Max drives 85 km at an average speed of 110 km/h and then 7 km at 70 km/h without stopping.

Viefleur [7K]

5x4=20 i think this is the right one

6 0

3 years ago

The ground temperature at an airport is 16 °C. The temperature decreases by 5.6 °C for every increase of 1 kilometer above the g

love history [14]

0 km = 16

1km = 16 - 5,6 = 10,4

2km = 10,4 - 5,6 = 4,8

3km = 4,8 - 5,6 = -0,8

4km = -0,8 - 5,6 = -6,4

5km = -6,4 - 5,6 = -12

3 0

2 years ago

Y is directly related to X, and Yis 81 when X is 27.<br> The constant of variation is

Arisa [49]

Answer:

3

Step-by-step explanation:

A direct variation is of the form

y = kx

We know x and y so we can find k

81 = k *27

Divide each side by 27

81/27 = 27k/27

3=k

The constant of variation is 3

5 0

3 years ago

Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study wa

Sladkaya [172]

Answer:

The null hypothesis is $H_0: \mu = 2.5$ .

The alternative hypothesis is $H_1: \mu > 2.5$

Step-by-step explanation:

Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years.

At the null hypothesis, we test if the mean is of 2.5 years, that is:

$H_0: \mu = 2.5$

A study was then done to see if the mean time has increased in the new century.

At the alternative hypothesis, we test if the mean has increased, that is, if it is above 2.5 years. So

$H_1: \mu > 2.5$

8 0

3 years ago

Suppose that a researcher is designing a survey to estimate the proportion of adults in your state who oppose a proposed law tha

irinina [24]

Answer:

$n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401$

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

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

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

Solution to the problem

The margin of error for the proportion interval is given by this formula:

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

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

The margin of error desired for this case is $ME= \pm 0.02$ equivalent to 2% points

For this case we need to assume a confidence level, let's assume 95%. And since we don't have prior estimation for the population proportion of interest the best value to do an approximation is $\hat p =0.5$

In order to find the critical value we need to take in count that we are finding the margin of error for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.96$

Now we have all the values needed and if we replace into equation (b) we got:

$n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401$

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

5 0

3 years ago