The slope of the lawn and passing through the points (1,3) and (5,13)

-3x - 6 = 2x - 14x - 6

Harman [31]

Answer:

x=0

Step-by-step explanation:

3 0

3 years ago

A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa

Norma-Jean [14]

Answer:

a) Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

b) $df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

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 true mean is higher than 15 at 5% of signficance.

c) Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation

$\bar X=18.3$ represent the sample mean

$s=3.72$ represent the sample standard deviation

$n=25$ sample size

$\mu_o =15$ represent the value that we want to test

$\alpha=0.05$ 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)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic 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".

Part b: P-value and conclusion

The first step is calculate the degrees of freedom, on this case:

$df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

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 true mean is higher than 15 at 5% of signficance.

Part c

Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0

3 years ago

Find the quotient of 2,300 and (0.4 · 10-8). In your final answer, include all of your calculations.

BigorU [14]

The first number in this item can be also expressed as 2.3 x 10³. Now, dividing the numbers 2.3 and 0.4 will give us an answer of 5.75. There exists a rule which states that when two similar numbers are divided only having different exponents, their exponents should be subtracted. The final answer to this item is therefore,
5.75 x 10¹¹

5 0

3 years ago

Which explanation provides the best real-world scenario of the graph?

Naddika [18.5K]

The last one is the only one that makes sense according to the standard position function. -16t^2 is the pull of gravity on an object in free fall, and the height is 120 feet above the ground. Hopefully that's what you need since there's no graph we can refer to

5 0

3 years ago

Read 2 more answers

Mia is "helping" her mom pick up 27 toys that are strewn on the floor. Mia's mom puts five toys into the toy box every 40 second

dedylja [7]

Answer:

8 min

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5 0

2 years ago