A race car travels 100 feet in 5 seconds, at this rate of speed, how many feet will the race car travel in a minute

2 answers:

6 0

There are 60 seconds in one minute, and 60/5=12, so the race car would travel 100 feet every 5 seconds 12 times in one minute. You can simply multiply 100 by 12 to get your answer.
100x12=1200
The race car travels 1,200 feet in one minute.

crimeas [40]3 years ago

4 0

60 sec in a min, 60/5 = 12, therefore 100 x 12= 1200 ft per min

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The desired percentage of SiO2 in a certain type of aluminous cement is 5.5. To test whether the true average percentage is 5.5

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Answer:

z=-3.25

$p_v =2*P(z$

Step-by-step explanation:

1) Data given and notation

$\bar X=5.24$ represent the mean production for the sample

$\sigma=0.32$ represent the sample standard deviation for the sample

$n=16$ sample size

$\mu_o =5.5$ represent the value that we want to test

$\alph=0.05a$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean production is different from 5.5 tons, the system of hypothesis would be:

Null hypothesis: $\mu =5.5$

Alternative hypothesis: $\mu \neq 5.5$

If we analyze the info given we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{5.24-5.5}{\frac{0.32}{\sqrt{16}}}=-3.25$

P-value

Since this is a two side test the p value would be:

$p_v =2*P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the production its significant different compared to the desired percentage of SiO2 of 5.5 at 5% of signficance.