(c) Find a solution of the differential equation that satisfies the initial condition y(1) = 5. (d) Find a solution of the diffe
rential equation that satisfies the initial condition y(5) = 1.
1 answer:
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Answer:
Reject the null hypothesis.
Step-by-step explanation:
n1 = 16
n2 = 21
S.V1 = 5.8
S.V2 = 2.4
= 0.05
= Population Variance 1 ≤ Population Variance 2
= Population Variance 1 > Population Variance 2
Test statistic value = 5.8 / 2.4 = 2.417
Degrees of freedom is n - 1
15 and 20
Critical value is = 2.2
2.417 > 2.2
we reject the null hypothesis as the critical value is greater than the test statistic.
Answer:
Q-12 The vertical length of the top part of elevator to ground is 0.8484 meters .
Q -14 The glide runway path is 24,752.47 feet .
Step-by-step explanation:
Q - 12
Given as :
The diagonal length of the elevator = e = 1.2 meters
The vertical length of the top part of elevator to ground = x meters
The angle made by elevator with ground = Ф = 45°
<u>Now, According to figure</u>
Sin angle =
Or, Sin Ф =
Or, Sin 45° =
Or, 0.707 × 1.2 meters = x
∴ x = 0.8484 meters
So, The vertical length of the top part of elevator to ground = x = 0.8484 meters
Hence,The vertical length of the top part of elevator to ground is 0.8484 meters . Answer
Q-14
Given as:
The altitude of decent of plane = H = 10,000 feet
The angle of descend = Ф = 22°
Let the glide runway path = y feet
<u>Now, According to question</u>
Tan angle =
Or, Tan Ф =
Or, Tan 22° =
Or, 0.4040 =
Or, y =
∴ y = 24,752.47
So,The glide runway path = y = 24,752.47 feet
Hence, The glide runway path is 24,752.47 feet . Answer