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MAXImum [283]
2 years ago
10

How would i anser this select all the ratios equivalent to 4:6

Mathematics
1 answer:
Natali5045456 [20]2 years ago
6 0
If we try to simplify the ratio or try to multiply both by any number, we will get the equivalent ratio


Like :

\frac{4}{6}  \\


Divide ( simplify ) by 2


\frac{ \frac{4}{2} }{ \frac{6}{2} }  \\  \\  \\  =  >  \frac{2}{3}




Or multiply by any number,



\frac{4}{6}  \\  \\  \\  =  >  \frac{4 \times 2}{6 \times 2}  \\  \\ \\    =  >  \frac{8}{12}





From the example, 2 / 3 and 8 / 12 are equivalent ratios to 4 / 6
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3 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\}

(c) Name: Binomial distribution

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
2 years ago
Ashton is depositing money into a checking account. After 3 months there is $120 in the account. After 6 months, there is $240 i
Talja [164]

The constant rate of change of the account is $40 or Increasing by $40 per month.

Step-by-step explanation:

Consider the provided information.

Joanne is depositing money into a bank account. After 3 months there is $120 in the account. After 6 months there is $240 in the account.

Rate of change is known as how one quantity change in relation to other.

The rate of change can be calculated as:

y2-y1/x2-x1  

Now use the above formula to calculated the rate of change.

240 - 120/6-3

120/3

40

Hence, the constant rate of change of the account is $40 or Increasing by $40 per month.

8 0
2 years ago
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