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3 years ago

A rectangular room is 6.8 meters longer than it is wide, and perimeter is 29.6 meters. Find the dimensions of the room.

Mathematics

1 answer:

GuDViN [60]3 years ago

8 0

Answer:

dimensions = 10.8 X 4

Step-by-step explanation:

Breadth/ width of room = x

Length of room = 6.8 + x

perimeter = 2 ( L + B )

29.6 = 2 ( 6.8 + x + x )

29.6/2 = 6.8 + 2x

14.8 - 6.8 = 2x

8 = 2x

8/2=x

4=x

Length of room = 6.8 + x = 6.8 + 4 = 10.8 m

Breadth/ width of room = x = 4m

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2 years ago

What is the diameter of a circle with area of 113.04 square cm. use 3.14? pls fast

galina1969 [7]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

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3 years ago

HELP PLEASE NO LINKS OR FILES THEY WILL BE REPORTED

Mariana [72]

Answer:

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Step-by-step explanation:

Volume = weight times height times length = 12 times 5 times 3 = 180.

3 0

3 years ago

3) Determne the largest multiple of 9 less than 200

BaLLatris [955]

Answer:

i think 22

Step-by-step explanation:

9*22=198

8 0

3 years ago