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2 years ago

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A basket contains the following pieces of fruit: 3 apples, 4 oranges, 4 bananas, 2 pears, and 3 peaches. Jonas picks a fruit at

random and does not replace it. Then Beth picks a fruit at random. What is the probability that Jonas gets an orange and Beth gets a peach?

1 answer:

yuradex [85]2 years ago

8 0

Answer:

Probability that Jonas gets an orange is 1/4 and Beth gets a peach is 3/16.

Step-by-step explanation:

Here,

The total number of fruits n(S) = 3 + 4 + 4 + 2 + 3

= 16

The total number of oranges n(O) = 4

The total number of peas n(P) = 3

Now,

P(O) = n(O)/n(S)

= 4/16

= 1/4

Again,

P(P)= n(P)/n(S)

= 3/16

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Solve the system of linear equations by eliminations x+5y=-2; 5x+y=14

Arada [10]

I’ve attached my work….
Hope it helps!

8 0

3 years ago

A stadium has 50,000 seats. Seats sell for $28 in Section A, $24 in Section B, and $20 in Section C. The number of seats in Sect

Pepsi [2]

Answer:

Section A holds 25,000 seats, section B holds 14,500 seats and section C holds 10,500 seats.

Step-by-step explanation:

From the information given you can write the following equations:

A+B+C=50,000 (1)

28A+24B+20C=1,258,000 (2)

A=B+C (3)

A= number of seats in section A

B= number of seats in section B

C= number of seats in section C

You can replace (3) in (1) and (2) to get two equations:

B+C+B+C=50,000

2B+2C=50000

28(B+C)+24B+20C=1,258,000

28B+24B+28C+20C=1,258,000

52B+48C=1,258,000

The two equations are:

2B+2C=50000 (4)

52B+48C=1,258,000 (5)

You can isolate B in (4):

2B=50,000-2C

B=(50,000/2)-(2C/2)

B=25,000-C

Now, you can replace B in (5):

52(25,000-C)+48C=1,258,000

1,300,000-52C+48C=1,258,000

1,300,000-1,258,000=4C

42,000=4C

C=42,000/4

C=10,500

Now, you can replace the value of C in B=25,000-C:

B=25,000-10,500

B=14,500

Finally, you can replace the values of B and C in A=B+C to find A:

A=14,500+10,500

A=25,000

According to this, the answer is that section A holds 25,000 seats, section B holds 14,500 and section C holds 10,500.

6 0

4 years ago

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

$2\cdot a \cdot c \ge 3$

$a\cdot c \ge \frac{3}{2}$

But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Example: $a = 2$ , $c = 2$

$b = \sqrt{1 + 2\cdot (2)\cdot (2)}$

$b = 3$

4 0

3 years ago

A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting

Soloha48 [4]

Answer:

Therefore the only statement that is not true is b.)

Step-by-step explanation:

There employees are 6 secretaries, 5 consultants and 4 partners in the firm.

a.) The probability that a secretary wins in the first draw

= $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees} = \frac{6}{15}$

b.) The probability that a secretary wins a ticket on second draw. It has been given that a ticket was won on the first draw by a consultant.

p(secretary wins on second draw | consultant wins on first draw)

= $\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}$

= $\frac{\frac{5}{15} \times \frac{6}{14}}{\frac{5}{15} } = \frac{6}{14}$ .

The probability that a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw = $\frac{6}{15}$ is not true.

The probability that a secretary wins on the second draw = $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secretaries \hspace{0.1cm} remaining } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} remaining} = \frac{6 - 1}{15 - 1} = \frac{5}{14}$

c.) The probability that a consultant wins on the first draw =

$\frac{number \hspace{0.1cm} of \hspace{0.1cm} consultants \hspace{0.1cm} } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} } = \frac{5 }{15} = \frac{1}{3}$

d.) The probability of two secretaries winning both tickets

= (probability of a secretary winning in the first draw) × (The probability that a secretary wins on the second draw)

= $\frac{6}{15} \times \frac{5}{14} = \frac{1}{7}$

Therefore the only statement that is not true is b.)

5 0

3 years ago

The radius of a sphere is 6 units.

tino4ka555 [31]

Answer: The radius of a sphere is 6 units

Step-by-step explanation:

3 0

4 years ago