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3 years ago

The length of a rectangular room is 7.4m and it's width is 5.5m. Find the area of the room.

Mathematics

1 answer:

oee [108]3 years ago

7 0

Area of a rectangle is
$length\times width$
(7.4×5.5)m
=40.7m
therefore the area of the room is 40.7m

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1 + 1? <br> .......……………………

masya89 [10]

Hmm difficult one but I think ima have to go with 2?????

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3 years ago

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Write the least common multiple of 120 and 80

Olin [163]

The least common multiple would be 240.
120(2)=240 AND
80(3)=240

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3 years ago

the length of the base of an isosceles triangle is 12 inches. Each base angle measures 50. which is a feasible plan to find the

kifflom [539]

To get the congruent angles we shall use the sine rule since we have the length of at least 1 side and the values of the angles.
Base length=12 inches
Base angles=50°
Third angle=180-(100)=80°
thus applying the sine rule we shall have:
a/sin A=b/sin B=c/sin C
thus we can get the congruent side as follows:
a/sin 50= 12/sin 80
a=(12 sin 50)/sin 80
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3 years ago

AB and BC share point B to form AC. If AC=7 and BC=5, then what is the length of AB?

AnnZ [28]

$|AC|=7;\ |BC|=5\\\\|AC|=|AB|+|BC|\\\\|AB|+5=7\ \ \ \ \ |subtract\ 5\ from\ both\ sides\\\\\boxed{|AB|=2}$

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3 years ago

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Two sections of a class took the same quiz. Section A had 15 students who had a mean score of 80, and Section B had 20 students

Nezavi [6.7K]

Answer: A. 85.7

Step-by-step explanation:

Given : Two sections of a class took the same quiz.

Section A had 15 students who had a mean score of 80, and Section B had 20 students who had a mean score of 90.

We know that , $\text{Mean}=\dfrac{\text{Sum of observations}}{\text{No. of observations}}$

Then , for section A :

$\text{Mean score }=\dfrac{\text{Sum of scores in sec A}}{\text{No. of students}}$

$\Rightarrow\ 80=\dfrac{\text{Sum of scores in sec A}}{15}\\\\\Rightarrow\ \text{Sum of scores in sec A}=80\times15=1200$

Similarly in Section B, $\text{Sum of scores in sec B}=90\times20=1800$

Total scores = Sum of scores in sec A+Sum of scores in sec B

=1200+1800=3000

Total students = Students in sec A +Students in sec B

=15+20=35

Now , the mean score for all of the students on the quiz = $\dfrac{\text{Total score}}{\text{Total students}}$

$=\dfrac{3000}{35}=85.7142857143\approx85.7$

Hence, the approximate mean score for all of the students on the quiz = 85.7

Thus , the correct answer is option A. 85.7.

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2 years ago