B - a b equals 2 1/3 a equals 1 1/2

Which of the following statements is false. the sum of two rational numbers is always rational.

djyliett [7]

Answer:

true

Step-by-step explanation:

6 0

3 years ago

Directions: Using the digits 1 to 9, at most one time each, fill in the boxes to find the largest or smallest possible values fo

12345 [234]

9514 1404 393

Answer:

smallest: 8x -3 = 4; 1y +9 = 2. total = -49/8

largest: 1x -9 = 8; 2y +3 = 7. total = 19

Step-by-step explanation:

If we use variables to represent the box contents, we can write ...

ax -b = c
dy +e = f

Then the values of x and y are ...

x = (c +b)/a

y = (f -e)/d

For positive integer values of the variables, x will always be positive, and y may or may not be negative.

<h3>Smallest sum</h3>

For the sum to be the smallest, we must have x be as small as possible and the ratio (f-e)/d be as negative as possible.

x will be small for large 'a' and for (c+b) small. For y to be as negative as possible, we want 'd' and 'f' small and 'e' large. Best results are obtained for

8x -3 = 4 ⇒ x = 7/8
1y +9 = 2 ⇒ y = -7

For these coefficients, the sum is -6 1/8 = -49/8.

(note that the values of 'b' and 'c' can be swapped with no net effect)

<h3>Largest sum</h3>

For the sum to be the largest, we must have x as large as possible: (b+c) large and 'a' small. At the same time we must have y be positive and as large as possible: (f-e) positive and large, 'd' small. Best results are obtained for

1x -9 = 8 ⇒ x = 17
2y +3 = 7 ⇒ y = 2

For these coefficients, the sum is 19. Again, 'b' and 'c' can be swapped with no effect.

_____

<em>Additional comment</em>

These extreme values are verified by examination of the 60,480 possible permutations of the coefficients.

8 0

3 years ago

Read 2 more answers

Josie sold 790 tickets to a local car show for a total of $4,390.00. A ticket for a Child costs $4.00 and adult ticket costs $7.

Papessa [141]

Let's define the following variables first.

A = number of tickets sold for adults

C = number of tickets sold for children

From the question, we can say that or form the following equations:

1. A + C = 790 tickets

2. $7A + $4C = $4, 390

The first equation can also be written as A = 790 - C. We can use this equation and replace "A" in the second equation.

$7(790-C)+4c=4,390$

From that, we can solve "C" by solving the equation formed above.

$\begin{gathered} \text{Distribute 7 to the terms inside the parenthesis.} \\ 7(790)-7(C)+4C=4,390 \\ 5,530-7C+4C=4,390 \\ \text{Subtract 5,530 on both sides of the equation.} \\ -7C+4C=-1140 \\ -3C=-1140 \\ \text{Divide -3 on both sides.} \\ C=380 \end{gathered}$

Therefore, 380 tickets for children were sold.

Since there are 790 tickets in total that are sold and 380 tickets for children were sold, we can say that 410 tickets for adult was sold.

$\begin{gathered} A=790-C \\ A=790-380 \\ A=410 \end{gathered}$

3 0

9 months ago

An unbiased coin is tossed 15 times. In how many ways can the coin land tails either exactly 8 times orexactly 5 times?

Kitty [74]

Answer:

Probability to get tails exactly 8 times or exactly 5 times is 0.29

Step-by-step explanation:

No of ways the coins lands tails exactly 8 times

P(8) = 15C8 × $0.5^{8}$ × $0.5^{15-8}$

No of ways the coin lands tails exactly 5 times

P(5) = 15C5 × $0.5^{5}$ × $0.5^{15-5}$

Probability to get tails exactly 8 times or 5 times

P(8)+P(5) = 15C8 × $0.5^{15}$ + 15C5 × $0.5^{15}$

P = $0.5^{15}$ ( 15C8 + 15C5 )

P = $0.5^{15}$ ( ( $\frac{15!}{8!(15-8)!}$ ) + ( $\frac{15!}{5!(15-5)!}$ ) )

P = $0.5^{15}$ ( 6435 + 3003 )

P = $0.5^{15}$ ( 9438)

P = $0.5^{14}$ ( 4719)

P = $\frac{4719}{16384}$

P = 0.29

8 0

3 years ago

A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o

balu736 [363]

Answer:

a

$P(X \ge 1) = 0.509$

b

$P(X \ge 1) = 0.6807$

Step-by-step explanation:

From the question we are told that

The number of students in the class is N = 20 (This is the population )

The number of student that will cheat is k = 3

The number of students that he is focused on is n = 4

Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally probability mass function is mathematically represented as

$P(X = x) = \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}$

Here C stands for combination , hence we will be making use of the combination functionality in our calculators

Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

Here

$P(X \le 0) = \frac{ ^{3} C_0 * ^{20 - 3} C_{4- 0}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ ^{3} C_0 * ^{17} C_{4}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ 1 * 2380}{ 4845}$

$P(X \le 0) = 0.491$

Hence

$P(X \ge 1) = 1 - 0.491$

$P(X \ge 1) = 0.509$

Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

$P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]$