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3 years ago

If speed varies inversely as the time it takes to drive and Kris takes 5 hours driving at 55 mph, what speed will Martin need to

Mathematics

1 answer:

galina1969 [7]3 years ago

3 0

Answer:

55mph

None of the option is correct

Step-by-step explanation:

Let v be the speed and t as the time taken. If speed varies inversely as the time it takes to drive, then v ∝ 1/t.

v = k/t where k is the constant of proportionality.

IF it takes Kris 5 hours when driving at 55 mph, then v = 55mph when t = 5 hours.

Substituting this values into the formula above;

55 = k/5

k = 55*5

k = 275mp/hr²

To calculate the speed it will Martin to drive for 5 hours, we will substitute k = 275 and t = 5 into the original equation v = k/t

v = 275/5

v = 55 mph

<em>Hence, martin will also need to drive at 55mph to take 5 hours</em>

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Slav-nsk [51]

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----------------

Extra side notes:

If angle 1 was a right angle, then that would mean angle GEF would have to be 0 degrees; however the diagram shows this isn't the case.

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3 years ago

The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

Alexandra [31]

Answer:

$\bar x = 260.1615$

$\sigma = 70.69$

The confidence interval of standard deviation is: $53.76$ to $103.25$

Step-by-step explanation:

Given

$n =20$

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

$\bar x = 260.1615$

$\bar x = 260.16$

Solving (b): The standard deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$