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Andreyy89
2 years ago
6

The total cost to rent a rowboat is $18 times the number of hours the boat is used. Write an equation to model this situation if

C equals total cost and H equals number of hours?
Answer choices -
C-18=h

H=18c

C=h/18

C=18h
Mathematics
2 answers:
AveGali [126]2 years ago
4 0
If the total cost (C) is $18 multiplied by the hours, it would be C=18h
elena-s [515]2 years ago
4 0

Answer:

C=18h

Step-by-step explanation:

Let

C------> represent the total cost

h-----> represent the number of hours

we know that

C=18h -----> linear direct variation that represent the situation

In this problem the constant of proportionality k is equal to the slope m of the line

so

m=18\frac{\$}{hour}

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Solve t/12=4 what is the solution for t?
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Step-by-step explanation:

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At 3pm, the length of the shadow of a thin vertical pole standing on level ground is the same as the height of the pole. A while
aliya0001 [1]

Answer: The height of the pole is 175.97 ( approx )

Step-by-step explanation:

Let the height of the pole is x cm,

Also, let the angle of elevation of the sun to the pole at 3 pm is \theta

Thus, by the question,

tan\theta = \frac{\text{ The height of the pole}}{\text{ The height of the shadow}}

\implies tan\theta = \frac{x}{x}    ( at 3pm the height of shadow = height of the pole)

\implies tan\theta = 1

\implies \theta = 45^{\circ}

Again, according to the question,

When the angle of elevation is   (\theta - 12^{\circ}),

The height of shadow = x + 95,

\implies tan(45-12)^{\circ}=\frac{x}{x+95}

\implies tan 33^{\circ}=\frac{x}{x+95}

\implies tan 33^{\circ}x + 95\times tan33^{\circ}=x

\implies tan 33^{\circ}x - x = - 95\times tan33^{\circ}

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4 0
3 years ago
-0.06m = 7.2 what is m?
Nata [24]
-0.06m=7.2
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3 0
2 years ago
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The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with me
lisov135 [29]

Answer:

a) <u>0.4647</u>

b) <u>24.6 secs</u>

Step-by-step explanation:

Let T be interval between two successive barges

t(t) = λe^λt where t > 0

The mean of the exponential

E(T) = 1/λ

E(T) = 8

1/λ = 8

λ = 1/8

∴ t(t) = 1/8×e^-t/8   [ t > 0]

Now the probability we need

p[T<5] = ₀∫⁵ t(t) dt

=₀∫⁵ 1/8×e^-t/8 dt

= 1/8 ₀∫⁵ e^-t/8 dt

= 1/8 [ (e^-t/8) / -1/8 ]₀⁵

= - [ e^-t/8]₀⁵

= - [ e^-5/8 - 1 ]

= 1 - e^-5/8 = <u>0.4647</u>

Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>

<u></u>

b)

Now we find t such that;

p[T>t] = 0.95

so

t_∫¹⁰ t(x) dx = 0.95

t_∫¹⁰ 1/8×e^-x/8 = 0.95

1/8 t_∫¹⁰ e^-x/8 dx = 0.95

1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t  = 0.95

- [ e^-x/8]¹⁰_t = 0.96

- [ 0 - e^-t/8 ] = 0.95

e^-t/8 = 0.95

take log of both sides

log (e^-t/8) = log (0.95)

-t/8 = In(0.95)

-t/8 = -0.0513

t = 8 × 0.0513

t = 0.4104 (min)

so we convert to seconds

t = 0.4104 × 60

t = <u>24.6 secs</u>

Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>

6 0
2 years ago
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