A 20-ounce candle is expected to burn for 60 hours. A 12-ounce candle is expected to burn for 36 hours. Assuming the variables a

Answer:

The resulting value is 2.381

Step-by-step explanation:

Using the information presented in the question, we will calculate the evidence supporting the professor's hypothesis

Given that:

x₁ = 74,

n₁ = 36

s₁ = 8

x₂ = 68

n₂ = 36

s₂ = 10

The hypotheses can be outlined as:

The critical value is t₃₆+₃₆-₂,₀.₀₁ = t₇₀,₀.₀₁

thus,

t₃₆+₃₆-₂,₀.₀₁ = t₇₀,₀.₀ = 2.381

This implies a range of -2.381 to 2.381

Thus, we can support the professor's assertion.

Answer:

Width of the rectangular prism

To determine the volume of the rectangular prism, we utilize the following formula:

v = lwh

Where:

v = volume

l = length

w = width

h = height

Now,

The variables provided in the problem include:

v = 138.24 cubic inches

h =  9.6 inches

l = 3.2 inches

w = ?

By substituting these values into the volume formula, we have:

138.24 = 3.2 * w * 9.6

138.24 = 30.72 * w

w = 138.24/30.72

w = 4.5

⇒ Therefore, the width of the rectangular prism measures 4.5 inches.

For this scenario, we can visualize that all points form a triangle. The three vertices are at the pitcher's mound, home plate, and the location where the outfielder catches the ball. We know two sides of the triangle and the angle that lies between these two sides.

<span>Using the cosine law, we can find the unknown third side. The formula to apply is:</span>

c^2 = a^2 + b^2 - 2ab cos θ

Where:

a = 60.5 ft

b = 195 ft

θ = 32°

Substituting the provided values results in:

c^2 = (60.5)^2 + (195)^2 - 2(60.5)(195) cos(32)

c^2 = 3660.25 + 38025 - 20009.7

c^2 = 21,675.56

c = 147.23 ft

<span>Thus, the distance the outfielder throws the ball towards home plate is approximately 147.23 ft.</span>

Answer:

a) 2.60x10^-7

b) 5.31x10^-5

c) 2.19x10^-3

Step-by-step explanation:

X=the number of hits

Probability can be determined by taking the number of successful outcomes and dividing it by the total number of possible outcomes.

Now

a) P(X=6)=P({1, 1, 1, 1, 1, 1})=6/40*5/39*4/38*3/37*2/36*1/35=2.60x10^-7

b) P(X=5)=P({0, 1, 1, 1, 1, 1})+P({1, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 1, 0}), and each scenario carries the same probability

P(X=5)=6P({1, 1, 1, 1, 1, 0})=6*(6/40*5/39*4/38*3/37*2/36*34/35)=5.31x10^-5

c) P(X=4)=P({0, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 0, 0}), where each case has an identical probability.

Answer:

II AB = CD

Step-by-step explanation:

|AB| = |CD|