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dolphi86 [110]
2 years ago
7

Marie wants to buy fencing and sod for her new triangular flower bed. One of the legs of the garden is 7 ft longer than the othe

r and the hypotenuse is 1 ft longer. If sod cost $12.72 per sqft and fencing cost $4.23 per ft. How much would it cost Marie to purchase enough sod and fencing?
Mathematics
1 answer:
djyliett [7]2 years ago
3 0

let's suppose x is the shortest leg of the triangle

The perimeter of the flower bed is 3x+8 ft

the surface to be coverd by sod would be x×(x+7)/2 sqft

(x+8)(x+8)=(x+7)^2+x^2

x'2+16x+64=x'2+14x+49+x^2

x^2-2x=15

x×(x-2)=15

x=5

(15+8)×$4.23=$97.29 for fencing

32.5sqft×$12.72=$413.4 for the sod

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Identify the graphed linear equation.<br> A) y=5x+2<br> B) y=5x-2<br> C) y=-5x+2<br> D) y=-5x-2
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Read 2 more answers
Problemas de razonamiento división de números decimales. Ayer Susana se fue de viaje a visitar a unos familiares. Recorrió 135,7
schepotkina [342]

Usando las relaciones entre velocidad, distancia y tiempo, se encuentra que ella condujo a una velocidad media de 90,5 km/h.

--------------------------

La <u>velocidad </u><u>es la distancia dividida por el tiempo</u>, por lo que:

v = \frac{d}{t}

  • Total de 135,75 km, o sea, d = 135,75
  • Llego en 1,5 horas, o sea, t = 1,5

La velocidad es:

v = \frac{d}{t} = \frac{135,75}{1,5}

División de decimales, o sea, seguimos multiplicando los números por 10 hasta que ninguno sea decimal:

v = \frac{135,75}{1,5} = \frac{1357,5}{15} = \frac{13575}{150} = 90,5

Ella condujo a una velocidad media de 90,5 km/h.

Un problema similar es dado en brainly.com/question/24558377

4 0
1 year ago
"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen
Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }]

Where

p' is the sample proportion/point estimator of the population proportion

Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} } is the margin of error (d) of the confidence interval.

Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648

So

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }

(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}

n*(d*Z_{1-\alpha /2})^2= p'(1-p')

n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}

n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}

n= 147.28 ≅ 148 boxes.

I hope it helps!

3 0
3 years ago
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