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Elden [556K]
3 years ago
13

The cost of fertilizing a lawn is​ $0.25 per square foot. find the cost to fertilize the triangular lawn whose base is left pare

nthesis 8 plus startroot 19 endroot right parenthesis8+19 feet and altitude is startroot 76 endroot76 feet.
Mathematics
1 answer:
blsea [12.9K]3 years ago
7 0
Let's convert the verbal descriptions into mathematical expressions to illustrate the problem more clearly. The base of the triangle is (8+√19) and the altitude or height is √76 . Then, the area of the triangle is

A=(1/2)(b)(h) = (1/2) (8+√19)(√76) = 19+8√19 square units

Then, you multiply this by $0.25 to determine the total cost. The answer would be $13.5.
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Let W be the subspace of R5 spanned by the vectors w1, w2, w3, w4, w5, where
Gwar [14]
To find W⊥, you can use the Gram-Schmidt process using the usual inner-product and the given 5 independent set of vectors. 

<span>Define projection of v on u as </span>
<span>p(u,v)=u*(u.v)/(u.u) </span>
<span>we need to proceed and determine u1...u5 as: </span>
<span>u1=w1 </span>
<span>u2=w2-p(u1,w2) </span>
<span>u3=w3-p(u1,w3)-p(u2,w3) </span>
<span>u4=w4-p(u1,w4)-p(u2,w4)-p(u3,w4) </span>
<span>u5=w5-p(u4,w5)-p(u2,w5)-p(u3,w5)-p(u4,w5) </span>

<span>so that u1...u5 will be the new basis of an orthogonal set of inner space. </span>

<span>However, the given set of vectors is not independent, since </span>
<span>w1+w2=w3, </span>
<span>therefore an orthogonal basis cannot be found. </span>
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3 years ago
Population growth is an example of _____________?
olga2289 [7]

Answer:

The answer is 1. an exponential pattern ,friend (:

4 0
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HELPPPP PLEASEEEeeeeeee
sergij07 [2.7K]

Answer:

I think it is non proportional

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Read 2 more answers
Find the GCF of <br><img src="https://tex.z-dn.net/?f=%20%7B6x%7D%5E%7B4%7D%20%20-%20%20%7B24x%7D%5E%7B3%7D%20" id="TexFormula1"
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6 0
2 years ago
The American Society of PeriAnesthesia Nurses (ASPAN; www.aspan.org) is a national organization serving nurses practicing in amb
Zigmanuir [339]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

114, 261, 319, 183,654,313,58,98,335,324,52,125,342,66,321,893,690,798,201,74,632,216,76,155,161,304,530,1110,93,631,75,344,264,1120,122,45,304,192,316,63

USING CALCULATOR :

The mean(m) of the data: = ΣX/n

n = sample size = 40

m = 12974/40

m = 324.4

Median = ((n + 1)/2) th term = 41/2 = (20 + 21)th term / 2

= (261 + 264) / 2 = 262.5

Standard deviation (s) = sqrt(Σ(x - m)²/n))

s = 281. 74

B.) Coefficient of skewness (Ks) :

3(mean - median) / standard deviation

3(324.4 - 262.5) / 281.74

= 0.66

Mild positive skewness Given a positive skew Coefficient value.

(d-1) Are there any outliers?

Yes

Four outliers

(d-2) What are the limits for outliers? (Round your answers to 1 decimal place. Negative amounts should be indicated by a minus sign.)

Lower bound = Q1 - (1.5 * IQR)

Upper bound = Q3 + (1.5 * IQR)

IQR = Q3 - Q1

Q3 = upper quartile = 343 ; Q1 = 106 ; Q2 = 262.5

IQR = 343 - 106 = 237

Lower bound = 106 - (1.5 * 237) = - 249.5

Upper bound = 343 + (1.5 * 237) = 698.5

-249.5 ≤ X ≤ 698.5

4 0
3 years ago
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