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sweet [91]
2 years ago
14

The length of a rectangular floor is 4 feet longer than its width w. The area of the floor is 525 ft^2. A) Write a quadratic equ

ation in terms of w that represents the situation. B) What are the dimensions of the floor? Show your work.​
Mathematics
1 answer:
mina [271]2 years ago
4 0

Answer:

x^21x+25x-525-0

x^21x+25x-525-0xx^2 - 3.7_) +5^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25ensiah193

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Sholpan [36]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Listed below are head injury measurements from small cars that were tested in crashes. The measurements are in​ "hic," which is a measurement of a standard​ "head injury​ criterion," (lower ​ "hic" values correspond to safer​ cars). The listed values correspond to cars​ A, B,​ C, D,​ E, F, and​ G, respectively.

514 541 302 400 507 406 369

Find the

a.​ mean,

b.​ median,

c.​ midrange,

d. mode for the data.

Also complete parts e. and f.

e. Which car appears to be the safest?

f. Based on these limited results, do small cars appear to have about the same risk of head injury in a crash?

Answer:

a) Mean = 434.14

b) Median = 406

c) Midrange = 421.5

d) Mode = 0

e) Car C appears to be the safest

f) The small cars does not appear to have about the same risk of head injury in a crash.

Step-by-step explanation:

We are given the head injury measurements from small cars that were tested in crashes.

The measurements are in​ "hic," which is a measurement of a standard​ "head injury​ criterion.

The listed values are;

A = 514

B = 541

​C = 302

D = 400

​E = 507

F = 406

G = 369

a)​ Mean

The mean of the measurements is given by

Mean = Sum of measurements/ Number of measurements

Mean = (514 + 541 + 302 + 400 + 507 + 406 + 369)/7

Mean = 3039/7

Mean = 434.14

b)​ Median

Arrange the measurements in ascending order (low to high)

302, 369, 400, 406, 507, 514, 541

The median is given by

Median = (n + 1)/2

Median = (7 + 1)/2

Median = 8/2

Median = 4th

Therefore, the 4th measurement is the median that is 406

Median = 406

c)​ Mid-range

The midrange is given by

Midrange = (Max + Min)/2

The maximum measurement in the data set is 541

The minimum measurement in the data set is 302

Midrange = (541 + 302)/2

Midrange = 843/2

Midrange = 421.5

d)​ Mode for the data

The mode of the data set is the most repeated measurement.

302, 369, 400, 406, 507, 514, 541

In the given data set we don't have any repeated measurement therefore, there is no mode or we can say the mode of this data set is 0.

Mode = 0

e) Which car appears to be the safest?

Since we are given that the measurements are in​ "hic," which is a measurement of a standard​ "head injury​ criterion," (lower ​ "hic" values correspond to safer​ cars)

The lowest hic value corresponds to car C that is 302

Therefore, car C appears to be the safest among other cars.

f) Based on these limited results, do small cars appear to have about the same risk of head injury in a crash?

302, 369, 400, 406, 507, 514, 541

As you can notice, the hic values differ a lot from each other therefore, we can conclude that the small cars does not appear to have about the same risk of head injury in a crash.

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