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3 years ago

Joe drove 538 miles in 9 hours 15 minutes what was his average speed per hour

1 answer:

7 0

To calculate any speed, the equation is speed=distance/time
Basically, you first need to take your 9 hours and 15 minutes and combine them into a number with a decimal value.
So, 15*4=60
Sound familiar? Good, because 25*4=100, so they are the same here.
So, we can say 9 hours and 15 minutes is equal to 9.25
Take this now: Speed=538/9.25
You get 58.1621, which you can round to 58.2
This is, of course, your final answer, so Joe drove at 58.16 mph

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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:

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

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

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

So, the confidence interval of the standard deviation is:

$\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} }$ to $\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }$

$70.69 * \sqrt{\frac{20 - 1}{32.852}$ to $70.69 * \sqrt{\frac{20 - 1}{8.907}$

$70.69 * \sqrt{\frac{19}{32.852}$ to $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

8 0

3 years ago

9 9/15 - 2 13/15 = <br> i have know idea what it is plz help

maria [59]

Step-by-step explanation:

plase see it my step work solution

5 0

3 years ago

Please help!! How do you do number 1?

alisha [4.7K]

Do a ratio tall over shadow. How tall is tree / shadow. How tall is person / shadow then cross multiply. X/7.5 =5/3. 37.5=3x. Divide by 3 answer 12.5

6 0

3 years ago

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NikAS [45]

Yes, most definitely
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3 0

3 years ago

Create your own real-world example of a relation that is a function. Domain: The set of Range: The set of

ivanzaharov [21]

Answer:

C(t)=5000 -10t

Step-by-step explanation:

There are many examples in the real world of relationships that are functions.

For example, imagine a tank full of water with a capacity of 5000 liters, this tank has a small hole, by which 10 liters of water are lost every hour.

If we call C the amount of water in the tank as a function of time, then we can write the following equation for C:

$C(t)=C_{0}-at\\$

Where:

C (t): Amount of water in the tank as a function of time

$C_{0}$ : Initial amount of water in the tank at time t = 0

a: amount of water lost per hour

t: time in hours

Then the equation is:

$C(t)=5000 -10t$

The graph of C (t) is a line of negative slope. This relation is a function since for each value of t there is a single value of C.

Its domain is the set of all positive real numbers t between [0,500]

Because the time count starts at t = 0 when the tank is full and ends at t = 500 when empty

Its Range is the set of all positive real numbers C between [0,5000] Because the amount of water in the tank can never be less than zero or greater than 5000Litres

7 0

3 years ago

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