See the picture attached to better understand the problem
we know that
in the right triangle ABC
tan 64°=AB/AC------> AB=AC*tan 64°-----> AB=x*tan 64°---> equation 1
in the right triangle ABD
tan 43°=AB/DA----> AB=DA*tan 43°---> AB=(240+x)*tan 43°---> equation 2
equate equation 1 and equation 2
x*tan 64°-=(240+x)*tan 43°---->x*tan 64=240*tan 43+x*tan 43
x*[tan 64-tan 43]=240*tan 43-----> x=240*tan 43/[tan 64-tan 43]
x=200.22 ft
AB=x*tan 64----> AB=200.22*tan 64-----> AB=410.51 ft
the answer is410.51 ft
Answer: 14/25 = 56/100
11/20 = 55/100
14/25 > 11/20
Step-by-step explanation:
The least common multiple of 20 and 25 is 100.
(100 = 20*5, 25*4)
So we can take the common denominator 100.
14*4/25*4 =56/100. ∴14/25 = 56/100
11*5/20*5 =55/100. ∴11/20 = 55/100
56>55
∴56/100 > 55/100
∴14/25 > 11/20
Hope it's good;)
Answer:
Step-by-step explanation:
Hello!
The objective of this experiment is to test if two different foam-expanding agents have the same foam expansion capacity
Sample 1 (aqueous film forming foam)
n₁= 5
X[bar]₁= 4.7
S₁= 0.6
Sample 2 (alcohol-type concentrates )
n₂= 5
X[bar]₂= 6.8
S₂= 0.8
Both variables have a normal distribution and σ₁²= σ₂²= σ²= ?
The statistic to use to make the estimation and the hypothesis test is the t-statistic for independent samples.:
t=
a) 95% CI
(X[bar]_1 - X[bar]_2) ± *
Sa²= = = 0.5
Sa= 0.707ç
(4.7-6.9) ± 2.306*
[-4.78; 0.38]
With a 95% confidence level you expect that the interval [-4.78; 0.38] will contain the population mean of the expansion capacity of both agents.
b.
The hypothesis is:
H₀: μ₁ - μ₂= 0
H₁: μ₁ - μ₂≠ 0
α: 0.05
The interval contains the cero, so the decision is to reject the null hypothesis.
<u>Complete question</u>
a. Find a 95% confidence interval on the difference in mean foam expansion of these two agents.
b. Based on the confidence interval, is there evidence to support the claim that there is no difference in mean foam expansion of these two agents?
Answer: I’m not really sure what the correct answer is if you could help that would be great so let me know
Step-by-step explanation:
Given:
Circular racetrack with a diameter of 1/2 mile
Find: how far does a car travel in one lap around the track?
We need to find the circumference of the racetrack.
Circumference is multiplying pi to the diameter of the racetrack.
Circumference = 3.14 * 1/2 mile
Circumference = 3.14/2 mile
Circumference = 1.57 miles rounded to the nearest tenth is 1.60 miles.