A bowl holds 3/10 cup of oil when it is 2/5 full. Which statement best describes the quotient of 3/10 divided by 2/5?

1 answer:

muminat3 years ago

5 0

If 3/10 cup is 2/5 full, then the bowl holds:

(3/10) / (2/5)

(3/10)/(2/5) = 3/10 * 5/2 = 15/20 = 3/4

The bowl holds 3/4 of a cup

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A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = $\frac{AB}{BC} = \frac{h}{x}$

Taking inverse trigonometric function, α = tan⁻¹ ( $\frac{h}{x}$ ) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

$\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}$

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and $\frac{dh}{dt}$ = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

$\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512$

= 0.12 (Rounding up to two decimal places)

So, the rate of change of the angle of elevation is 0.12 radians/second.

5 0

3 years ago

John is interested in purchasing a multi-office building containing five offices. The current owner provides the following proba

Elenna [48]

Answer:

Option B.

Step-by-step explanation:

The given table is:

Number of Lease Offices : 0 1 2 3 4 5

Probability : 5/18 1/4 1/9 1/18 2/9 1/12

The expected probability is

Expected probability = $\sum_{i=0}^5 x_{i}p(x_i)$

Expected probability = $0p(0)+1P(1)+2P(2)+3P(3)+4P(4)+5P(5)$

Expected probability = $0\cdot (\frac{5}{18})+1\cdot (\frac{1}{4})+2\cdot (\frac{1}{9})+3\cdot (\frac{1}{18})+4\cdot (\frac{2}{9})+5\cdot (\frac{1}{12})=\frac{35}{18}$