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3 years ago

The length of a rectangle is 15 feet. The width is 6 feet. What is the perimeter of the

Mathematics

2 answers:

weqwewe [10]3 years ago

7 0

The perimeter is 42 feet because 15+15+6+6 ( all the sides) adds up to the perimeter

Pavlova-9 [17]3 years ago

5 0

we know that, the opposite sides of a rectangle are equal so u can simply 2 times the sum ogf length and breadth..

in your question,

length(l)=15 ft

breadth(b)=6 ft

therefore,

perimeter(p)=2(l+b)

=2(15+6)

=2×21

=42 ft

hope this helps!!!

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What number is 40% of 720

Zarrin [17]

You would have to set you problem up like
x 40
____=_____
720 100

you would multiply 40 by 720 then you would divide 28,800 by 100 your answer will be 288.

3 0

2 years ago

The length of the rectangle is 4cm less than its width . what are the dimensions of the rectangle if its area is 140

melomori [17]

Let the width of the rectangle be x
Then length =x-4
Area=x(x-4)=140
x²-4x=140
x²-4x-140=0
x²-14x+10x-140=0
x(x-14)+10(x-14)=0
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Therefore,length=14 cm and width=10 cm

8 0

3 years ago

The sum of 4 consecutive integers is 354.<br> What is the second number in this sequence?

Ulleksa [173]

354 / 4 = 88.5

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87 + 88 + 89 +90 =

2nd number is 88

8 0

3 years ago

Which product of prime polynomials is equivalent to 36x3 – 15x2 – 6x??

eimsori [14]

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4 0

2 years ago

Read 2 more answers

The article "Occurrence and Distribution of Ammonium in Iowa Groundwater" (K. Schilling, Water Environment Research, 2002:177–18

Papessa [141]

Answer: 95% confidence interval for the difference between the proportions would be (1.31, 1.39).

Step-by-step explanation:

Since we have given that

Number of alluvial wells = 349

Number of quaternary wells = 143

Number of alluvial wells that had concentrations above 0.1 = 182

Number of quaternary wells that had concentrations above 0.1 = 112

Average of alluvial wells = 0.27

Standard deviation = 0.4

Average of quaternary wells = 1.62

Standard deviation =1.70

So, 95% confidence interval gives

alpha = 5% level of significance.

$\dfrac{\alpha}{2}=2.5\%\\\\z_{\frac{\alpha}{2}}=1.96$

So, 95% confidence interval becomes,

$(1.62-0.27)\pm 1.96\sqrt{\dfrac{0.4^2}{349}+\dfrac{1.7^2}{143}}\\\\=1.35\pm 1.96\times 0.020\\\\=(1.35-0.040,1.35+0.040)\\\\=(1.31,1.39)$

Hence, 95% confidence interval for the difference between the proportions would be (1.31, 1.39).

6 0

3 years ago