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____ [38]
3 years ago
9

How is 12.375 written in expanded form

Mathematics
1 answer:
poizon [28]3 years ago
8 0
10.000+2.000+0.3+0.07+0.005
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$500 at 4% for 3 years <br> :FIND THE INTEREST EARNED
mixer [17]

Answer:60$

Step-by-step explanation:

3 0
3 years ago
What is the cost if a $1,200 washing machine after a discount if 1/5 the original price
Fantom [35]
1/5 (or 20%) of $1,200 is $240
$1,200 - $240 = $960
The cost of a $1,200 washing machine after a discount of 1/5 (20%) is $960
3 0
3 years ago
if your average stride length is 2.6 feet how many strides will it take you to walk to grandparents house
IceJOKER [234]

The distance from your house to your grandparents house is 3.67 miles or 19377.6 ft.

So the total number of strides needed would be:

strides needed = 19377.6 ft / (2.6 ft / stride)

<span>strides needed = 7,452.92 = 7,453 strides</span>

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
Can someone please help me
Dafna1 [17]
<h3>Answer: Choice D)  3n^2</h3>

"mono" means "one". I often think of "monorail" which means "one rail" to help remember this. So "monomial" means "one term". This reduces our choices to either C or D, as they show one term each. Choices A and B are ruled out as these are binomials, showing two terms each.

Choice C shows a cubic monomial since the exponent here is 3. So the degree is 3. We can rule out choice C.

Choice D has a 2nd degree monomial because the exponent is 2. The leading coefficient is 3 as this is the number to the left of the variable term. All of choice D fits with the description of "A monomial of the 2nd degree with leading coefficient of 3"

8 0
3 years ago
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