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2 years ago

A state seal, which is round, hangs in the capitol building. it has a radius of 2 feet. what is the seal's circumference?

Mathematics

1 answer:

Arte-miy333 [17]2 years ago

3 0

Answer:

12.56637ft

Step-by-step explanation:

Formula: c=2π*r

c=2π*2 = 12.56637ft

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What is the equation of a line, in general form, with a point (-3, 0) and an undefined slope?. Choices: x + 3 = 0. y + 3 = 0. x

andreev551 [17]

<span> Undefined slope is when the line is vertical, and the slope is infinite (technically);
Any y value is irrelevant in this kind of line, so in slope intercept, the equation is x = -3.;
Putting it in standard form, you get x + 3 = 0;
The first choice is correct.</span>

6 0

3 years ago

Read 2 more answers

Which equation has no solution?

Alex

Answer:

B. the absolute value of 5 minus 3 x equals negative 8

Step-by-step explanation:

The absolute value is never negative as per definition.

A. |2x - 1| = 0, has solution
B. |5 - 3x| = - 8, no solution
C. |-x - 3| = 5, has solution
D. |-x + 9| = 0, has solution

Correct choice is B

4 0

3 years ago

Read 2 more answers

4x + y = 3 -2x + 3y = -19 what's the solution plz help

Rainbow [258]

The solution to given system of equations is x = 2 and y = -5

<em><u>Solution:</u></em>

<em><u>Given the system of equations are:</u></em>

4x + y = 3 ---------- eqn 1

-2x + 3y = -19 ---------- eqn 2

We have to find the solution to above system of equations

<em><u>We can solve the system by substitution method</u></em>

From eqn 1,

4x + y = 3

Isolate y to one side

y = 3 - 4x ----------- eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

-2x + 3(3 - 4x) = -19

-2x + 9 - 12x = -19

Combine the like terms

-14x = -19 - 9

-14x = -28

Divide both sides of equation by -14

<em><u>Substitute x = 2 in eqn 3</u></em>

y = 3 - 4(2)

y = 3 - 8

Thus the solution is x = 2 and y = -5

5 0

3 years ago

The vertices of an isosceles triangle are a(3, 6), b(7, 2), and c(4, 3). what is the equation of the triangle's line of symmetry

Paul [167]

Based on the calculations, the equation of the triangle's line of symmetry is equal to: B. y = x - 1.

<h3>How to determine the equation of the triangle's line of symmetry?</h3>

First of all, we would determine the midpoint where the line of symmetry passes through points A and B:

Midpoint at point A = (3 + 7)/2 = 10/2 = 5.

Midpoint at point B = (6 + 2)/2 = 8/2 = 4.

Point D = (5, 4)

Mathematically, the standard equation of a line is given by y = ax + b.

D(5, 4); 4 = 5a + b ....equation 1.

C(4, 3); 3 = 4a + b ....equation 2.

Subtracting eqn. 2 from eqn. 1, we have:

1 = a

Next, we would find the value of b:

4 = 5a + b

4 = 5(1) + b

b = 4 - 5

b = -1.

Therefore, the equation of the triangle's line of symmetry is given by:

y = ax + b

y = x(1) + (-1)

y = x - 1.

Read more on line of symmetry here: brainly.com/question/10152876

#SPJ1

5 0

2 years ago

Based on the number of voids, a ferrite slab is classified as either high, medium, or low. Historically, 5% of the slabs are cla

AnnyKZ [126]

Answer:

(a) Name: Multinomial distribution

Parameters: $p_1 = 5\%$ $p_2 = 85\%$ $p_3 = 10\%$ $n = 20$

(b) Range: $\{(x,y,z)| x + y + z=20\}$

Parameters: $p_1 = 5\%$ $n = 20$

$(d)\ E(x) = 1$ $Var(x) = 0.95$

$(e)\ P(X = 1, Y = 17, Z = 3) = 0$

$(f)\ P(X \le 1, Y = 17, Z = 3) =0.07195$

$(g)\ P(X \le 1) = 0.7359$

$(h)\ E(Y) = 17$

Step-by-step explanation:

Given

$p_1 = 5\%$

$p_2 = 85\%$

$p_3 = 10\%$

$n = 20$

$X \to$ High Slabs

$Y \to$ Medium Slabs

$Z \to$ Low Slabs

Solving (a): Names and values of joint pdf of X, Y and Z

Given that:

$X \to$ Number of voids considered as high slabs

$Y \to$ Number of voids considered as medium slabs

$Z \to$ Number of voids considered as low slabs

Since the variables are more than 2 (2 means binomial), then the name is multinomial distribution

The parameters are:

$p_1 = 5\%$ $p_2 = 85\%$ $p_3 = 10\%$ $n = 20$

And the mass function is:

$f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z$

Solving (b): The range of the joint pdf of X, Y and Z

Given that:

$n = 20$

The number of voids (x, y and z) cannot be negative and they must be integers; So:

$x + y + z = n$

$x + y + z = 20$

Hence, the range is:

$\{(x,y,z)| x + y + z=20\}$

Solving (c): Names and values of marginal pdf of X

We have the following parameters attributed to X:

$p_1 = 5\%$ and $n = 20$

Hence, the name is: Binomial distribution

Solving (d): E(x) and Var(x)

In (c), we have:

$p_1 = 5\%$ and $n = 20$

$E(x) = p_1* n$

$E(x) = 5\% * 20$

$E(x) = 1$

$Var(x) = E(x) * (1 - p_1)$

$Var(x) = 1 * (1 - 5\%)$

$Var(x) = 1 * 0.95$

$Var(x) = 0.95$

$(e)\ P(X = 1, Y = 17, Z = 3)$

In (b), we have: $x + y + z = 20$

However, the given values of x in this question implies that:

$x + y + z = 1 + 17 + 3$

$x + y + z = 21$

Hence:

$P(X = 1, Y = 17, Z = 3) = 0$

$(f)\ P{X \le 1, Y = 17, Z = 3)$

This question implies that:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) + P(X = 1, Y = 17, Z = 3)$

Because

$0, 1 \le 1$ --- for x

In (e), we have:

$P(X = 1, Y = 17, Z = 3) = 0$

So:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) +0$

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)$

In (a), we have:

$f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z$

So:

$P(X=0; Y=17; Z = 3) = \frac{20!}{0! * 17! * 3!} * (5\%)^0 * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20!}{1 * 17! * 3!} * 1 * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20!}{17! * 3!} * (85\%)^{17} * (10\%)^{3}$

Expand

$P(X=0; Y=17; Z = 3) = \frac{20*19*18*17!}{17! * 3*2*1} * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20*19*18}{6} * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = 20*19*3 * (85\%)^{17} * (10\%)^{3}$

Using a calculator, we have:

$P(X=0; Y=17; Z = 3) = 0.07195$

So:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)$

$P(X \le 1, Y = 17, Z = 3) =0.07195$

$(g)\ P(X \le 1)$

This implies that:

$P(X \le 1) = P(X = 0) + P(X = 1)$

In (c), we established that X is a binomial distribution with the following parameters:

$p_1 = 5\%$ $n = 20$

Such that:

$P(X=x) = ^nC_x * p_1^x * (1 - p_1)^{n - x}$

So:

$P(X=0) = ^{20}C_0 * (5\%)^0 * (1 - 5\%)^{20 - 0}$

$P(X=0) = ^{20}C_0 * 1 * (1 - 5\%)^{20}$

$P(X=0) = 1 * 1 * (95\%)^{20}$

$P(X=0) = 0.3585$

$P(X=1) = ^{20}C_1 * (5\%)^1 * (1 - 5\%)^{20 - 1}$

$P(X=1) = 20 * (5\%)* (1 - 5\%)^{19}$

$P(X=1) = 0.3774$

So:

$P(X \le 1) = P(X = 0) + P(X = 1)$

$P(X \le 1) = 0.3585 + 0.3774$

$P(X \le 1) = 0.7359$

$(h)\ E(Y)$

Y has the following parameters

$p_2 = 85\%$ and $n = 20$

$E(Y) = p_2 * n$

$E(Y) = 85\% * 20$

$E(Y) = 17$

8 0

3 years ago