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3 years ago

6

A grab-bag contains 30 packages worth $.65 each, 10 packages $.60 cents each, and 15 packages worth $.30 each. How much should t

he game owner charge to make it a fair game?

1 answer:

maria [59]3 years ago

5 0

Answer:

$0.40

I had the same question and it was $0.40

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5/17 multiplied by 3/8

tino4ka555 [31]

5/17 × 3/8 = 15/ = 15/136

6 0

3 years ago

Read 2 more answers

Find the 20th term in the sequence, where the first term is -2 and the common ratio is-2.

geniusboy [140]

Answer:

Step-by-step explanation:

(1)

a1=-2

r=-2

$a_{20}=(-2)(-2)^{20-1}=-2(-2)^{19}=2^{20}=1048576\\(2)\\a_{1}=1000\\r=\frac{200}{1000} =\frac{1}{5} \\a_{8}=1000(\frac{1}{5} )^{8-1}=1000(.2)^7=0.0128\\(3)\\a_{1}=2\\a_{3}=50\\a_{2}=\sqrt{2*50} =10\\r=\frac{10}{2} =5$

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2 years ago

Pls answer this I need it quick

Olenka [21]

Answer:

Step-by-step explanation:

5

7 0

3 years ago

The area of the city of Vancouver is 115 km^2. if the shape of Vancouver is approximately a square, how wide is the city of Vanc

Dmitry [639]

Answer:

The wide of the city is 10.7km

Step-by-step explanation:

To solve this we first have to know the formula to calculate the area of a square

a = area = 115 km²

s = side

a = s * s

now we replace with the known values

115 km² = s * s

115 km² = s²

√(115 km²) = s

10.724km = s

Round to the nearest tenth

s = 10.724km = 10.7km

The wide of the city is 10.7km

5 0

3 years ago

Find, correct to four decimal places, the length of the curve of intersection of the cylinder 16x2 + y2 = 16 and the plane x + y

Yuri [45]

Let the curve C be the intersection of the cylinder

$16x^2+y^2=16$

and the plane

$x+y+z=1$

The projection of C on to the x-y plane is the ellipse

$16x^2+y^2=16$

To see clearly that this is an ellipse, le us divide through by 16, to get

$\frac{x^2}{1}+ \frac{y^2}{16}=1$

or

$\frac{x^2}{1^2}+ \frac{y^2}{4^2}=1$ ,

We can write the following parametric equations,

$x=cos(t), y=4sin(t)$

for

$0\le t \le 2\pi$

Since C lies on the plane,

$x+y+z=1$

it must satisfy its equation.

Let us make z the subject first,

$z=1-x-y$

This implies that,

$z=1-sin(t)-4cos(t)$

We can now write the vector equation of C, to obtain,

$r(t)=(cos(t),4sin(t),1-cos(t)-4sin(t))$

The length of the curve of the intersection of the cylinder and the plane is now given by,

$\int\limits^{2\pi}_0 {|r'(t)|} \, dt$

But

$r'(t)=(-sin(t),4cos(t),sin(t)-4cos(t))$

$|r'(t)|=\sqrt{(-sin(t))^2+(4cos(t))^2+(sin(t)-4cos(t))}$

$\int\limits^{2\pi}_0 {\sqrt{2sin^2(t)+32cos(t)-8sin(t)cos(t)} }\, dt=24.08778184$

Therefore the length of the curve of the intersection intersection of the cylinder and the plane is 24.0878 units correct to four decimal places.

6 0

3 years ago