All categories

3 years ago

A fair-sided coin is tossed 18 times.

Mathematics

1 answer:

Lady bird [3.3K]3 years ago

5 0

Most likely to be letter C

You might be interested in

Mayra has 3 cats. Luis has twice as many cats as Mayra.

kompoz [17]

1. 3x=y
3+y=z
2. 1/2y=x
7x=z

8 0

2 years ago

Jasmine purchased a laptop worth $1,500 in 2007. It loses its value by %32 each year. What is the value of the laptop in 2010? R

Lunna [17]

Answer:

$471.65

Step-by-step explanation:

-The value of the laptop after each year is calculated by the formula:

$P_t=P_o(1-d)^t$

where:

$P_t$ is the value at time t

$P_o$ is the initial value

$d$ is the rate of depreciation

$t$ is the time

#We substitute the values to solve for the value after 3 years(2010-2007=3):

$P_t=P_o(1-d)^t\\\\=1500(1-0.32)^3\\\\=\$471.65$

Hence, the value of the laptop in 2010 is $471.65

3 0

3 years ago

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

Someone Please Help Quickly! Its an Easy Question

Paul [167]

Answer:

$y=\frac{2}{3}x-6$

Step-by-step explanation:

Use the slope-intercept form:

$y=mx+b$

m is the slope and b is the y-intercept. Looking at the graph, you can find the y-intercept. The y-intercept is the point where x equals 0:

$b=-6$

$y-intercept=(0,-6)$

To find the slope, take any two points from the line:

$(6,-2)(3,-4)$

Use the slope formula for when you have two points:

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{rise}{run}$

The rise over run is the change in the y-axis over the change of the x-axis. Insert the appropriate values:

$(6(x_{1}),-2(y_{1})\\(3(x_{2}),-4(y_{2})$

$\frac{-4-(-2)}{3-6}$

Simplify parentheses (two negatives makes a positive):

$\frac{-4+2}{3-6}$

$\frac{-2}{-3}$

Simplify (two negatives make a positive):

$m=\frac{2}{3}$

The slope is $\frac{2}{3}$ and the y-intercept is $-6$ . Insert these into the equation:

$y=\frac{2}{3}x-6$

Finito.

7 0

3 years ago

Which equation is a line that is parallel to the x-<br> axis and passes through the point (5,2)?

Natali [406]

Answer:

y = 2

Step-by-step explanation:

Since the line is parallel to the x-axis,

the gradient, m = 0

From the point, we know that

x = 5

y = 2

So y = 2 is the line that parallel to x-axis

4 0

3 years ago