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3 years ago

How many seconds are in 76 minutes

Mathematics

2 answers:

Kipish [7]3 years ago

5 0

Answer:

4560

Step-by-step explanation:

Well, first we need to understand how different time formats (for lack of a better word) are related in terms of seconds.

1 minute has 60 seconds

I hour has 3600 seconds

1 day has 86400 seconds

1 week has 604800 seconds

Now, with this information at hand, you can find out how many seconds are in X minutes or X days or X months or a year (and so on).

As for your question, we know that one minute has 60 seconds therefore 76 minutes will have (76x60) seconds

AlexFokin [52]3 years ago

3 0

In one minute there are 60 seconds so to convert minutes into seconds multiply 76 by 60!

76 * 60

4560

In 76 minutes there are 4,560 seconds

Hope this helped!

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Answer:

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

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Step-by-step explanation:

Data given and notation

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X=115 represent the adults with the medication was effective

$\hat p=\frac{115}{180}=0.639$ estimated proportion of adults with the medication was effective

$p_o=0.65$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is less than 0.65.:

Null hypothesis: $p \geq 0.65$

Alternative hypothesis: $p < 0.65$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

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Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

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It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

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