Answer: option c
Step-by-step explanation: with a two-tailed hypothesis she can find the way to go deep on the study, she will be able to increase the statistical power. She will be able to separate in specifics groups.
Answer:
89.1° or -1.4°
Step-by-step explanation:
1. Location:
You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.
The French troops are about 50 m below you and 1.2 km distant.
2. Finding the firing angle
Data:
R = 1200 m
u = 600 m/s
h = -50 m (the height of the target)
a = 9.8 m/s²
We have two conditions.
Horizontal distance
(1) 1200 = 600t cosθ
Vertical distance
(2) -50 = 600t sinθ - 4.9t²
Divide each side of (1) by 600cosθ.

Substitute (3) into (2)

Recall that
(5) sec²θ = 1/cos²θ = tan²θ + 1
Substitute (5) into (4)

Set up a quadratic equation

Solve for θ
Use the quadratic formula.
tanθ = 61.249 or -0.025
θ = arctan(61.249) = 89.1° or
θ = arctan(-0.025) = -1.4°
Answer:
(- 4, - 12 ) , (4, 12 )
Step-by-step explanation:
Given the 2 equations
y = 3x → (1)
y = x² + 3x - 16 → (2)
Substitute y = x² + 3x - 16 into (1)
x² + 3x - 16 = 3x ( subtract 3x from both sides )
x² - 16 = 0 ( add 16 to both sides )
x² = 16 ( take the square root of both sides )
x = ±
= ± 4
Substitute these values into (1) for corresponding values of y
x = - 4 : y = 3 × - 4 = - 12 ⇒ (- 4, - 12 )
x = 4 : y = 3 × 4 = 12 ⇒ (4, 12 )
The speed of wind and speed of plane in still air are 23 and 135
km/h respectively.
<u>Step-by-step explanation:</u>
Let the speed of wind and speed of plane in still air are w and p km/h respectively.
The effective speed on onward journey was
................(1)
The effective speed on return journey was
..............(2)
Adding equation (1) and equation (2) we get,
⇒ 
⇒ 
⇒ 
Putting value of
in
we get:
⇒ 
⇒ 
⇒ 
Therefore ,The speed of wind and speed of plane in still air are 23 and 135
km/h respectively.
Answer:
a) f(-6) = 37
b) f(8) = 65
Step-by-step explanation:
f(-6) = (-6)² + 1 = 36 + 1 = 37
f(8) = 8²+ 1 = 64 + 1 = 65