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umka2103 [35]
4 years ago
12

Which of the following expressions is equal to -×^2-81?

Mathematics
1 answer:
Vsevolod [243]4 years ago
8 0
-x^2-81=-(x^2+81)=-(x^2-(9i)^2)=-(x+9i)(x-9i)=(-x-9i)(x-9i)\to C.
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The length of an English essay is 180 lines long with an average of 15 words per line If it is written with an average of 10 wor
Ivenika [448]

Answer:

18 lines would be needed to complete the essay.

Step-by-step explanation:

To solve this you need to divide the amount of words per line by the number of lines.

6 0
3 years ago
HELP Some friends decided to equally split the cost of the apples to make applesauce. The expression 18b5 represents how much mo
lara [203]
<span>B) The cost per bushel of apples</span>
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3 years ago
Read 2 more answers
5 feet<br> 4 feet<br> What is the perimeter?<br> 16 feet<br> 14 feet<br> 18 feet
brilliants [131]

perimiter means all the way around, and im assuming this is a 4-sided shape, so 5+5+4+4 = 18 ft.

8 0
3 years ago
We are conducting a hypothesis test to determine if fewer than 80% of ST 311 Hybrid students complete all modules before class.
Juli2301 [7.4K]

Answer:

z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124  

Null hypothesis:p\geq 0.8  

Alternative hypothesis:p < 0.8  

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

p_v =P(Z  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis:p\leq 0.8  

Alternative hypothesis:p > 0.8

p_v =P(Z>2.124)=0.013  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

\hat p=\frac{97}{110}=0.882 estimated proportion of students who completed the modules before class

p_o=0.8 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v{/tex} represent the p value (variable of interest)  2) Concepts and formulas to use  We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%:  Null hypothesis:[tex]p\geq 0.8  

Alternative hypothesis:p < 0.8  

When we conduct a proportion test we need to use the z statistic, 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.

3) Calculate the statistic  

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

z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124  

4) Statistical decision  

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 assumed \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  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis:p\leq 0.8  

Alternative hypothesis:p > 0.8

p_v =P(Z>2.124)=0.013  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

5 0
4 years ago
Jacobs &amp; Johnson, an accounting firm, employs 15 accountants, of whom 9 are CPAs. If a delegation of 4 accountants is random
Artemon [7]
1st acc = 6/14 chance 
<span>2nd acc = 5/13 </span>
<span>3rd acc = 4/12 </span>

<span>6/14 * 5/13 * 4/12 = 0.0549450549 </span>

<span>round to 3 decimal places ==> 0.055
Please put me on brainlest if its correct! Thank you!</span>
7 0
3 years ago
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