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3 years ago

PLEASE HELP ME ASAP!!!! ILL MARK BRIANLIEST AND GIVE EXTRA POINTS

Mathematics

2 answers:

vivado [14]3 years ago

7 0

Answer:

4. 7/9 (b - 27) > 49/81

=> 7/9b - 21 > 49/81

=> 7/9b - 21 > 7/9

=> 7/9b -21 + 21 > 7/9 + 21

=> 7/9b > 7/9 + 189/9

=> 7/9b > 196/9

=> 7/9b / 7/9 > 196/9 / 7/9

=> To divide fractions, take the reciprocal of the divisor fraction and then multiply.

=> b > 196/9 x 9/7

=> b > 1764 / 63

=> b > 28

4th answer is b > 28

5. This question doesn't make any sense.

=> 11w - 8w greater or equal to 14w

=> 3w greater or equal to 14w.

If we make "w" as any number, this equation will not come true as 14 is larger than 3.

This answer may be correct the equation is like:

3w is smaller or equal to 14w.

kumpel [21]3 years ago

3 0

Answer:

b > 250/9 and w ≥ 0

Step-by-step explanation:

7/9 (b - 27) > 49/81

First, simplify both sides of the inequality:

7/9b - 21 > 49/81

Add 21 to both sides

7/9b - 21 +21 > 49/81 +21

Multiply both sides by 9/7

9/7 x 7/9b > 49/81 x 9/7

b > 1750/81

Simplify

b > 250/9

11w - 8w ≥ 14w

Simplify both sides once again

3w ≥ 14w

Subtract 14w from both sides

3w - 14w ≥ 14w - 14w

-11w ≥ 0

Divide both sides by -11w

w ≥ 0

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Bob and Anna are planning to meet for lunch at Sally's Restaurant, but they forgot to schedule a time. Bob and Anna are each goi

Viktor [21]

Answer:

The expected cost is $8.75

Step-by-step explanation:

Given

$Time = \{1pm, 2pm, 3pm, 4pm\}$

$C_1 = \$5$ --- If Bob and Anna meet

$C_2 = \$10$ --- If Bob and Anna do not meet

Required

The expected cost of Bob's meal

First, we list out all possible time both Bob and Anna can select

We have:

$(Bob,Anna) = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3)$ $,(3,4),(4,1),(4,2),(4,3),(4,4)\}$

$n(Bob, Anna) = 16$

The outcome of them meeting at the same time is:

$Same\ Time = \{(1,1),(2,2),(3,3),(4,4)\}$

$n(Same\ Time) = 4$

The probability of them meeting at the same time is:

$Pr(Same\ Time) = \frac{n(Same\ Time)}{n(Bob,Anna)}$

$Pr(Same\ Time) = \frac{4}{16}$

$Pr(Same\ Time) = \frac{1}{4}$

The outcome of them not meeting:

$Different = \(Bob,Anna) = \{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2)$

$,(3,4),(4,1),(4,2),(4,3)\}$

$n(Different) = 12$

The probability of them meeting at the same time is:

$Pr(Different) = \frac{n(Different)}{n(Bob,Anna)}$

$Pr(Different) = \frac{12}{16}$

$Pr(Different) = \frac{3}{4}$

The expected cost is then calculated as:

$Expected = C_1 * P(Same) + C_2 * P(Different)$

$Expected = \$5 * \frac{1}{4} + \$10 * \frac{3}{4}$

$Expected = \frac{\$5}{4} + \frac{\$30}{4}$

Take LCM

$Expected = \frac{\$5+\$30}{4}$

$Expected = \frac{\$35}{4}$

$Expected = \$8.75$

The expected cost is $8.75

4 0

3 years ago

Simplify. x^2-3x-18/x+3

andre [41]

Answer:

1. Option C is correct

2. Option A is correct

3. Option C is correct

4. Option B is correct

5. Option D is correct

Step-by-step explanation:

1. x^2-3x-18/x+3

Factorize the numerator

x^2-6x+3x-18/x+3

x(x-6)+3(x-6)/x+3

(x+3)(x-6)/x+3

x-6

x-6 where x≠6

Option C is correct.

2. x-2/x^2+4x-12

Factorizing the denominator

x-2/x^2+6x-2x-12

x-2/x(x+6)-2(x+6)

x-2/(x-2)(x+6)

1/x+6

1/x+6 where x≠-6

Option A is correct.

3. 5x^3/7 x^3+x^4

5x^3/7x^3+x^4

5x^3/x^3(7+x)

5/7+x

Option C is correct

5/7+x where x≠-7

4. Simplify x/6x-x^2

x/6x-x^2

x/x(6-x)

1/6-x

Option B is correct

1/6-x where x≠6

5. 2/3a * 2/a^2

Multiplying both terms

4/3a^3

Option D is correct.

4/3a^3 where a≠0

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3 years ago

−50/54−40%+1.65

irga5000 [103]

The answer is - 0.53

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4 years ago

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Margaret [11]

Hello

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