Answer:
Area = 20 ft²
Step-by-step explanation:
Area of a thrombus
½ × d1 × d2
½ × 5 × 8
20
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So we can only find x if other two angles of triangle is given.
In angle B as u can see boxed shape angle is there then it probably means it's a 90° angle.
We know sum of angles of triangle = 180°
{Reson :-
There's a formula to fund sum of angles of figure
- (n-2) × 180
- (3 -2) × 180
- 1 × 180
- 180°
}
Now Let's proceed
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Fill all the given values

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
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
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
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Value of x = 35°
~BrainlyVIP ⸙
Answer:
H0: μd=0 Ha: μd≠0
t= 0.07607
On the basis of this we conclude that the mean weight differs between the two balances.
Step-by-step explanation:
The null and alternative hypotheses as
H0: μd=0 Ha: μd≠0
Significance level is set at ∝= 0.05
The critical region is t ( base alpha by 2 with df=5) ≥ ± 2.571
The test statistic under H0 is
t = d/ sd/ √n
Which has t distribution with n-1 degrees of freedom
Specimen A B d = a - b d²
1 13.76 13.74 0.02 0.004
2 12.47 12.45 0.02 0.004
3 10.09 10.08 0.01 0.001
4 8.91 8.92 -0.01 0.001
5 13.57 13.54 0.03 0.009
<u>6 12.74 12.75 -0.01 0.001</u>
<u>∑ 0.06 0.0173</u>
d`= ∑d/n= 0.006/6= 0.001
sd²= 1/6( 0.0173- 0.006²/6) = 1/6 ( 0.017294) = 0.002882
sd= 0.05368
t= 0.001/ 0.05368/ √6
t= 0.18629/2.449
t= 0.07607
Since the calculated value of t= 0.07607 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the mean weight differs between the two balances.