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saveliy_v [14]
2 years ago
12

the remains of an ancient ball court include a rectangular playing alley with a perimeter of about 48 M. the length of the alley

is 5 times the width. find the length and the width of the playing alley​
Mathematics
1 answer:
Firdavs [7]2 years ago
8 0

Answer:

Length is 20 m and width is 4 m.

Step-by-step explanation:

Given:

Perimeter of the rectangular alley, P=48\textrm{ m}

Length is 5 times the width.

Let width be x.

So, as per question,

Length,l = 5x

Now, perimeter of rectangle is given as:

P=2(l+b)

Plug in 48 for P, 5x for l and x for b.

48=2(5x+x)\\48=2(6x)\\48=12x\\x=\frac{48}{12}=4

Therefore, width is 4 m.

Length is 5x=5\times 4=20 m.

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You would use the 0.50 as the input. And sin más your adding .15 as a fine, you would use 0.15.

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Frankie wants to build a path from one corner of his yard to the opposite corner. His yard measures 20 ft. x 32 ft. What will be
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Frankie's yard is a rectangle with the longer side = 32ft and the shorter side = 20ft. You want to find the length of the diagonal going through the rectangle. In essence, you basically have a triangle with two sides, one 32ft, the other 20ft, and you're looking for the hypotenuse.

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adam bought a box of fruit that weighed 6 3/8 kilograms. If he bought a second box that weighed 7 25 kilograms, what is the comb
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Box 1 = 6 & 3/8 kilograms.
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Lindsay opened a lemonade stand on her street. It costs her $15.00 to buy the supplies each day, and she sells lemonade for $2.0
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A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec
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Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

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The information provided is:

Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18

The sample size is, <em>n</em> = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191

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The null hypothesis is:

<em>H</em>₀: The usual load is 13 credits, i.e. <em>μ</em> = 13.

Assume that the significance level of the test is, <em>α</em> = 0.05.

Construct a (1 - <em>α</em>) % confidence interval for population mean to check the claim.

The (1 - <em>α</em>) % confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\times SE

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As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

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P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

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