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maks197457 [2]
3 years ago
6

According to Ohm’s law, which combination of units is the same as the unit for resistance? volt ÷ ampere ampere × volt volt + am

pere ampere ÷ volt
Physics
2 answers:
sergiy2304 [10]3 years ago
8 0

Answer:

The answer is: volt ÷ ampere

Explanation:

Hope this helps. :)

If it's right, can you mark me as Brainliest.

Maksim231197 [3]3 years ago
6 0

Answer:

volt ÷ ampere

Explanation:

The mathematical form of Ohms law is given by :

V = IR

Where V is voltage

I is current

R is resistance

R=\dfrac{V}{I}

The unit of voltage is volt and that of current is ampere

Unit of resistance :

R=\dfrac{\text{volt}}{\text{ampere}}

So, volt ÷ ampere is the same as the unit of resistance. Hence, the correct option is (a).

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A machine is designed to fill jars with 16 ounces of coffee. A consumer suspects that the machine is not filling the jars comple
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Answer:

a) Null hypothesis:  \mu \geq 16  

Alternative hypothesis :\mu  

b) t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.77  

p_v =P(t_{(7)}  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

c) We can conclude that the mean is significantly less than 16 ounces at 10% of significance.  so then the consumer suspect is correct.

Explanation:

Data given and notation  

\bar X=15.6 represent the mean for the sample

s=0.3 represent the sample standard deviation  

n=8 sample size  

\mu_o =16 represent the value that we want to test  

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

Is a one tailed left test.  

What are H0 and Ha for this study?  

Null hypothesis:  \mu \geq 16  

Alternative hypothesis :\mu  

The degrees of freedom on this case are:

df=n-1=8-1=7

Compute the test statistic

The statistic for this case is given by:  

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

t-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:  

t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.77  

Give the appropriate conclusion for the test

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

p_v =P(t_{(7)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

We can conclude that the mean is significantly less than 16 ounces at 10% of significance.  so then the consumer suspect is correct.

3 0
3 years ago
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