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4 years ago

11

The price of a gallon of unleaded gas has dropped to $2.83 today. Yesterday's price was$2.89 . Find the percentage decrease. Rou

nd to the nearest tenth of a percent.

2 answers:

alexandr1967 [171]4 years ago

5 0

Decrease = initial price - final price

% decrease = [ (initial price - final price) / initial price ] *100

% decrease =[ (2.89 - 2.83) / 2.89] * 100 = 2.08, which can be rounded to 2.1

Answer: 2.1 %

Yanka [14]4 years ago

3 0

2.83 to 2.89's percentage of decrease is....
this is how you calculate this. You first use 2.89 minus 2.83. The answer is 0.06. Then use 0.06 to divide with 2.89. The anser is 0.020.For it to turn into a percentage number, times 0.020 with a hundred. And the answer is 2. So it is a 2 percent decrease. I didn't know how to round it though.

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3 years ago

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

vlabodo [156]

Answer:

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

Step-by-step explanation:

Bernoulli's equation,

$P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2$

P₁ = P₂= atmospheric presser

$\rho$ = density

$\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2$ [since P₁ = P₂]

$\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)$

$\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2$

$\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2$

$\Rightarrow v^2_2- v^2_1=2g h$ [ $h_1-h_2=h$ ]

Here $v_1\approx 0$

$\Rightarrow v^2_2=2g h$

$\therefore v_2=\sqrt {2gh$

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole $v_2$ = $\sqrt {2\times 9.8\times 10.4}$ m/s

=14.28 m/s.

Given, the rate of flow from the leak is $2.53\times 10^{-3}$ $m^3/min$

$=\frac{2.53\times 10^{-3}}{60}$ $m^3/s$

Let the diameter of the hole be d.

Then the cross section area of the hole is $=\pi (\frac d2)^2$

We know that,

The rate of flow = Cross section area × speed

$\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28$

$\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}$

$\Rightarrow d= 1.94 \times 10^{-3}$

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

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3 years ago