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Luden [163]
3 years ago
5

What’s the answer to this ?

Mathematics
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

59.44 units²

Step-by-step explanation:

The area of each triangular section can be computed from the side length (5 units) and the central angle (72°) as ...

... section area = (1/2)sin(72°)·5² ≈ 12.5·0.951057 ≈ 11.8882 . . . units²

Then 5 such sections will have an area of ...

... pentagon area = 5·11.8882 units² = 59.4410 units²

_____

<em>Comment on central angle</em>

The central angles total 360°. There are 5 equal sections, so the central angle for each of them is 360°/5 = 72°.

<em>Comment on area formula</em>

It can be occasionally handy to know that the area of a triangle can be computed from the lengths of adjacent sides and the angle between them. The formula for sides of length "a" and "b" and angle β is ...

... A = (1/2)ab·sin(β)

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I need help please either discord me the answer at F4deG0d#0988 or do it on here cause this is important.
algol13

Answer:

x= 12.5

So

2x = 25

4x = 50

6x = 75

Step-by-step explanation:

line CG is 180 degrees, so 12x + 30 = 180

x = 12.5

7 0
3 years ago
The number of entrees purchased in a single order at a Noodles &amp; Company restaurant has had an historical average of 1.7 ent
SOVA2 [1]

Answer:

t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744    

p_v =P(t_{(47)}>2.744)=0.0043  

If we compare the p value and the significance level given \alpha=0.02 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=2.1 represent the mean

s=1.01 represent the sample standard deviation

n=48 sample size  

\mu_o =1.7 represent the value that we want to test

\alpha=0.02 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 1.7, the system of hypothesis would be:  

Null hypothesis:\mu \leq 1.7  

Alternative hypothesis:\mu > 1.7  

If we analyze the size for the sample is > 30 but 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".  

Calculate the statistic

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

t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=48-1=47  

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

p_v =P(t_{(47)}>2.744)=0.0043  

Conclusion  

If we compare the p value and the significance level given \alpha=0.02 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.  

4 0
3 years ago
Read 2 more answers
What is the common ratio for the geometric sequence below, written as a fraction?
MaRussiya [10]
5/8 hope this helps!!
3 0
2 years ago
Read 2 more answers
5.<br>Find the sum of all two - digit natural numbers which are divisible by 3.​
kykrilka [37]

Answer:

1665

Step-by-step explanation:

This is AP, where:

  • First term is: 12= 3*4
  • Last term is: 99= 3*33
  • Common difference is: 3
  • Number of terms is: 33- 3= 30 (because the 4th term is included)
  • S=1/2*30(12+99)= 1665

The sum of all two - digit natural numbers which are divisible by 3 is 1665

3 0
3 years ago
There are 126 Year 7 pupils in a tennis club, out of a total of 168 pupils.
Ivenika [448]

Answer:

75% of students are year 7

Step-by-step explanation:

divide 126 by 168 which equals 75% of year 7 pupils

7 0
2 years ago
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