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Inessa05 [86]
3 years ago
14

Freezer A measures 1 ft times 1 ft times 5 ft and sells for ​$300. Freezer B measures 1.5 ft times 1.5 ft times 4 ft and sells f

or ​$600. Which freezer is the better buy in terms of dollars per cubic​ foot?
Mathematics
1 answer:
Sladkaya [172]3 years ago
7 0

Answer:

Freezer A is better option as price per cubic foot is less for Freezer A

Step-by-step explanation:

Data provided in the question:

Measure of Freezer A = 1 ft × 1 ft × 5 ft

Measure of Freezer B = 1.5 ft × 1.5 ft × 4 ft

Price of Freezer  A = $300

Price of Freezer  B = $600

Now,

Volume of Freezer A = 1 ft × 1 ft × 5 ft

= 5 ft³

Volume of Freezer B = 1.5 ft × 1.5 ft × 4 ft

= 9 ft³

Now,

Price per cubic foot for Freezer A = $300 ÷ 5 ft³

= $60/ft³

Price per cubic foot for Freezer B = $600 ÷ 9 ft³

= $66.67/ft³

Hence,

Freezer A is better option as price per cubic foot is less for Freezer A

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Two​ researchers, Jaime and​ Mariya, are each constructing confidence intervals for the proportion of a population who is​ left-
Irina-Kira [14]

Answer:

Jaime's wrong, becuase the distance(absolute value) from the point estimate to the lower bound is different than he distance from the upper bound to the point estimate.

Step-by-step explanation:

The distance(absolute value) from the point estimate to the lower bound must be the same as the distance from the upper bound to the point estimate.

The point estimate is 0.14.

Jaime

Jaime's interval has a lower bound of 0.049 and an upper bound of 0.191

upper - point = 0.191 - 0.14 = 0.051

point - lower = 0.14 - 0.049 = 0.091

Jaime's wrong, becuase the distance(absolute value) from the point estimate to the lower bound is different than he distance from the upper bound to the point estimate.

Mariya

Just to check.

Mariya's interval has a lower bound of 0.079 and an upper bound of 0.201.

upper - point = 0.201 - 0.14 = 0.061

point - lower = 0.14 - 0.079 = 0.061

Mariya has the same distances, so it is correct.

6 0
3 years ago
A circle has a radius of 12 cm. If the radius of the circle is increased by a factor of 7, how many times larger will the circle
-Dominant- [34]

Answer:

The circumference is also 7 times larger

12 cm radius :2*pi*12= 75.43

12*7=84

84cm radius: 2*pi*84= 528

To check if its indeed 7 times larger: 528/75.43 and you'll get 7

If youre looking for the difference between the two: 528-75.43= 452.57

4 0
3 years ago
(a) Show that a differentiable function f decreases most rapidly at x in the direction opposite the gradient vector, that is, in
Sophie [7]

Answer:

Step-by-step explanation:

\text{Show that a differentiable function f decreases most rapidly at x in the }

\text{direction opposite the gradient vector, that is, in the direction of} -\bigtriangledown f(x)\text{. Let}\  \theta \ \text{be the angle between} \bigtriangledown f(x) \  \text{and unit vector u. Then } D_u f = \mathbf{|\bigtriangledown f| \  cos  \ \theta }}

\text{Since the minimum value of} \ \  \mathbf{cos   \ \theta} \  \ is \mathbf{-1} \  \text{occuring \ for \ 0} \le \ \theta \ < 2x,  \\ \\ when  \ \theta = \mathbf{\pi} , \text{the mnimum value of} \  D_uf  \ is} -|\bigtriangledown f|,  \text{occuring when the direction of u is } \\ \\  \ \mathbf{the \ opposite \  of} \  \text{the direction of }  \ \bigtriangledown f (assuming \ \bigtriangledown f\ is \  not \ zero)

b) \text{From part A:}

If \ f(x,y) = x^4y -x^2y^2 \ \  decreases \ fastest \ at \ the \point \ (2,-5)\\ \\ F(x,y) = x^4y -x^2y^3 \\ \\ f_x = \dfrac{df}{dx}= \dfrac{d}{dx}(x^4y-x^2y^3)  \\ \\ f_x = \dfrac{df}{dx}= y4x^3 -2y^3x  \\ \\ For(2,-5) \\ \\ f_x = (-5)4(2)^3 -2(-5)^3(2) \\ \\ \mathbf{ f_x = 340}

However; f_y = \dfrac{df}{dy} = \dfrac{d}{dy}(x^4y - x^2y^3) \\ \\ f_y = x^4 -3x^2y^2 \\ \\  Now, for (2, -5)\\ \\f_y = (2)^4 -3(2)^2(-5)^2 \\ \\ f_y = -284

So; \bigtriangledown = < 340,-284> \text{this is the direction of fastest decrease}

6 0
3 years ago
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