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Anika [276]
4 years ago
7

Students have organized

Mathematics
1 answer:
Olenka [21]4 years ago
4 0

Answer:

Part A. At least 6 hours

Part B.  In less than 2.5 hours Elijah will be behind Mercedes

Part C.  In more than 2.5 hours Elijah will be ahead Aubrey

Step-by-step explanation:

D = distance

v =speed

t = time

Formula connecting D, v and t:

D=v\cdot t

Part A.

Steve's speed: v=3.5\ mph

Distance: at least 21 miles

Time: unknown, so

3.5\cdot t\ge 21\\ \\35t\ge 210\ [\text{Multiplied by 10}]\\ \\t\ge \dfrac{210}{35}\\ \\t\ge \dfrac{30}{5}\\ \\t\ge 6

It would take Steve at least 6 hours to walk at least 21 mi on Day 1.

Part B.

Mercedes's speed: v_M=2.4\ mph

Elijan's speed: v_E=3.2\ mph

Elijan's Distance walked: D_E miles

Mercedes's Distance walked: D_M miles

Time: x hours

Mercedes is 2 miles ahead, so

D_E=3.2x\\ \\D_M=2.4x+2

Elijan will be behind when

D_E

In 2.5 hours Elijan will catch up Mercedes, and in less than 2.5 hours Elijah will be behind Mercedes.

Part C.

Aubrey's speed: v_M=3\ mph

Elijan's speed: v_E=3.2\ mph

Elijan's Distance walked: D_E miles

Aubrey's Distance walked: D_A miles

Time: x hours

At the beginning of Day 3, Elijah starts walking at the marker for Mile 42,  and Aubrey starts walking at the marker for Mile 42.5.

D_E=42+3.2x\\ \\D_A=42.5+3x

Elijan will be ahead of Aubrey when

D_E>D_A\\ \\42+3.2x> 42.5+3x\\ \\3.2x-3x>42.5-42\\ \\0.2x>0.5\\ \\2x>5\ [\text{Multiplied by 10}]\\ \\x>\dfrac{5}{2}\\ \\x>2.5\ hours

In 2.5 hours Elijan will catch up Aubrey, and in more than 2.5 hours Elijah will be ahead Aubrey.

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Answer:

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Step-by-step explanation:

We are given the following in the question:

Event: A standard pair of six-sided dice is rolled

A: rolling a sum less than 7

Sample space:

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

A:

{(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1)}

Formula:

\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}

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0.4167 is the probability of rolling a sum less than 7.

8 0
3 years ago
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