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Sauron [17]
3 years ago
6

The area of a rectangle is 20 square inches. If the length is 4 inches less than 6 times the width, then find the dimensions of

the rectangle. Round off your answers to the nearest hundredth.
Mathematics
1 answer:
Korolek [52]3 years ago
6 0

Answer:

dimensions are 2.19 inches by 9.14 inches

Step-by-step explanation:

Here, we are interested in knowing the dimensions of the rectangle

Now, let the width of the rectangle be x, from the question, we can write;

6x-4 as the length

Area of rectangle = l * b

x(6x-4) = 20

6x^2 - 4x = 20

divide through by 2

3x^2-2x = 10

3x^2 -2x -10 = 0

using quadratic formula;

x = -b ± √(b^2 - 4ac)/2a

From the quadratic equation, a = 3, b = -2 and c = -10

substituting these values, we have

x = 2 ± √(-2)^2 -4(3)(-10))/6

x = 2 ± √(124)/6

x = 2 + √(124)/6 or 2 - √(124)/6

x = (2+11.14)/6 or (2-11.14)/6

x = 2.19 or -1.52

Since width cannot be negative, we choose 2.19 alone

So the length is 6(2.19) -4 = 9.14

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Answer:

  about 10.24 ft

Step-by-step explanation:

The formula for the volume of a cylinder is ...

  V = πr²h . . . . where h is the height and r is the radius

The formula for the volume of a sphere is ...

  V = (4/3)πr³ = πr²·(4/3r) . . . . equivalent to a cylinder of height 4/3r

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We have a cylinder of height 3d = 3(2r) = 6r. It has half a sphere on top, so the equivalent height of that is (1/2)·(4/3r) = 2/3r.

Then our total volume is equivalent to a cylinder with radius r and height (6 2/3)r = (20/3)r. That is, ...

  22,500 ft³ = πr²·(20/3)r = (20π/3)r³

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The radius of the silo should be about 10.24 feet.

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A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr
Marianna [84]

Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men      Women

The sample mean : 24.51      22.69

Sample standard deviation : 4.48    3.86

Sample size : 35    40

<em>Let </em>\mu_1<em> = mean number of times men order take-out dinners in a month.</em>

<em />\mu_2<em> = mean number of times women order take-out dinners in a month</em>

(a) So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0     {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, H_A : \mu_1-\mu_2\neq 0     {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviation;

                      T.S. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t__n_1_-_n_2_-_2

where, \bar X_1 = sample mean for men = 24.51

\bar X_2 = sample mean for women = 22.69

s_1 = sample standard deviation for men = 4.48

s_2 = sample standard deviation for women = 3.86

n_1 = sample of men = 35

n_2 = sample of women = 40

Also,  s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} }  =  \sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2}  }{35+40-2} } = 4.16

So, <u>test statistics</u>  =  \frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40}  } }  ~ t_7_3

                              =  1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

                     P-value = P( t_7_3 > 1.89) = 0.0331

So, P-value for two tailed test is = 2 \times 0.0331 = <u>0.0662</u>

4 0
3 years ago
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