We appreciate your visit to 48 What is the probability that a number selected at random from the first 100 prime numbers is even A tex frac 99 100 tex. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Below is a detailed solution for each part of the question.
–––––––––––––––
Problem 48
We are asked for the probability that a number selected at random from the first 100 prime numbers is even. Notice that among prime numbers only one even prime exists, namely 2. Since there is only 1 even prime among the 100 primes, the probability is
[tex]$$
\text{Probability} = \frac{1}{100} = 0.01.
$$[/tex]
–––––––––––––––
Problem 49
A committee consists of 6 men and 4 women (a total of 10 members). Two members are chosen at random. We need the probability that both selected members are women.
1. First, calculate the total number of ways to choose 2 out of 10:
[tex]$$
\binom{10}{2} = 45.
$$[/tex]
2. Next, calculate the number of ways to choose 2 women out of 4:
[tex]$$
\binom{4}{2} = 6.
$$[/tex]
3. Therefore, the probability that both members selected are women is
[tex]$$
\text{Probability} = \frac{6}{45} = \frac{2}{15}.
$$[/tex]
–––––––––––––––
Problem 50
Let the current ages of the husband, wife, and child be [tex]$H$[/tex], [tex]$W$[/tex], and [tex]$C$[/tex], respectively.
Three years ago the average age was 27. Hence, the sum of their ages three years ago was:
[tex]$$
3 \times 27 = 81.
$$[/tex]
This gives the equation:
[tex]$$
(H-3) + (W-3) + (C-3) = 81 \quad \Longrightarrow \quad H + W + C = 90.
$$[/tex]
Also, five years ago the average age of the wife and child was 20. So, the sum of their ages was:
[tex]$$
2 \times 20 = 40.
$$[/tex]
That is,
[tex]$$
(W-5) + (C-5) = 40 \quad \Longrightarrow \quad W + C = 50.
$$[/tex]
Subtracting, we find the husband’s age:
[tex]$$
H = 90 - (W+C) = 90 - 50 = 40.
$$[/tex]
Thus, the present age of the husband is 40 years.
–––––––––––––––
Problem 51
We need to evaluate the expression:
[tex]$$
287^2 + 269^2 - 2(287 \times 269).
$$[/tex]
Notice that the expression resembles the expansion of a square:
[tex]$$
(a-b)^2 = a^2 + b^2 - 2ab.
$$[/tex]
Let [tex]$a = 287$[/tex] and [tex]$b = 269$[/tex]. Then,
[tex]$$
(287-269)^2 = (18)^2 = 324.
$$[/tex]
Thus, the value of the expression is 324.
–––––––––––––––
Problem 52
A piece of machinery has a current value of Birr 100,000, and it loses 25% of its value each year. This means that after each year it is worth 75% of its previous value. After 4 years, its value will be:
[tex]$$
\text{Value after 4 years} = 100\,000 \times (0.75)^4.
$$[/tex]
Calculating [tex]$(0.75)^4$[/tex] gives approximately 0.31640625, so
[tex]$$
\text{Value after 4 years} \approx 100\,000 \times 0.31640625 = 31\,640.625.
$$[/tex]
–––––––––––––––
Problem 53
We are given that [tex]$x$[/tex] and [tex]$y$[/tex] are integers satisfying:
[tex]$$
3x+9 < 0 \quad \text{and} \quad -4y-16 > 0.
$$[/tex]
Solve the first inequality:
[tex]\[
3x+9 < 0 \quad \Longrightarrow \quad 3x < -9 \quad \Longrightarrow \quad x < -3.
\][/tex]
Thus, [tex]$x$[/tex] is a negative integer less than [tex]$-3$[/tex].
Now solve the second inequality:
[tex]\[
-4y-16 > 0 \quad \Longrightarrow \quad -4y > 16 \quad \Longrightarrow \quad y < -4,
\][/tex]
after dividing by [tex]$-4$[/tex] (remembering to reverse the inequality). So, [tex]$y$[/tex] is also a negative integer.
Since both [tex]$x$[/tex] and [tex]$y$[/tex] are negative, their product [tex]$xy$[/tex] will be positive (because the product of two negative numbers is positive). Therefore, it is always true that:
[tex]$$
xy \text{ is positive.}
$$[/tex]
–––––––––––––––
Problem 54
We need to compare the following two quantities for real number [tex]$x$[/tex]:
Quantity A:
[tex]$$
\sqrt{4x^2+4x+1},
$$[/tex]
Quantity B:
[tex]$$
2x+1.
$$[/tex]
Notice that the expression under the square root is a perfect square:
[tex]$$
4x^2+4x+1 = (2x+1)^2.
$$[/tex]
Thus, we have:
[tex]$$
\sqrt{4x^2+4x+1} = \sqrt{(2x+1)^2} = |2x+1|.
$$[/tex]
Now, consider the two possible cases for [tex]$2x+1$[/tex]:
1. If [tex]$2x+1 \ge 0$[/tex], then [tex]$|2x+1| = 2x+1$[/tex]. In this case, Quantity A equals Quantity B.
2. If [tex]$2x+1 < 0$[/tex], then [tex]$|2x+1| = -(2x+1)$[/tex], which is a positive number while [tex]$2x+1$[/tex] is negative. Thus, Quantity A is greater than Quantity B.
From both cases, we see that Quantity A is always greater than or equal to Quantity B.
–––––––––––––––
Summary of Answers
48. The probability that a randomly selected prime is even is
[tex]$$\frac{1}{100}.$$[/tex]
49. The probability that both selected members are women is
[tex]$$\frac{2}{15}.$$[/tex]
50. The present age of the husband is 40 years.
51. The expression evaluates to 324.
52. The value of the machinery after 4 years is approximately Birr 31,640.625.
53. For the given conditions, the product [tex]$xy$[/tex] is always positive.
54. Quantity A ([tex]$\sqrt{4x^2+4x+1}$[/tex]) is always greater than or equal to Quantity B ([tex]$2x+1$[/tex]).
These are the step-by-step explanations and the corresponding answers to the questions.
–––––––––––––––
Problem 48
We are asked for the probability that a number selected at random from the first 100 prime numbers is even. Notice that among prime numbers only one even prime exists, namely 2. Since there is only 1 even prime among the 100 primes, the probability is
[tex]$$
\text{Probability} = \frac{1}{100} = 0.01.
$$[/tex]
–––––––––––––––
Problem 49
A committee consists of 6 men and 4 women (a total of 10 members). Two members are chosen at random. We need the probability that both selected members are women.
1. First, calculate the total number of ways to choose 2 out of 10:
[tex]$$
\binom{10}{2} = 45.
$$[/tex]
2. Next, calculate the number of ways to choose 2 women out of 4:
[tex]$$
\binom{4}{2} = 6.
$$[/tex]
3. Therefore, the probability that both members selected are women is
[tex]$$
\text{Probability} = \frac{6}{45} = \frac{2}{15}.
$$[/tex]
–––––––––––––––
Problem 50
Let the current ages of the husband, wife, and child be [tex]$H$[/tex], [tex]$W$[/tex], and [tex]$C$[/tex], respectively.
Three years ago the average age was 27. Hence, the sum of their ages three years ago was:
[tex]$$
3 \times 27 = 81.
$$[/tex]
This gives the equation:
[tex]$$
(H-3) + (W-3) + (C-3) = 81 \quad \Longrightarrow \quad H + W + C = 90.
$$[/tex]
Also, five years ago the average age of the wife and child was 20. So, the sum of their ages was:
[tex]$$
2 \times 20 = 40.
$$[/tex]
That is,
[tex]$$
(W-5) + (C-5) = 40 \quad \Longrightarrow \quad W + C = 50.
$$[/tex]
Subtracting, we find the husband’s age:
[tex]$$
H = 90 - (W+C) = 90 - 50 = 40.
$$[/tex]
Thus, the present age of the husband is 40 years.
–––––––––––––––
Problem 51
We need to evaluate the expression:
[tex]$$
287^2 + 269^2 - 2(287 \times 269).
$$[/tex]
Notice that the expression resembles the expansion of a square:
[tex]$$
(a-b)^2 = a^2 + b^2 - 2ab.
$$[/tex]
Let [tex]$a = 287$[/tex] and [tex]$b = 269$[/tex]. Then,
[tex]$$
(287-269)^2 = (18)^2 = 324.
$$[/tex]
Thus, the value of the expression is 324.
–––––––––––––––
Problem 52
A piece of machinery has a current value of Birr 100,000, and it loses 25% of its value each year. This means that after each year it is worth 75% of its previous value. After 4 years, its value will be:
[tex]$$
\text{Value after 4 years} = 100\,000 \times (0.75)^4.
$$[/tex]
Calculating [tex]$(0.75)^4$[/tex] gives approximately 0.31640625, so
[tex]$$
\text{Value after 4 years} \approx 100\,000 \times 0.31640625 = 31\,640.625.
$$[/tex]
–––––––––––––––
Problem 53
We are given that [tex]$x$[/tex] and [tex]$y$[/tex] are integers satisfying:
[tex]$$
3x+9 < 0 \quad \text{and} \quad -4y-16 > 0.
$$[/tex]
Solve the first inequality:
[tex]\[
3x+9 < 0 \quad \Longrightarrow \quad 3x < -9 \quad \Longrightarrow \quad x < -3.
\][/tex]
Thus, [tex]$x$[/tex] is a negative integer less than [tex]$-3$[/tex].
Now solve the second inequality:
[tex]\[
-4y-16 > 0 \quad \Longrightarrow \quad -4y > 16 \quad \Longrightarrow \quad y < -4,
\][/tex]
after dividing by [tex]$-4$[/tex] (remembering to reverse the inequality). So, [tex]$y$[/tex] is also a negative integer.
Since both [tex]$x$[/tex] and [tex]$y$[/tex] are negative, their product [tex]$xy$[/tex] will be positive (because the product of two negative numbers is positive). Therefore, it is always true that:
[tex]$$
xy \text{ is positive.}
$$[/tex]
–––––––––––––––
Problem 54
We need to compare the following two quantities for real number [tex]$x$[/tex]:
Quantity A:
[tex]$$
\sqrt{4x^2+4x+1},
$$[/tex]
Quantity B:
[tex]$$
2x+1.
$$[/tex]
Notice that the expression under the square root is a perfect square:
[tex]$$
4x^2+4x+1 = (2x+1)^2.
$$[/tex]
Thus, we have:
[tex]$$
\sqrt{4x^2+4x+1} = \sqrt{(2x+1)^2} = |2x+1|.
$$[/tex]
Now, consider the two possible cases for [tex]$2x+1$[/tex]:
1. If [tex]$2x+1 \ge 0$[/tex], then [tex]$|2x+1| = 2x+1$[/tex]. In this case, Quantity A equals Quantity B.
2. If [tex]$2x+1 < 0$[/tex], then [tex]$|2x+1| = -(2x+1)$[/tex], which is a positive number while [tex]$2x+1$[/tex] is negative. Thus, Quantity A is greater than Quantity B.
From both cases, we see that Quantity A is always greater than or equal to Quantity B.
–––––––––––––––
Summary of Answers
48. The probability that a randomly selected prime is even is
[tex]$$\frac{1}{100}.$$[/tex]
49. The probability that both selected members are women is
[tex]$$\frac{2}{15}.$$[/tex]
50. The present age of the husband is 40 years.
51. The expression evaluates to 324.
52. The value of the machinery after 4 years is approximately Birr 31,640.625.
53. For the given conditions, the product [tex]$xy$[/tex] is always positive.
54. Quantity A ([tex]$\sqrt{4x^2+4x+1}$[/tex]) is always greater than or equal to Quantity B ([tex]$2x+1$[/tex]).
These are the step-by-step explanations and the corresponding answers to the questions.
Thanks for taking the time to read 48 What is the probability that a number selected at random from the first 100 prime numbers is even A tex frac 99 100 tex. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
