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Alona [7]
3 years ago
12

Help please! I have choral practice...

Mathematics
2 answers:
Maslowich3 years ago
7 0
<span>This polygon is composed of a right triangle and <span>a parallelogram.

The area of right angle :
A_\Delta=\frac{1}{2}\cdot15\cdot8=\frac{1}{2}\cdot120=\boxed{60\ (cm^2)}

The area of the </span></span><span>parallelogram</span>:
A_P=15\cdot(13-8)=15\cdot5=\boxed{75\ (cm^2)}

The area of the polygon is equal:A=A_\Delta+A_P

Therefore, the answer is:
\boxed{\boxed{A=60\ cm^2+75\ cm^2=135\ cm^2}}
zalisa [80]3 years ago
7 0

Slice the picture in half, across the middle dotted line.

The top half is a triangle.  Its base is 15cm and its height is 8cm .
Area of a triangle is (1/2) (base x height) = (1/2) (120cm²) = 60 cm².

The bottom half is a parallelogram.  Its base is 15cm and its height is 5cm.
Area of a parallelogram is (base x height) = (15cm x 5cm) = 75 cm².

Add the two pieces together.  (60cm²) + (75 cm²) = <em>135 cm²

</em>
Breathe from the diaphragm, enunciate, and watch the conductor.<em>


</em>
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inn [45]

Answer:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

df = n-1=34-1=33

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

Step-by-step explanation:

Information provided

\bar X=12.2 represent the sample mean fould against

s=1.6 represent the sample standard deviation

n=34 sample size  

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)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is higher than 11.5 fouls per game:  

Null hypothesis:\mu \leq 11.5  

Alternative hypothesis:\mu > 11.5  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

The statistic is given by:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

P value

The degreed of freedom are given by:

df = n-1=34-1=33

Since is a one side test the p value would be:  

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

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2 years ago
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