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4 years ago

5

A container is filled with blueberries 1/6 of blueberries is poured equally into two bowls what fraction of the blueberries in e

ach bowl

1 answer:

Harrizon [31]4 years ago

3 0

1/12

1/6 divided by 2 is 1/12

You might be interested in

I need this for my notes so i can know what to do. What is the length of AC? And Length of AB?

andre [41]

Answer:

AC 5

AB = 3

Step-by-step explanation:

AC = distance between the points

C-A = 3 - -2 = 3+2 = 5

The distance is 5

AB = distance between the points

B-A = 1 - -2 = 1+2 = 3

The distance is 3

8 0

3 years ago

Read 2 more answers

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

3 years ago

Resuelve las siguientes ecuaciones: -x^2 - x =0 -X^2 + 12x = 0 3X^2 - 9x = 0 X^2 - 2x - 3 = 0

vaieri [72.5K]

Responder:

<em>a) 1 </em>

<em>b) 12 </em>

<em>c) 3 </em>

<em>d) -1 y 3 </em>

Explicación paso a paso:

Dadas las siguientes ecuaciones;

a) x²-x =0

x² = 0+x

x² = x

x = 1

b) -x²+12x = 0

-x² = -12x

x² = 12x

x = 12

c) 3x² - 9x = 0

Suma 9x a ambos lados

3x²-9x+9x = 0+9x

3x² = 9x

3x = 9

x = 9/3

x = 3

d) x²-2x-3 = 0

x²-3x+x-3 = 0

x(x-3)+1(x-3) = 0

(x+1)(x-3) = 0

x+1 = 0 y x-3 = 0

x = -1 y 3

4 0

3 years ago

Hannah has to make 25 gallons of punch for a potluck. The punch is made of soda and fruit drink. The cost of the soda is $1.79 p

Rus_ich [418]

Answer:

Hannah needs 10 gallons of soda and 15 gallons of fruit drink

Step-by-step explanation:

Let

x -----> the number of gallons of soda needed

y ----> the number of gallons of fruit drink needed

we know that

$x+y=25$ -----> equation A

$1.79x+2.49y=2.21(25)$

$1.79x+2.49y=55.25$ -----> equation B

Solve the system by graphing

Remember that the solution of the system of equations is the intersection point both graphs

The solution is the point (10,15)

see the attached figure

therefore

Hannah needs 10 gallons of soda and 15 gallons of fruit drink

3 0

3 years ago

Solve. 2x2 + 14x = −24 A) x = 2, x = 3 B) x = 3, x = 4 C) x = −2, x = −3 D) x = −4, x = −3

Aleksandr [31]

Answer:

<em>Answer is option d</em><em>)</em><em> </em><em>x</em><em> </em><em>=</em><em> </em><em>-4</em><em> </em><em>,</em><em> </em><em>x</em><em> </em><em>=</em><em> </em><em>-3</em>

Step-by-step explanation:

$given \: equation \: is \\ 2 {x}^{2} + 14x = - 24 \\ 2 {x}^{2} + 14x + 24 = 0 \: \\ dividing \: this \: equation \: by \: 2 \: we \: get \\ {x}^{2} + 7x + 12 = 0 \\ by \: factorization \\ (x + 4)(x + 3) = 0 \\ x = - 4 \: \: \: \:( or) \: \: \: x = - 3$

<em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em> </em>

4 0

3 years ago