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3 years ago

8

There are 12 green balls and (x+2) yellow balls in a box. If the probability of selecting a green ball at random is 2/5, find th

e value of x.

1 answer:

Marysya12 [62]3 years ago

8 0

Answer:

x = 16

Step-by-step explanation:

The probability of obtaining a green ball is

P(green) = $\frac{12}{12+x+2}$ = $\frac{12}{x+14}$ and

P(green) = $\frac{2}{5}$ , hence equating gives

$\frac{12}{x+14}$ = $\frac{2}{5}$ ( cross- multiply )

2(x + 14) = 60 ( divide both sides by 2 )

x + 14 = 30 ( subtract 14 from both sides )

x = 16

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Answer quick for my hw pls

Readme [11.4K]

Answer:

day 14 , 450 dollars

Step-by-step explanation:

We called x is the number of days

Zippy Rent-a-car: 25x + 100

Speedy Rent-a-car: 30x + 30

25x + 100 = 30x + 30

25x -30x = 30 - 100

-5x = -70

x = 14

5 0

3 years ago

Read 2 more answers

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

What is the smallest unit of measurement?

kvasek [131]

The smallest unit of measure is the millimeter

Hope this helps ;)

3 0

3 years ago

Read 2 more answers

lana [24]

Answer:

so what are we trying to figure out?

Step-by-step explanation:

4 0

3 years ago

I do no understand this

vovangra [49]

The equation of a line for the given slope is $y=\frac{1}{2} x+1$ .

Solution:

Given data:

Slope (m) = $\frac{1}{2}$

Point = (–2, 0)

$x_1=-2, y_1=0$

To find the equation of the given line in the point-slope form:

<u>Point-slope formula:</u>

$y-y_1=m(x-x_1)$

Substitute the given values in the formula.

$$y-0=\frac{1}{2} (x-(-2))$

$$y=\frac{1}{2} (x+2)$

$$y=\frac{1}{2} x+\frac{2}{2}$

$$y=\frac{1}{2} x+1$

The equation of a line for the given slope is $y=\frac{1}{2} x+1$ .

5 0

3 years ago