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evablogger [386]
3 years ago
6

A taxi driver from a certain company charges $ 150 for stopping and loading passengers and then $ 30 for every kilometer travele

d.
a) Determine the linear function that relates the fare and the kilometers
tours.
b) How much should a person who travels 20km pay?

Let x be the number of kilometers traveled by a person in the taxi.
Mathematics
1 answer:
Xelga [282]3 years ago
8 0

Answer:

a) linear function =  150 + 30x represents 1km

    and 150 + 60x represents 2km We can therefore remove the ones and even factor 15 and 3 into

3(5+1x)

b)  y = 2/3x + 225

Can be factored from 2/3 (30x+75)  and simplified to  2/3(3x+15/2)

Step-by-step explanation:

if 150 stop and 30 for each mile

we show zero on each axis and show the y intercept at 150

The X axis should determine time or distance travelled

So if showing distance we show miles in 10,20,30,40,

and costs 0,30,60,90 but if 150 applies to any passengers entering vehicle and all journeys then we start graph at $ cost = 150 and go up by 30 showing 150.180.210.240  for the data for the y axis/

Therefore 20k is 2/3 km and we can show 2/3 as our multiple.

2/3 of 225 is 150 to find we flip 3/2 x 150 = 225

y = 2/3 ( 30x + 225) and we have our equation for b

y = 2/3x + 225

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11111nata11111 [884]

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Answer:

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Step-by-step explanation:

It should be no surprise that the shortest allowable length is the nominal length less the full allowed tolerance:

  11 1/4 - 1/8 = 11 1/8 . . . inches

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Likewise, the longest allowable length is the nominal length plus the maximum tolerance:

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Home Videos Inc. surveys 450 households and finds that the mean amount spent for renting or buying videos is P135 a month and th
adell [148]

SOLUTION:

Case: Hypothesis testing

Step 1: Null and Alternative hypotheses

\begin{gathered} H_0:\mu=P127.50 \\ H_1:\mu\leq P127.50 \end{gathered}

Step 2: T-test analysis

\begin{gathered} t=\frac{\hat{x}-\mu}{\frac{s}{\sqrt{n}}} \\ t=\frac{135-127.5}{\frac{75.25}{\sqrt{450}}} \\ t=2.144 \end{gathered}

Step 3: t-test with the significance level

\begin{gathered} t_{\alpha}=? \\ \alpha=0.05 \\ From\text{ }tables \\ t_{0.05}=1.654 \end{gathered}

Step 4: Comparing

t>t_{\alpha}

So tail to reject the null hypothesis. There is enough evidence at a 0.05 level of significance to claim that the mean spent is greater than P127.50.

Final answer:

Yes, there is evidence sufficient to conclude that the mean amount spent is greater than P127.50 per month at a 0.05 level of significance.

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Pretty simple.
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