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IgorLugansk [536]

3 years ago

15

The perimeter of a rectangle is 86. The length of the rectangle is one more than twice the width. Determine the length and width

of the rectangle.

1 answer:

nordsb [41]3 years ago

6 0

There are two equation of the system

(1) The perimeter

2(w + l) = p

2(w + l) = 86

(2) The length and the width

2w = l + 2

From the second equation

2w = l + 2

l + 2 = 2w

l = 2w - 2

Subtitute l with 2w - 2 in the first equation to find the width

2(w + l) = 86

2(w + 2w - 2) = 86

2(3w - 2) = 86

6w - 4 = 86

6w = 90

w = 15

Now, find the length by subtituting w with 15

l = 2w - 2

l = 2(15) - 2

l = 30 - 2

l = 28

The dimension of the rectangle

length = 28 cm

width = 15 cm

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a building with a height of 32m casts a shadow that is 20m long. a person standing casts a shadow that is 1.2 m long how tall is

wlad13 [49]

Most of the information's required for solving the question is already given in the question.
Height of the building that casts a shadow of 20 m = 32 m
Then
Height of the man that casts a shadow of 1.2 m = (32/20) * 1.2 meter
= 3.2 * 1.2 meter
= 3.84 meter
So the actual height of the person casting a shadow of 1.2 meter is 3.84 meters. I hope that the procedure used for solving the problem is easy enough for you to understand. You can definitely use this method in future for solving problems of similar type without requiring any additional help from outside.

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4 years ago

Find the radius of the cylinder when volume is 304 cm^3 and height is 10 cm

aleksandr82 [10.1K]

Answer:

<h2>3.11 cm</h2>

solution,

Volume of cylinder=304 cm^3

height=10 cm

Radius=?

Now,

$volume = \pi {r}^{2} h \\ or \: 304 = 3.14 \times {r}^{2} \times 10 \\ or \: 304 = 31.4 \times {r}^{2} \\ or \: {r}^{2} = \frac{304}{31.4} \\ or \: {r}^{2} = 9.68 \\ or \: r = \sqrt{9.68} \\ or \: r = \sqrt{ {(3.11)}^{2} } \\ r = 3.11 \: cm$

Hope this helps..

Good luck on your assignment..

7 0

3 years ago

What percent of 105 is 7.35

Shtirlitz [24]

Answer: 7

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batterie

Fittoniya [83]

Answer:

The confidence interval is $25.16 < \mu < 26.85$

Step-by-step explanation:

From the question we are given a data set

25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

The mean of the this sample data is

$\= x = \frac{\sum x_i}{n}$

where is the sample size with values n = 10

$\= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}$

$\= x = 26$

The standard deviation is evaluated as

$\sigma = \sqrt{\frac{\sum (x-\= x)}{n} }$

substituting values

$= \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }$

$\cdot \ \cdot \ \cdot \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }$

$\sigma = 1.625$

The confidence level is given as 90% hence the level of significance is calculated as

$\alpha = 100 -90$

$\alpha =10$ %

$\alpha = 0.10$

Now the critical values of $\frac{\alpha }{2}$ is obtained from the normal distribution table as

$Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining the critical values of $\frac{\alpha }{2}$ instead of $\alpha$ is because we are considering two tails of the area under the normal curve

The margin of error is evaluated as

$MOE = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

$MOE = 1.645 * \frac{1.625 }{\sqrt{10} }$

$MOE = 0.845$

The 90%, two sided confidence interval is mathematically evaluated as

$\= x - MOE < \mu < \= x + MOE$

$26 - 0.845 < \mu < 26 + 0.845$

$25.16 < \mu < 26.85$

Given that the lower and the upper limit is greater than 25 then we can assure the manufactures that the battery life exceeds 25 hours

5 0

3 years ago

The width of a rectangle is half as long as the

Dmitrij [34]

Length=20ft
width=10ft

4 0

3 years ago

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