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Vlad1618 [11]
3 years ago
15

How do I find the perimeter of a triangular pyramid with a base edge 6

Mathematics
1 answer:
givi [52]3 years ago
5 0
Perimeter=4(6)= 24 
because you have 4 sides and each side must be equal in a triangular prism.
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Pls help me i need to do all my math for to day and i dont know how to do it <br><br> -11x(x+12)
Vesna [10]

Answer:

If you want to simplify, -11x^{2}-132x

Step-by-step explanation:

4 0
3 years ago
Two trains begin in Milford and end in Pinkerton, which is 300 miles away. Train A leaves Milford at 10:00 a.m. and travels at a
Dvinal [7]

Answer:

A. Train A

Step-by-step explanation:

Hope this helps!!

7 0
3 years ago
132% of what number is 25 nearest tenth
trapecia [35]

Answer:

100%

Step-by-step explanation:

3 0
3 years ago
In doing so, you collect a random sample of 50 salespersons employed by his company, which is thought to be representative of sa
Deffense [45]

Answer:

z=\frac{0.36 -0.45}{\sqrt{\frac{0.45(1-0.45)}{50}}}=-1.279  

p_v =P(z

And we can use the following code to find it  "=NORM.DIST(-1.279,0,1,TRUE)"

Step-by-step explanation:

Assuming this complete problem: "The CEO of a software company is committed to expanding the proportion of highly qualified women in the organization's staff of salespersons. He believes that the proportion of women in similar sales positions across the country is less than 45%. Hoping to find support for his belief, he directs you to test

H0: p .45 vs H1: p < .45.

In doing so, you collect a random sample of 50 salespersons employed by his company, which is thought to be representative of sales staffs of competing organizations in the industry. The collected random sample of size 50 showed that only 18 were women.

Compute the p-value associated with this test. Place your answer, rounded to 4 decimal places, in the blank. For example, 0.3456 would be a legitimate entry."

1) Data given and notation

n=50 represent the random sample taken

X=18 represent the number of women in the sample selected

\hat p=\frac{18}{50}=0.36 estimated proportion of women in the sample

p_o=0.45 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v 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 proportion of women is less than 0.45:  

Null hypothesis:p\geq 0.45  

Alternative hypothesis:p < 0.45  

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{0.36 -0.45}{\sqrt{\frac{0.45(1-0.45)}{50}}}=-1.279  

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 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

And we can use the following code to find it  "=NORM.DIST(-1.279,0,1,TRUE)"

4 0
3 years ago
How do you factor b^2-81
Naddik [55]
The middle number is 0 and the last is -81Factoring means we want something like(b+_)(b+_)We need two numbers that add together to get and Multiply to get -81which are 9 and -9:<span><span>9+-9 = 0 and </span><span>9*-9 = -81</span></span>
Fill in the blanks in
(b+_)(b+_)
with 9 and -9 to get...<span><span>(<span>b+9</span>)</span><span>(<span>b−9</span>)</span></span>
The answer is : (b+9)(b−9)
<span>
</span>
7 0
3 years ago
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