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d1i1m1o1n [39]
3 years ago
9

Fifty-five percent of the members of the junior varsity swim team wear glasses. Of the team memebers who do not wear glasses, 30

percent of them are in the 10th grade. To the nearest whole percent, what is the probability that a randomly chosen member of the JV swim team does not wear glasses and is in the 10th grade?.
Mathematics
1 answer:
GarryVolchara [31]3 years ago
8 0

the probability that a randomly chosen member of the JV swim team does not wear glasses and is in the 10th grade is 14%.

Given, the members of the junior varsity swim team wear glasses is 55%.

And the members of junior varsity swim team who do not wear glasses and in 10th grade is 30%.

Let the total no. of members be x.

So, the members of the junior varsity swim team wear glasses (55%),will be

0.55x.

The total number of members not wearing glasses, will be

x - 0.55 x = 0.45 x.

Now, the no. of members not wearing glasses and in 10th grade will be,

\frac{0.45x\times30}{100} =0.135x

We know that the probability of an event E, P(E) will be,

P(E)=\frac{No.\ of Favourable\ outcomes}{total \ no. \ of \ outcomes}

So, the probability for no. of members not wearing glasses and in 10th grade will be,

P(E)=\dfrac{0.135x}{x}

P(E)=0.135\\\\P(E)=13.5 \%

Hence the nearest whole percent of probability is 14%.

For more details on probability follow the link:

brainly.com/question/795909

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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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4 years ago
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