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zzz [600]
2 years ago
14

Geraldo gets paid the same amount for each hour he works. The graph shows the amount he is paid, y, as it relates to the number

of hours he works, x. Yesterday, when he worked 6 hours, he was paid $84.
Geraldo’s Earnings
On a coordinate plane, a graph titled Geraldo's Earnings has number of hours on the x-axis and wages in dollars on the y-axis. The line goes through points (0, 0) and (6, 84).


Which ordered pair could represent the number of hours and amount Geraldo is paid today at the same hourly wage?
(3, 42)
(4, 60)
(5, 62)
(8, 100)

Mathematics
1 answer:
Dima020 [189]2 years ago
4 0

Answer:

(3, $42)

\

Step-by-step explanation:

Geraldo gets paid the same hourly rate from day to day.  This rate is the slope of the line representing his earnings.  It is m = rise/run, which here is

m = $84/(6 hr), or m = $14/hr.  Thus, the equation representing his earnings is y = ($14/hr)x.

The ordered pair (3, $42) satisfies this equation.  None of the other three do.

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The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. T
Anit [1.1K]

Answer:

a) 25

b) 67

c) 97

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample. In this problem, \sigma = 0.25

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

M = z*\frac{\sigma}{\sqrt{n}}

0.1 = 1.96*\frac{0.25}{\sqrt{n}}

0.1\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.1}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}

n = 24.01

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

M = z*\frac{\sigma}{\sqrt{n}}

0.06 = 1.96*\frac{0.25}{\sqrt{n}}

0.06\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.06}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}

n = 66.7

Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

M = z*\frac{\sigma}{\sqrt{n}}

0.05 = 1.96*\frac{0.25}{\sqrt{n}}

0.05\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.05}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}

n = 96.04

Rounding up, 97

8 0
1 year ago
A turtle and a rabbit engage in a footrace over a distance of 4.00 km. The rabbit runs 0.500 km and then stops for a 90.0-min na
Oxana [17]

Answer:

(a) 2.29 km/h

(b) 9 km/h

Step-by-step explanation:

For part (a) you have to apply<em> the average speed formula</em>, which is defined by:

v=\frac{d}{t}

where d is the total distance traveled and t is the total time needed.

d=\frac{4.00}{1.75}=2.29 km/h

For part (b) you have to calculate the running time (T) , which is the total time of the race minus the nap time:

The nap time in hours is:

90/60 = 1.5 h (because there are 60 minutes in one hour)

The running time is:

T= 1.75 - 1.5 = 0.25 h

Let t1 represent the time before the nap and t2 the time after the nap:

t1+t2 = T

t1+t2 = 0.25

You have to apply the formula d=vt before and after the nap:

-Before the nap, the distance traveled was 0.50 km

0.50 = v1t1

-Afer the nap, the distance traveled was 3.50 km

3.50=v2t2

But v2=2v1 (because after the nap the rabbit runs twice as fast)

You have to solve the system of equations:

t1=0.25-t2 (I)

v1t1=0.50 (II)

2v1t2=3.50 (III)

Replacing (I) in (II)

v1(0.25-t2)=0.50

Applying distributive property and solving:

0.25v1-v1t2=.050

For (III) you have that v1t2=3.50/2=1.75. Hence:

0.25v1-1.75=0.50

Solving for v1:

v1 = 9 km/h

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Sidana [21]
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Zepler [3.9K]
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2 years ago
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dexar [7]

Answer:

       X     =      8    x    28 =   224

                   

      Nandita  earns $8 last month by babysitting.

Step-by-step explanation:

                       225

divided by

                   <u>       28    </u>

                            8

4 0
1 year ago
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