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rusak2 [61]
2 years ago
15

A parking garage charges $5.50 to park for 4 hours and $7.75 to park for 7 hours. If the cost is a linear function of the number

of hours parked, what is the cost to park for 3 hours?
A) $2.25
B) $2.50
C) $3.25
D) $4.75
Mathematics
2 answers:
Oksana_A [137]2 years ago
7 0
Y=mx+b
m=slope
b=yint

two points
cost=y
(4,5.5) and (7,7.75)
slope=(y2-y1)/(x2-x1)
slope=(7.75-5.5)/(7-4)=2.25/3=0.75

y=0.75x+b
find b
(4,5.5)
x=4
y=5.5
5.5=0.75(4)+b
5.5=3+b
minus 3 fromboth sides
2.5=b

y=0.75x+2.5
3 hours
y=0.75(3)+2.5
y=2.25+2.5
y=4.75

answer is D
Jobisdone [24]2 years ago
4 0
Slope = (7.75 - 5.50)/(7 - 4) = 2.25/3 = 0.75
The equation described by the function is
y - 5.5 = 0.75(x - 4)
y = 0.75x - 3 + 5.5 = 0.75x + 2.5
For x = 3, y = 0.75(3) + 2.5 = 2.25 + 2.5 = 4.75
Therefore, it costs $4.75 to park for 3 hours.
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Answer:

Option B is the correct answer.

Step-by-step explanation:

Jen decides to read 100 pages the first day and 50 pages each day thereafter.

Number of pages read by Jen per day = 50

Number of pages read by Jen on first day = 100

Ariel's progress on reading the book is represented by the linear function y = 40x + 80, where y is the total number of pages read after x days.

Number of pages read by Ariel per day = 40

Number of pages read by Ariel on first day = 40 + 80 = 120

Option A:

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Wrong

Option B:

Ariel reads 10 pages per day less than Jen.

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Option C:

Ariel read 20 pages less the first day than Jen read.

Wrong

Option D:

The reading rate each day for Jen and Ariel is the same.

Wrong

Option B is the correct answer.

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First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

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The amount of time a passenger waits at an airport check-in counter is random variable with mean 10 minutes and standard deviati
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Answer:

(a) less than 10 minutes

= 0.5

(b) between 5 and 10 minutes

= 0.5

Step-by-step explanation:

We solve the above question using z score formula. We given a random number of samples, z score formula :

z-score is z = (x-μ)/ Standard error where

x is the raw score

μ is the population mean

Standard error : σ/√n

σ is the population standard deviation

n = number of samples

(a) less than 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Therefore, the probability that the average waiting time waiting in line for this sample is less than 10 minutes = 0.5

(b) between 5 and 10 minutes

i) For 5 minutes

x = 5 μ = 10, σ = 2 n = 50

z = 5 - 10/2/√50

z = -5 / 0.2828427125

= -17.67767

P-value from Z-Table:

P(x<5) = 0

Using the z table to find the probability

P(z ≤ 0) = P(z = -17.67767) = P(x = 5)

= 0

ii) For 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Hence, the probability that the average waiting time waiting in line for this sample is between 5 and 10 minutes is

P(x = 10) - P(x = 5)

= 0.5 - 0

= 0.5

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2 years ago
A force of 500.0 is represented graphically with its tail at the origin and the tip pointed in a direction 30.0° above the posit
trapecia [35]

Answer:

F^{'}=(250\sqrt{3},250 })

Step-by-step explanation:

We have F´ =500 and \alpha=30º, so x and y components:

F´ = (F_{x} , F_{y}) this is

F_{x} = F^{'} *Cos\alpha

F_{y} = F^{'} *Sin\alpha  

F_{x} = F`*Cos\alpha =500*Cos30=500*\frac{\sqrt{3} }{2} =250\sqrt{3}

F_{y}=F*Sin\alpha  =500*Sin30=500*\frac{1}{2}=250;

Finally

F' = (250\sqrt{3} , 250)

3 0
2 years ago
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