ANSWER:
n = 1.05n divided by 105 times 100
OR
n = (the number of students at the END of the school year) divided by 105 times 100
WORKING OUT:
1.05n = 651 students
1n = 651/105 times 100 = ANSWER
I believe it is 3.75 miles. 5 2/5=5.4 and 1 11/25=1.44. 5.4/1.44=3.75.
Answer:
2n +10= 140
4n- 40= 80
2n= 3n - 20
Step-by-step explanation:
it is the am progression or sum progression
Answers:
- C) x = plus/minus 11
- B) No real solutions
- C) Two solutions
- A) One solution
- The value <u> 18 </u> goes in the first blank. The value <u> 17 </u> goes in the second blank.
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Explanations:
- Note how (11)^2 = (11)*(11) = 121 and also (-11)^2 = (-11)*(-11) = 121. The two negatives multiply to a positive. So that's why the solution is x = plus/minus 11. The plus minus breaks down into the two equations x = 11 or x = -11.
- There are no real solutions here because the left hand side can never be negative, no matter what real number you pick for x. As mentioned in problem 1, squaring -11 leads to a positive number 121. The same idea applies here as well.
- The two solutions are x = 0 and x = -2. We set each factor equal to zero through the zero product property. Then we solve each equation for x. The x+2 = 0 leads to x = -2.
- We use the zero product property here as well. We have a repeated factor, so we're only solving one equation and that is x-3 = 0 which leads to x = 3. The only root is x = 3.
- Apply the FOIL rule on (x+1)(x+17) to end up with x^2+17x+1x+17 which simplifies fully to x^2+18x+17. The middle x coefficient is 18, while the constant term is 17.
Answer: Our required probability is 
Step-by-step explanation:
Since we have given that
Number of coins = 3
Number of coin has 2 heads = 1
Number of fair coins = 2
Probability of getting one of the coin among 3 = 
So, Probability of getting head from fair coin =
Probability of getting head from baised coin = 1
Using "Bayes theorem" we will find the probability that it is the two headed coin is given by
Hence, our required probability is
No, the answer is not
