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deff fn [24]
3 years ago
13

The school field trip needs at least 1 chaperone for every 12 students. The school has 121 students who want to go on the field

trip. If c represents the number of chaperones, and the school does not yet have enough to hold the field trip, how many chaperones might there be? (multiple choice)

Mathematics
1 answer:
nordsb [41]3 years ago
7 0

Answer:

Step-by-step explanation:

The answer is A: <em>c<11</em>

This is because the required amount of chaperones is 11 because the total number of students (121) divided by the required number of students per chaperone (12) is results in 10 remainder 1, and 11 required chaperones.

If the school does not have have enough chaperones that means they must have less than 11 chaperones, so the answer is c (the number of chaperones) is less than 11.

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When a jug is half-filled with marbles, it weighs 2.6 kg. If the jug weighs 4 kg when full, find the weight of the empty jug.
Fittoniya [83]

Answer:

1.2 kg

Step-by-step explanation:

filling the jug up half way adds 1.4 kg to the weight.

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What is the sum of 6 and 7/8+ 4 and 5/8
Reptile [31]
THE ANSWER IS 11 AND 1/2


so explanation
6 and 7/8+4 and 5/8=6+7/8+4+5/8=6+4+7/8+5/8=10+(7+5)/8=10+12/8=10+1+4/8=11+4/8=11+1/2


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Please help me find X.
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The correct answer is 54.3.
4 0
3 years ago
2. Let A, B and c be three towers. The bearing of B from A is 120° and angle BC is 68º. The distance from Tower A to Tower B is
g100num [7]

(i) The bearing of B from A and angle x are supplementary, then:

\begin{gathered} 120\degree+x=180\degree \\ x=180\degree-120\degree \\ x=60\degree \end{gathered}

(ii)

From the above diagram, the bearing angle of A from B is:

120\degree+180\degree=300\degree

(iii)

From the above diagram, the bearing angle of C from B is:

180\degree+52\degree=232\degree

(iv)

Applying the law of cosines with the sides AB = 145m and CB = 240 m, and the angle ABC = 68°, the distance AC is:

\begin{gathered} AC^2=AB^2+CB^2-2\cdot AB\cdot CB\cdot cos\left(\angle ABC\right) \\ AC^2=145^2+240^2-2\cdot145\cdot240\cdot cos(68\degree) \\ AC^2=21025+57600-69600\cdot cos(68\operatorname{\degree}) \\ AC^2=52552.3811 \\ AC=\sqrt{52552.3811} \\ AC\approx229\text{ m} \end{gathered}

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1 year ago
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