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lilavasa [31]
3 years ago
6

Han earns $33.00 fo babysitting 4 hours At this rate how much will he earn if he babysits for 7 hours

Mathematics
2 answers:
sertanlavr [38]3 years ago
7 0

Answer:

$57.75

Step-by-step explanation:

There is a direct proportionality relationship between the amount he earns and the time he spends babysitting.

As the number of hours increases, the time increases at a fixed rate.

If the time is t and the amount earned f

then,

f ∝ t

33 = k(4) where k is the constant of proportionality

k = 33/4

f = 33/4 (t)

The amount he will earn if he babysits for 7 hours

= (33/4) × 7

= $57.75

cupoosta [38]3 years ago
3 0

Answer:

$57.75

Step-by-step explanation:

$33.00 divided by 4 is $8.25 per hour. Then you multiply 7 time $8.25 which gives you the correct answer of $57.75.

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The circle P has a center at (0, 0) and a point on the circle at (0, 4). If it is dilated by a factor of 4, what is the distance
galina1969 [7]

Answer:

A. 32

Step-by-step explanation:

If the center is (0, 0) and a point is (0, 4) then the distance from the center to that point is 4 units. That distance is the radius. If you are dilating by a factor of 4, multiply the radius by 4 and you get 16. The new radius is 16 and the diameter= radius*2.

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2 years ago
Given a joint PDF, f subscript X Y end subscript (x comma y )equals c x y comma space 0 less than y less than x less than 4, (1)
ioda

(1) Looks like the joint density is

f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0

In order for this to be a proper density function, integrating it over its support should evaluate to 1. The support is a triangle with vertices at (0, 0), (4, 0), and (4, 4) (see attached shaded region), so the integral is

\displaystyle\int_0^4\int_y^4 cxy\,\mathrm dx\,\mathrm dy=\int_0^4\frac{cy}2(4^2-y^2)=32c=1

\implies\boxed{c=\dfrac1{32}}

(2) The region in which <em>X</em> > 2 and <em>Y</em> < 1 corresponds to a 2x1 rectangle (see second attached shaded region), so the desired probability is

P(X>2,Y

(3) Are you supposed to find the marginal density of <em>X</em>, or the conditional density of <em>X</em> given <em>Y</em>?

In the first case, you simply integrate the joint density with respect to <em>y</em>:

f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^x\frac{xy}{32}\,\mathrm dy=\begin{cases}\frac{x^3}{64}&\text{for }0

In the second case, we instead first find the marginal density of <em>Y</em>:

f_Y(y)=\displaystyle\int_y^4\frac{xy}{32}\,\mathrm dx=\begin{cases}\frac{16y-y^3}{64}&\text{for }0

Then use the marginal density to compute the conditional density of <em>X</em> given <em>Y</em>:

f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}\frac{2xy}{16y-y^3}&\text{for }y

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The diameter of a semicircle is 8 kilometers. What is the semicircle's radius?
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