All categories

2 years ago

If you leave a $10.50 tip for the waiter on a meal that cost $58.25, what percent tip did you leave?

Mathematics

2 answers:

Vinil7 [7]2 years ago

4 0

B.18% 58.25 x .18 = 10.485

7nadin3 [17]2 years ago

3 0

Answer:

Option B is correct

18% tip did you leave for the waiter.

Step-by-step explanation:

As per the statement:

If you leave a $10.50 tip for the waiter on a meal that cost $58.25.

Cost of a tip = $ 10.50

Cost of a meal = $58.25

We have to find the percent tip did you leave.

$\text{Percent of tip} = \frac{\text{Cost of tip}}{\text{Cost of a meal}} \times 100$

Substitute the given values we have;

$\text{Percent of tip} = \frac{10.50}{58.25}\times 100$

Simplify:

Percent of tip = 18.0257511 %

Therefore, 18% tip did you leave for the waiter.

You might be interested in

Q12.

SVEN [57.7K]

Answer:

Step-by-step explanation:

1l ........8.5 miles

x l .......6 miles

-----------------------

x=6*1/8.5

x=0.70 l

2*0.7=1.4 l petrol/day ( to work and come back home)

5*1.4=7 l/week ( 5 days works in a week)

7*1.26=8.82 L /week

8.82>8.5

The petrol costs more

So the answer is NO

6 0

3 years ago

Five and one half less than 3 times a number

Rama09 [41]

3x - 5 1/2 is the expression

5 0

3 years ago

Which three lengths CANNOT be the lengths of the sides of a triangle

kogti [31]

Answer:

I believe 21 cm

Step-by-step explanation:

4 0

3 years ago

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What

galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

$P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

$P(D_i)=\frac{1}{6}$ for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

$P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}$

$P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}$

$P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}$

$P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}$

$P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}$

$P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0$

Part A:

The probability that all of the balls selected are white:

$P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)$

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find $P(D_3|A)$

The data required is calculated above:

$P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

7 0

3 years ago

Someone help me find this answer

Vanyuwa [196]

Finding the perimetre: 7x+4x-3+x=12x-3

Good luck!

5 0

2 years ago