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pantera1 [17]
1 year ago
12

The Davidson family wants to expand its rectangular patio, which currently measures 15 ft by 12 ft. They want to extend the leng

th and width the same amount to increase the total area of the patio by 160 ft2. Which quadratic equation best models the situation?
A = lw

(15)(12) + (x)(x) = (15)(12) + 160
 (15x)(12x) = (15)(12) + 160
2(15 + x) + 2(12 + x) = (15)(12) + 160
(15 + x)(12 + x) = (15)(12) + 160
Mathematics
2 answers:
tiny-mole [99]1 year ago
8 0
You just need to do 15x12=180+160=340divided by 2
8_murik_8 [283]1 year ago
3 0

Answer:

Option 4 - (15+x)(12+x)=(15)(12)+160

Step-by-step explanation:

Given : The Davidson family wants to expand its rectangular patio, which currently measures 15 ft by 12 ft. They want to extend the length and width the same amount to increase the total area of the patio by 160 ft sq.

To find : Which quadratic equation best models the situation?  

Solution :

Dimension of rectangular patio is

Length = 15 ft.

Width = 12 ft.

Let x be the expansion of length and width of the rectangular patio.

So, The new length of the rectangular patio is (x+15).

The new width of the rectangular patio is (x+12).

We know, Area of rectangle = Length × Width

According to question,

They want to extend the length and width the same amount to increase the total area of the patio by 160 ft sq.

(12+x)(15+x)=160+15\times 12

180+12x+15x+x^2=160+180

180+27x+x^2=340

x^2+27x-160=0        

Therefore, The required quadratic equation is (12+x)(15+x)=160+15\times 12

Hence, Option 4 is correct.

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x = -0.3

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You can substitute y in 2x – 4y = 3, to solve for x.

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2x - 4(3x) = 3 (Simplify)

2x - 12x = 3 (Simplify)

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A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints
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Answer:

a)0.099834

b) 0

Step-by-step explanation:

To solve for this question we would be using , z.score formula.

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16.

a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams.

Standard Deviation = √variance

= √0.16 = 0.4

Standard deviation = 0.4

Mean = 21.37

x = 20.857

z = (x-μ)/σ

z = 20.857 - 21.37/0.4

z = -1.2825

P-value from Z-Table:

P(x<20.857) = 0.099834

b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams.

z score formula used = (x-μ)/σ/√n

x = 20.857

Standard deviation = 0.4

Mean = 21.37

n = 100

z = 20.857 - 21.37/0.4/√100

= 20.857 - 21.37/ 0.4/10

= 20.857 - 21.37/ 0.04

= -12.825

P-value from Z-Table:

P(x<20.857) = 0

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5 0
1 year ago
Harry put blue and green marbles in a bag. Exactly 3/4 of the marbles in the bag are blue. Which of the following could be the t
pantera1 [17]

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D

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Passengers of flyaway airlines can purchase tickets for either business class or economy class. On one particular flight there w
Slav-nsk [51]

Answer:

x = 8

y = 146

Error = 4.86%

Step-by-step explanation:

Number of business class passenger = x

and the economy class passenger = y

Total number of passengers = 154

x + y = 154 ------(1)

Cost of business class tickets and economy class tickets are €320 and €85 respectively.

Total amount received by airlines is €14970.

320x + 85y = 14970

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Multiply equation (1) by 17 and subtract from equation (2)

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x = 8

From equation (1),

8 + y = 154

y = 154 - 8

y = 146

Airline officer wrote down the amount received as €14270

Then difference from the actual amount received = 14970 - 14270

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% Error =  \frac{\text{Difference in amount}}{\text{Actual amount}}\times 100

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Therefore, x = 8 and y = 146

             and % error = 4.68%

8 0
2 years ago
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