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marin [14]
3 years ago
15

Find the surface area AND the volume of the triangular prism

Mathematics
2 answers:
Whitepunk [10]3 years ago
5 0
Surface Area: 1146.62
Volume: 2081.78
siniylev [52]3 years ago
5 0

Answer:

SA=1200 in squared. V=2160 in cubed.

Step-by-step explanation:

Let's start with the surface area. The area of of a triangle is 1/2 x b x h. But we have 2 congruent triangle on the prism, so we have this.

2 x 1/2 x b x h.

The 2 and 1/2 cancel so we are left with b x h.

The other sides are easy.

Find the area of the sides we know that have base times height.

16 x 30 + 16 x 9 + 30 x 9...

To find the last side, we need to find the hypotenuse.

To find it, we use the pythagorean theorem.

a²+b²=c²

Use the values of the right triangle's dimensions and plug them in.

16²+30²=c²

256+900=c²

c²=1156

c=34

now using this, we have

16 x 30 + 16 x 9 + 30 x 9 + 9 x 34.

Now solve!

SA = 1200 in squared

The volume is easy.

All we need is the area of the triangular side and multiply it by the depth, 9.

16 x 30 x 1/2 =

240

240 x 9=

2160. in cubed

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project.
Kobotan [32]

Answer:

t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457

p_v =P(t_{75}  

Comparing the p value with a significance level for example \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.  

Step-by-step explanation:

Data given and notation

\bar X_{A}=3.7 represent the mean for athletes  

\bar X_{NA}=4.1 represent the mean for non athletes  

s_{A}=1.1 represent the sample standard deviation for athletes  

s_{NA}=1.3 represent the sample standard deviation for non athletes

n_{A}=32 sample size for the group 2  

n_{NA}=45 sample size for the group 2  

\alpha=0.01 Significance level provided  

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the population mean for athletes is lower than the population mean for non athletes, the system of  hypothesis would be:  

Null hypothesis:\mu_{A}-\mu_{NA}\geq 0  

Alternative hypothesis:\mu_{A} - \mu_{NA}< 0  

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{A}-\bar X_{NA})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{NA}}{n_{NA}}}} (1)  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457

P value  

We need to find first the degrees of freedom given by:

df=n_A +n_{NA}-2=32+45-2=75

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

p_v =P(t_{75}  

Comparing the p value with a significance level for example \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.  

5 0
3 years ago
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