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3 years ago

What is the probability that a randomly selected candle is orange scented?

1 answer:

8 0

Since the factory produces equal numbers of 4 types of candles, we can find the number of candles it produces of any given type:

$\dfrac{8000 \,\textrm{candles}}{4 \,\textrm{types}} = 2000 \,\dfrac{\textrm{candles}}{\textrm{type}}$

We have found that the factory produces 2000 candles of each type. Since the probability of something happening is the number of favored outcomes divided by the number of total outcomes, we can find the probability of an orange scented candle being chosen are:

$P(\textrm{orange scented}) = \dfrac{2000}{8000} = \boxed{\dfrac{1}{4}}$

The answer is 1/4.

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Which of these best represents the rate of change of the temperature with respect to altitude, if the temperature is 10 degrees

Brilliant_brown [7]

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3 years ago

Multiply 5.46 × 12.9 = ____ PLS HELP FAST

dmitriy555 [2]

Answer:

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3 years ago

The Green Goober, a wildly unpopular superhero, mixes 3 liters of yellow paint with 5 liters of blue paint to make 8 liters of s

Degger [83]

Let

y------> the amounts (in liters) of yellow paint

b-----> the amounts (in liters) of blue paint

we know that

$\frac{y}{b}=\frac{3}{5}$

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2 years ago

Read 2 more answers

Simplify

Goryan [66]

Answer:

310

Step-by-step explanation:

Use the Order of Operations method. (P.E.M.D.A.S.)

Our expression: 3+2⋅4+62⋅5−1

Multiply 62 and 5, and 2 and 4:

3+8+310-1

Add and Subtract from left to right:

11+310-1

311-1

310

6 0

2 years ago

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

Answer-

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

Solution-

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

Now, equating this to 0

$(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

6 0

2 years ago