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3 years ago

if one out of every 12 students at a school share a locker,how many share a locker in a school of 456 students?

Mathematics

2 answers:

Scilla [17]3 years ago

5 0

1 out of 12 can be written as 1/12

now multiply 456 by 1/2

this is the same as dividing 456 by 12

456/12 = 38 students share a locker

Sidana [21]3 years ago

4 0

456/12=38

answer
38 share lockers

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A line segment has endpoints at (–1, 4) and (4, 1). Which reflection will produce an image with endpoints at (–4, 1) and (–1, –4

katrin2010 [14]

Answer:- A reflection of the line segment across the line y = –x .

Explanation:-

A reflection over the line y = -x, the x-coordinate and y-coordinate interchange their places and they are negated (the signs are changed).

Given :- A line segment has endpoints at (–1, 4) and (4, 1) such that it reflects produce an image with endpoints at (–4, 1) and (–1, –4).

(-1, 4)→(-4, 1) and

(4, 1)→(-1, -4)

Thus this shows a reflection of the line segment across the line y = –x.

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3 years ago

Read 2 more answers

At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more

I am Lyosha [343]

Answer:

The probability is $P( X \le 4 ) = 0.0054$

Step-by-step explanation:

From the question we are told that

The percentage that are on time is p = 0.76

The sample size is n = 11

Generally the percentage that are not on time is

$q = 1- p$

$q = 1- 0.76$

$q = 0.24$

The probability that no more than 4 of them were on time is mathematically represented as

$P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)$

=> $P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}$

$= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}$

$P( X \le 4 ) = 0.0054$

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3 years ago

Simplify 4 times square root 24 minus 4 square root 8 plus square root of 98

Mrrafil [7]

Answer:117.59

Step-by-step explanation: sorry if i am wrong

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2 years ago

How do you write an equation involving absolute value

sweet [91]

Ex: l -3 l = 3.

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THE ANSWER WILL ALWAYS BE POSITIVE (except...)

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3 years ago

Respond pls <br> To get some points

Ipatiy [6.2K]

<h2>Answer:</h2>

1/2

<h2>Step-by-step explanation:</h2><h3>Known :</h3>

The school football team has a match tomorrow
The probability of the team winning is 1/6
The probability of the team drawing is 2/6

<h3>Asked :</h3>

The probability of the team losing

<h3>Solution :</h3>

Probability of the team losing the match = 1 - 1/6 - 2/6

Probability of the team losing the match = 6/6 - 1/6 - 2/6

Probability of the team losing the match = 3/6

Probability of the team losing the match = 1/2

<h3>Conclusion :</h3>

The probability of the team losing the match is 1/2

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2 years ago