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Goshia [24]
3 years ago
5

The area of a rectangular wall of a barn is 144 square foot. Its length is 10 foot longer than the width. Find the length and wi

dth of the wall of the barn.
The width is ___ feet?
Mathematics
1 answer:
Nimfa-mama [501]3 years ago
8 0

9514 1404 393

Answer:

  • width: 8 ft
  • length: 10 ft

Step-by-step explanation:

The area is the product of length and width:

  A = LW

  144 = (w+10)(w)

  w^2 +10w -144 = 0 . . . . . . subtract 144 and put in standard form

  (w +18)(w -8) = 0 . . . . . . . . factor

  w = -18 or 8 . . . . . . . values that make the factors zero, -18 is extraneous

The width of the wall is 8 ft; the length is 18 ft.

_____

<em>Additional comment</em>

The equation is factored by looking for factors of -144 that have a sum of 10. We can choose from the different ways 144 can be factored:

  -144 = -1(144) = -2(72) = -3(48) = -4(36) = -6(24) = -8(18) = -9(16) = -12(12)

The respective sums are 143, 70, 45, 32, 18, 10, 7, 0.

So, the factors of -144 that have a sum of 10 are {-8, 18}. These are the constants that go into the binomial factors.

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Answer:

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Step-by-step explanation:

5 0
3 years ago
A rectangular solid has sides of 7 cm, 9 cm, and 11 cm. What is its surface area?
otez555 [7]

Answer:

Surface area is, 478 square centimeter

Step-by-step explanation:

Surface area (S)of rectangle is given by:

S = 2(lw+wh+hl)                 .....[1]

where,

l is the length

w is the width

h is the height of the rectangle respectively.

As per the statement:

A rectangular solid has sides of 7 cm, 9 cm, and 11 cm

⇒ l = 7 cm , w = 9 cm and h = 11 cm

Substitute in [1] we have;

S = 2(7 \cdot 9+9 \cdot 11+11 \cdot 7)

⇒S = 2(63+99+77) = 2 \cdot 239 = 478 cm^2

Therefore, the surface area is, 478 square centimeter

7 0
3 years ago
A college entrance exam company determined that a score of 2222 on the mathematics portion of the exam suggests that a student i
kolbaska11 [484]

Answer:

Yes, Students are scoring above 2222 on the math portion of the​ exam.

Step-by-step explanation:

We are given that a college entrance exam company determined that a score of 2222 on the mathematics portion of the exam suggests that a student is ready for​ college-level mathematics i.e., population mean is 22.

Null Hypothesis, H_0 : \mu = 22 {Students are scoring equal to 2222 on the math portion of the​ exam}

Alternate Hypothesis, H_0 : \mu > 22 {Students are scoring above 2222 on the math portion of the​ exam}

Also, a random sample of 150 students who completed this core set of courses results in a mean math score of 22.7 on the college entrance exam with a standard deviation of 3.93 i.e.;

Sample mean, Xbar = 22.7    and Sample standard deviation, s = 3.93

The Test statistics is given by;

                            Z = \frac{Xbar -\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

Test Statistics = \frac{22.7 -22}{\frac{3.93}{\sqrt{150} } } ~ t_1_4_9

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Since, we are not given with the significance level, so we assume it to be 5%. At 5% significance level t table gives critical value of 1.6578 at 149 degree of freedom. Since our test statistics is more than the critical value so we reject null hypothesis as our test statistics fall in the rejection region.

Therefore, we conclude that Students are scoring above 2222 on the math portion of the​ exam.

7 0
3 years ago
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Answer:

Step-by-step explanation:

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