All categories

2 years ago

Polly is a server at a

Mathematics

1 answer:

Anna [14]2 years ago

8 0

Answer:

They gave her a $6.12 tip

Step-by-step explanation:

You have to multiply the total ($34) by the Percent (18% = 0.18), and then you will have your final answer.

You might be interested in

John needs to paint one wall in his school. He knows that 1 can of paint covers an area of 24 square feet. John uses a meter sti

laila [671]

The complete question in the attached figure

step 1
find the area of the wall

[area of the wall]=[area of rectangle]+[area of triangle]
[area of the wall]=[3*2]+[3*1/2]----> 7.5 m²

1 m²-------> 10.7639 ft²

area of the wall------> 7.5*10.7639------> 80.73 ft²

we know that
1 can of paint covers an area of --------> 24 ft²
x---------------------------------------------> 80.73 ft²
x=80.73/24---------> x=3.36 can of paints

the answer is
T<span>he fewest number of cans is 4</span>

6 0

3 years ago

a garden is made up of two squares and a quarter circle. what are the perimeter and area of the garden

FrozenT [24]

The complete question in the attached figure

Part A) find the perimeter
[perimeter of the garden]=[perimeter square 1]+[perimeter a quarter circle]+[perimeter square 2]

[perimeter square 1]=5+5+5-----> 15 ft
[perimeter square 2]=5+5+5-----> 15 ft
[perimeter a quarter circle]=(2*pi*r)/4------> 2*pi*5/4-----> 7.85 ft

[perimeter of the garden]=[15]+[7.85]+[15]-------> 37.85 ft

the answer Part A) is
the perimeter of the garden is 37.85 ft

Part B) Find the area of the garden

[Area of the garden]=[Area square 1]+[Area a quarter circle]+[Area square 2]

[Area square 1]=5*5-----> 25 ft²
[Area square 2]=5*5-----> 25 ft²
[Area a quarter circle]=(pi*r²)/4------> pi*5²/4-----> 19.625 ft²

[Area of the garden]=[25]+[19.625]+[25]-------> 69.625 ft²

the answer Part B) is
the Area of the garden is 69.625 ft²

8 0

3 years ago

Production managers on an assembly line must monitor the output to be sure that the level of defective products remains small. T

lisabon 2012 [21]

Answer:

Factory owners would consider the Type I error worse, because the assembly process will be halted while it shouldn't have been and thus they will lose money.

8 0

3 years ago

In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and

Umnica [9.8K]

Answer:

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

<em>A little more detail</em>

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0

3 years ago

If A is the center of the circle, which of these are radii?

loris [4]

Answer:

D. A E

Step-by-step explanation:

BEACAUSE THE A IS NEAR TO THE E

6 0

2 years ago