We'll use standard labeling of right triangle ABC, C=90 degrees, legs a, b, hypotenuse c.
11.
Right triangle, cliff peak A, boat B, angle opposite cliff is B=28.9 deg. adjacent leg a=65.7 m, cliff height is leg b.
tan B = b/a
b = a tan B = 65.7 tan 28.9° = 36.3 m
12.
Similar story, boat at B, opposite b=3.5 m, rope c=12 m
sin B = b/c
B = arcsin b/c = arcsin (3.5/12) = 17.0°
13.
c=124 m, A=58°
sin A = a/c
a = c sin A = 124 sin 58 = 105.2 m
14.
That's a hypotenuse c=4-1.2 = 2.8 m to a height b=1.8m so
cos A = b/c
A = arccos b/c = arccos (1.8/2.8) = 50.0°
15.
Not a right triangle, an isosceles triangle. Half of it is a right triangle with hypotenuse one arm, c=9.8 cm and angle opposite half the base of B=62/2=31°. We're after d=2b:
sin B = b/c
b = c sin B
d = 2b = 2 c sin B = 2(9.8) sin 31 = 10.1 cm
Almost equilateral
Answer:

Step-by-step explanation:
you set each of them equal to zero,


and solve for 'm'
Answer:
Step-by-step explanation:
From the given information, we can compute the table showing the summarized statistics of the two alloys A & B:
Alloy A Alloy B
Sample mean

Equal standard deviation

Sample size

Mean of the sampling distribution is :

Standard deviation of sampling distribution:

Hypothesis testing.
Null hypothesis: 
Alternative hypothesis: 
The required probability is:
![P(\overline X_A - \overline X_B>4|\mu_A - \mu_B) = P\Big (\dfrac{(\overline X_A - \overline X_B)-\mu_{X_A-X_B}}{\sigma_{\overline x_A -\overline x_B}} > \dfrac{4 - \mu_{X_A-\overline X_B}}{\sigma _{\overline x_A - \overline X_B}} \Big) \\ \\ = P \Big( z > \dfrac{4-0}{1.2909}\Big) \\ \\ = P(z \ge 3.10)\\ \\ = 1 - P(z < 3.10) \\ \\ \text{Using EXCEL Function:} \\ \\ = 1 - [NORMDIST(3.10)] \\ \\ = 1- 0.999032 \\ \\ 0.000968 \\ \\ \simeq 0.0010](https://tex.z-dn.net/?f=P%28%5Coverline%20X_A%20-%20%5Coverline%20X_B%3E4%7C%5Cmu_A%20-%20%5Cmu_B%29%20%3D%20P%5CBig%20%28%5Cdfrac%7B%28%5Coverline%20X_A%20-%20%5Coverline%20X_B%29-%5Cmu_%7BX_A-X_B%7D%7D%7B%5Csigma_%7B%5Coverline%20x_A%20-%5Coverline%20x_B%7D%7D%20%3E%20%5Cdfrac%7B4%20-%20%5Cmu_%7BX_A-%5Coverline%20X_B%7D%7D%7B%5Csigma%20_%7B%5Coverline%20x_A%20-%20%5Coverline%20X_B%7D%7D%20%20%20%5CBig%29%20%5C%5C%20%5C%5C%20%3D%20P%20%5CBig%28%20z%20%3E%20%5Cdfrac%7B4-0%7D%7B1.2909%7D%5CBig%29%20%5C%5C%20%5C%5C%20%3D%20P%28z%20%5Cge%203.10%29%5C%5C%20%5C%5C%20%3D%201%20-%20P%28z%20%3C%203.10%29%20%5C%5C%20%5C%5C%20%5Ctext%7BUsing%20EXCEL%20Function%3A%7D%20%5C%5C%20%5C%5C%20%20%3D%201%20-%20%5BNORMDIST%283.10%29%5D%20%20%5C%5C%20%5C%5C%20%3D%201-%200.999032%20%5C%5C%20%5C%5C%200.000968%20%5C%5C%20%5C%5C%20%5Csimeq%20%200.0010)
This implies that a minimal chance of probability shows that the difference of 4 is not likely, provided that the two population means are the same.
b)
Since the P-value is very small which is lower than any level of significance.
Then, we reject
and conclude that there is enough evidence to fully support alloy A.