=7, f(n)=-4*f(n-1)+15, which is a type of recursive sequence in mathematics. In this sequence, each term is defined as four times the negative of the previous term added to 15. For any term f(n), the previous term is f(n-1). Therefore, if we want to find the next term after F(1), which is f(2), we substitute n=2 into the equation, giving us: f(2) = -4*f(2-1)+15 = -4*F(1)+15 = -4*7+15 = -28+15 = -13. Continuing this calculating method allows us to find subsequent terms in the sequence.
Learn more about Recursive Sequence here:
#SPJ2
Answers
Answer:
35
Step-by-step explanation:
To find this, we are going to use the combination formula as follows;
nCr = n!/(n-r)!r!
In this case, n = 7 and r = 3
Substituting these values;
7C3 = 7!/(7-3)! 3! = 7!/4!3! = 35
Answers
Since, each row has one more log than the row above it
so, this is arithematic sequence
We are given that
First row is
so,
Last row is
so,
Each row has one more log than the row above it
so,
now, we can find number of rows
we can plug values
we can solve for n
now, we can find total number of logs
now, we can plug values
So,
Number of logs in the pile are 119........Answer
The log in the pile is an illustration of arithmetic progression.
The number of logs in the pile is 119.
The first term of the progression is:
The last term is:
The common difference is:
First, we calculate the number of terms using:
So, we have:
Subtract 14 from both sides
Add 1 to both sides
The number of logs in the pile is calculated using the sum of n terms of an AP formula:
So, we have:
Hence, the number of logs in the pile is 119.
Read more about arithmetic progressions at: