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Nady [450]
3 years ago
6

A typical tip in a restaurant is​ 15% of the total bill. If the bill is ​$100​, what would the typical tip​ be

Mathematics
1 answer:
Natasha2012 [34]3 years ago
8 0

Answer:

15 dollars

Step-by-step explanation:

Total Bill = Cost of the Meal + tip

Total Bill = 100 + 15/100 * 100

Total Bill = 115

So the tip is 15 dollars.

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In triangle abc,angle a=3x+1,angle b=4x-17,and angle c=5x-20.what type of triangle is angle abc
lutik1710 [3]
(3x + 1) + (4x - 17) + (5x - 20) = 180
12x - 36 = 180
12x = 216
x = 18
angle a = 55
angle b = 55
angle c = 70
This is an acute isosceles triangle.
7 0
2 years ago
A taxi service charges an initial fee plus $1.80 per mile. How far can you travel for 14$
BigorU [14]

14 / 1.8 = 7.77777778

about 7.77777778 miles

but it depends on what the initail fee is

3 0
3 years ago
Please help! i need to find the mean
irina1246 [14]

Answer:

<h3>          The mean is 3.74</h3>

Step-by-step explanation:

\overline x =\dfrac{-3\cdot4+(-2)\cdot2+3\cdot1+4\cdot1+6\cdot4+8\cdot7}{4+2+1+1+4+7}\\\\\\\overline x =\dfrac{-12-4+3+4+24+56}{19}\\\\\\\overline x =\dfrac{71}{19}=3.7368421....\approx3.74

7 0
3 years ago
Labor Picture
oksian1 [2.3K]

Answer:

  9.9

Step-by-step explanation:

The table tells you numbers who "want" a job. The question asks you for numbers "looking for a job". Those may be different things. The question cannot be answered from the table, unless you make the assumption that those wanting a job are actually looking for a job.

  total looking = (working and looking) + (not working and looking)

  = 5.2 million + 4.7 million

  = 9.9 million . . . . looking for a job

7 0
3 years ago
The mean amount of money spent on lunch per week for a sample of 458458 college students is $43$ 43. If the margin of error for
a_sh-v [17]

Answer:

The 99% confidence interval for the mean amount of money spent on lunch per week for all college students is between $41.40 and $44.60.

Step-by-step explanation:

A confidence interval has two bounds, a lower bounds and an upper bound.

These bounds depend on the sample mean and the margin of error.

The lower bound is the sample mean subtracted by the margin of error.

The upper bound is the margin of error added to the sample mean.

In this problem, we have that:

The sample mean is $43.

The margin of error is $1.60 for a 99% confidence interval.

Lower bound: 43 - 1.60 = $41.40

Upper bound: 43 + 1.60 = $44.60

The 99% confidence interval for the mean amount of money spent on lunch per week for all college students is between $41.40 and $44.60.

6 0
3 years ago
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