Ask question

Ask question

All categories

3 years ago

5

Help pls help I will mark brainlist

2 answers:

Kay [80]3 years ago

7 0

Answer:

x = 20

y = 5

Step-by-step explanation:

<u>Adjacent angles of a parallelogram are supplementary</u> (add up to 180°).

So,

6x + 17 + 2x + 3 = 180

Combine like terms

6x + 2x + 17 + 3 = 180

8x + 20 = 180

Subtract 20 to both sides

8x = 180 - 20

8x = 160

Divide 8 to both sides

x = 160 / 8

x = 20

Now,

<u>Opposite sides of parallelogram are equal.</u>

So,

JK = ML

6y - 13 = 32 - 3y

Add 3y and 13 to both sides

6y + 3y = 32 + 13

9y = 45

Divide 9 to both sides

y = 45 / 9

y = 5

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807 </h3>

liberstina [14]3 years ago

7 0

Answer:

x = 20, y = 5

Step-by-step explanation:

Looks like we need to find the values of x and y.

Given is the parallelogram.

<u>Properties we'll use:</u>

Opposite sides are parallel and congruent
Adjacent angles are supplementary

<u>Find the value of y:</u>

JK = ML
6y - 13 = 32 - 3y
6y + 3y = 32 + 13
9y = 45
y = 5

<u>Find the value of x:</u>

∠J and ∠M are supplementary, so their sum is 180°.
6x + 17 + 2x + 3 = 180
8x + 20 = 180
8x = 160
x = 20

<em>If you need to find the side or angle measures just plug in the values of x or y.</em>

You might be interested in

Sophia and Charlie brought potato salad to the company cookout. Sophia brought 2 quarts and Charlie brought 4 pints. How many TO

yaroslaw [1]

16 cups of potato salad

3 0

4 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

3 years ago

Find the value of X x-5=7

ivolga24 [154]

Answer:

x = 12

Step-by-step explanation:

12 - 5 = 7

4 0

3 years ago

To ride an amusement park ride, your height must be at or above the line on the sign. For each set of information, write an inte

Rainbow [258]

Answer:

3ft tall yes 2ft and 12inches no

Step-by-step explanation:

hope this helps

7 0

3 years ago

can someone help me with order of operat ions?. (15+50 - 5)/12. please give stet by step answer for fifth grader.

pantera1 [17]

GEMDAS: Grouping (Parenthesis), Exponents, Multiplicatio/Divison (left to right), Addition/Subtraction (left to right)

I will walk you through this problem:
(15 + 50 - 5) / 12

So, 15 plus 50 is 65. 5 subtracted from 65 is 60.

So now, your equation looks like this:
60/12

60 divided by 12 is 5, because 5 multiplied by 12 is 60.

So your answer is...??

6 0

3 years ago