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3 years ago

Work out Ryan’s average speed for his total return journey detailing how you worked it out.

Mathematics

1 answer:

pickupchik [31]3 years ago

7 0

Answer:

For the return, i guess that you say in the interval between 15:40 and 15:45, that is the moment where he goes back to his home, and pass through the local shop.

Ok, in the graph we can see pairs (time, position)

The average speed can be obtained with two of these pairs, and the average speed is:

As = (position2 - position1)/(time2 - time1)

In this particular case, we have the pairs:

(15:40, 900m) and (15:45, 400m)

(i guess that the distance is in meters)

Then the average speed is:

As = (900m - 400m)/(15:45 - 15:40)

and 15:45 - 15:40 = 0.5 minutes.

As= 500m/5 min = 100 m/min

This means that the average speed in this return is 100 meters per minute

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X+10=1/3(5x+10) How do you solve this equation?

Nikitich [7]

Answer:

x = 10

Step-by-step explanation:

First, subtract 10 from both sides

Next, Simplify

Then, subtract 5/3x from both sides

After, simplify

Lastly, multiply both sides by 3

Then finally simplify for your final answer of x = 10

Hope this helps! :)

3 0

2 years ago

Read 2 more answers

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.