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3 years ago

Find the total cost of a $58 bill with a 20% tip

Mathematics

2 answers:

yaroslaw [1]3 years ago

8 0

the answer is $69.60

Sergeu [11.5K]3 years ago

7 0

Answer:

The total cost of a $58 bill with a 20% tip would be $69.60.

Step-by-step explanation:

To find the amount you must first find the value of 20% of 58. In order to do that, you can multiply them together (don't forget to turn 20% into a decimal first).

58 * .20 = 11.60

Now we take the tip amount and add it to the bill total for the total amount.

$58 + $11.60 = $69.60

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Current and Quick Ratios The Nelson Company has $1,250,000 in current assets and $500,000 in current liabilities. Its initial in

murzikaleks [220]

Answer:

Step-by-step explanation:

$1,250,000 in current assets (cr) and $500,000 in current liabilities(cl)

current ratio (cr)=1250000/500000=2.5

Δ note payable ( change in note payable NP)

minimum current=2.2

2.2=(1250000+ΔNP)/(500000+ΔNP)

2.2(500000+ΔNP)=1250000+ΔNP

1100000+2.2ΔNP=1250000+ΔNP

2,2ΔNP-ΔNP=1250000-1100000

1.2ΔNP=150000

ΔNP=150000/1.2=125000

Nelson's short-term debt (notes payable) increase without pushing its current ratio below 2.2=125000

assuming this amount used to increase the inventory

new inventory=335000+125000=360000

current asset= 1250000+125000=1375000

the new ratio=(1375000-360000)/500000+125000

new ratio = 1.624

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3 years ago

2.4x + 7.2 – 4.3 plz.

Zinaida [17]

Answer:

2.4x+2.9

Step-by-step explanation:

3 0

3 years ago

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Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

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3 years ago

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Reil [10]

Answer:

Step-by-step explanation:

8 is LESS THAN 1/2x

so youd take those numbers and multiply by 1/2

C, B, and A are all under 8, so that would be wrong

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3 years ago

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2.1 = -0.6m<br><br>What is the solution to m?

Anon25 [30]

Answer:

m=-3.5

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3 years ago

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