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Mademuasel [1]
3 years ago
14

Please help will give brainliest

Mathematics
1 answer:
garri49 [273]3 years ago
5 0

Answer:

b = 17

Step-by-step explanation:

For this triangle we have to

a=18.2\\B=62\°\\C=48\°

We want to find the length of b

We know that the sum of the internal angles of a triangle is 180 °

So

A + 62 +48=180\\\\A=180-62-48\\\\A=70\°

Now we use the sine theorem to find the length of b:

\frac{sin(A)}{a}=\frac{sin(B)}{b}=\frac{sin(C)}{c}

Then:

\frac{sin(A)}{a}=\frac{sin(B)}{b}\\\\b=\frac{sin(B)}{\frac{sin(A)}{a}}\\\\b=a*\frac{sin(B)}{sin(A)}\\\\b=(18.2)\frac{sin(62)}{sin(70)}\\\\b=17

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Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders

Step-by-step explanation:

The null hypothesis is that  the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders

H0:  μL > μs      against the claim Ha:  μL≤ μs

the alternate hypothesis is the mean days surfed for all long boarders isless or equal to  the mean days surfed for all short boarders (because long boards can go out in many different surfing conditions)

The test statistic is

t= x1- x2/  √s1/n1+ s2/n2

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Longboards

Mean

ˉx=∑x/n=4+8+9+4+9+7+9+6+6+11+15+13+16+12+10+12+18+20+15+10+15+19+21+9+22+19+23+13+12+10/30

=377/30

=12.5667

Longboard Variance S2=[∑dx²-(∑dx)²/n]/n-1

=[831-(-13)²/30]/29

=831-5.6333/29

=825.3667/29

=28.4609

Shortboard Mean

ˉx=∑x/n=6+4+6+6+7+7+7+10+4+6+7+5+8+9+4+15+13+9+12+11+12+13+9+11+13+15+9+19+20+11/30

=288/30

=9.6

Shortboard Variance S2=[∑x²-(∑x)²/n]/n-1

=[ 3270-(288)2/30]/29

=3270-2764.8/29

=505.2/29

=17.4207

2) Putting values in the test statistic

t=|x1-x2|/√S²1/n1+S²2/n2

t =|12.5667-9.6|/√28.4609/30+17.4207/30

t =|2.9667|/√0.9487+0.5807

t=|2.9667|/√1.5294

t=|2.9667|/1.2367

t=2.3989

3) Degree of freedom =n1+n2-2=30+30-2=58

4) The critical region is t ≤ t(0.05) (58) =1.6716

5) Since the calculated t= 2.4 does not fall in the critical region t(0.05) (58)  ≤ 1.6716 we do not reject H0.

The p-value is 0.008969. The result is significant at p <0 .05.

6 0
3 years ago
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