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4 years ago

Alex swims at an average speed of 45 m/min. How far does he swim in 1 min 24 sec? m

Mathematics

2 answers:

denpristay [2]4 years ago

3 0

Answer:

63 m

Step-by-step explanation:

We know the rate in 45 m / min

We need to change the time to min

1 min 24 sec

We know that 1 min = 60 sec so 24 seconds is 24/60 ths of a minute

1 min 24/60 min

1 2/5 min

Changing to decimal form

1.4 minutes

Now we multiply by the rate to get the distance

d = r *t

d = 45 m/ min * 1.4 min

d= 63 m

babymother [125]4 years ago

3 0

Answer:

I would first divide 45 meters by 60 seconds to know now far Alex swims every second:

45/60 = .75m/second.

Then I would multiply .75×24 seconds:

.75 ×24 = 18 meters in 24 seconds.

Add the 45 meters he swam in one minute to the 18 meters he swam in 24 seconds.

He swam 63 meters in 1 minute 24 seconds.

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3 years ago

A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A ran

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Answer:

a) The 95% confidence interval would be given (0.070;0.121).

b) "increase the sample size n"

"decrease za/2 by decreasing the confidence"

Step-by-step explanation:

Notation and definitions

$X=48$ number of vehicles classified as sports utility.

$n=500$ random sample taken

$\hat p=\frac{48}{500}=0.096$ estimated proportion of vehicles classified as sports utility vehicles.

$p$ true population proportion of vehicles classified as sports utility vehicles.

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

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

(a) Use a 95% confidence interval to estimate the proportion of sports utility vehicles in California. (Round your answers to three decimal places.)

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 95% confidence interval the value of $\alpha=1-0.5=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.096 - 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.070$

$0.096 + 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.121$

And the 95% confidence interval would be given (0.070;0.121).

We are confident (95%) that about 7.0% to 12.1% of vehicles in California are classified as sports utility .

(b) How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers. Select all that apply.)

For this case we just have two ways to increase the accuracy one is "increase the sample size n" since if we have a larger sample size the estimation would be more accurate. And the other possibility is "decrease za/2 by decreasing the confidence" because if we decrease the confidence level the interval would be narrower and accurate

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