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3 years ago

Can someone help me thankyou <3

Mathematics

1 answer:

grigory [225]3 years ago

6 0

Answer:

P(G)= 7/10

B, 1, P(not B)

1- 8/10, p(Y)= 2/10

Step-by-step explanation:

hope this helps

correct me if this is wrong

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The ratio of men to women working for a company is 5 to 7. If there are 126 women working for the company, what is the total num

SIZIF [17.4K]

Answer:

There are 216 employees

Step-by-step explanation:

First you divide 126 by 7 (number of women) and you get 18 then you times that by 5 (total number of men) and you get 90 finally you add 90 (men) and 126 (women) to get 216 employees in total.

8 0

3 years ago

Evaluate the expression below when x = -2. x2 − 8 x + 6

nordsb [41]

Solve for x over the real numbers:x = -2. x^2 - 8 x + 6
Hint: | Rewrite the right hand side of the equation.-2. x^2 - 8 x + 6 = -2 x^2 - 8 x + 6:x = -2 x^2 - 8 x + 6
Hint: | Move everything to the left hand side.Subtract -2 x^2 - 8 x + 6 from both sides:2 x^2 + 9 x - 6 = 0
Hint: | Write the quadratic equation in standard form.Divide both sides by 2:x^2 + (9 x)/2 - 3 = 0
Hint: | Solve the quadratic equation by completing the square.Add 3 to both sides:x^2 + (9 x)/2 = 3
Hint: | Take one half of the coefficient of x and square it, then add it to both sides.Add 81/16 to both sides:x^2 + (9 x)/2 + 81/16 = 129/16
Hint: | Factor the left hand side.Write the left hand side as a square:(x + 9/4)^2 = 129/16
Hint: | Eliminate the exponent on the left hand side.Take the square root of both sides:x + 9/4 = sqrt(129)/4 or x + 9/4 = -sqrt(129)/4
Hint: | Look at the first equation: Solve for x.Subtract 9/4 from both sides:x = sqrt(129)/4 - 9/4 or x + 9/4 = -sqrt(129)/4
Hint: | Look at the second equation: Solve for x.Subtract 9/4 from both sides:Answer: x = sqrt(129)/4 - 9/4 or x = -9/4 - sqrt(129)/4

3 0

4 years ago

Read 2 more answers

I have no idea where to start on either of these problems. Would someone explain the steps to me? My teacher hasn’t gone over th

bazaltina [42]

Answer:

Problem 9: -1/2

Problem 10: 1/5

Step-by-step explanation:

Problem 10: Label the given ln e^(1/5) as y = ln e^(1/5).

Write the identity e = e. Raise the first e to the power y and the second e to the power 1/5 (note that ln e^(1/5) = 1/5). Thus, we have:

e^y = e^(1/5), so that y = 1/5 (answer).

Problem 9: Let y = (log to the base 4 of) ∛1 / ∛8, or

y = (log to the base 4 of) ∛1 / ∛8, or

y = (log to the base 4 of) 1 /2

Write out the obvious:

4 = 4

Raise the first 4 to the power y and raise the second 4 to the power (log to the base 4 of) 1 /2. This results in:

4^y = 1/2. Solve this for y.

Note that 4^(1/2) = 2, so that 4^(-1/2) = 1/2

Thus, y = -1/2

3 0

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

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

Please help me

natka813 [3]

Answer:

The answer is Y=3x

when x is 0, y is 0

y=3x

Y= 3 X 0 = 0 (correct)

when x is 1, y is 3

y=3x

y=3 X 1= 3 (correct)

when x is 2 , y is 6

y=3x

y= 3 X 2= 6 (correct)

when x is 3, y is 9

y=3x

y= 3 X 3= 9 (correct)

when x is 4, y is 12

y=3x

y= 3 X 4= 12 ( Correct)

6 0

3 years ago