When randomly selecting​ adults, let M denote the event of randomly selecting a male and let B denote the event of randomly selecting someone with blue eyes. What does P(M|B) ​represent? Is P(M|B) the same as P(B|M)​?

Answer:

D. allows a person to borrow more money at a lower interest rate

carry on learning

Answer:

- 850

3

Step-by-step explanation:

650 - 700 - 800

650 - 1500

- 850

25 - 45 + 23

- 20 + 23

3

Answer:

.40 is the greatest .350 is the second greatest and last but not least .3456 is the lowest

Step-by-step explanation:

Answer:

9

[tex]3^{\frac{4}{2} }[/tex] = [tex]3^{2} =9[/tex]

Step-by-step explanation:

Answer:

The sum of squared errors for the regression equation is 62.

Step-by-step explanation:

The sum of squared errors can be computed as follows:

X            Y           Y* = 2 + 3X         Y - Y*           (Y - Y*)^2

0            5                  2                      3                    9

3            5                  11                     -6                  36

7            27               23                     4                    16

10          31                32                    -1                     1  

20         68               68                     0                   62

From the above, we have:

Error = Y -  Y*

Error^2 = (Y - Y*)^2

Sum of squared errors = Sum of Error^2 = Total of (Y - Y*)^2 = 62

Therefore, the sum of squared errors for the regression equation is 62.

Answer:

The quadrilateral is a rhombus

Step-by-step explanation:

Given

[tex]A = (2, 8)[/tex]

[tex]B = (3, 11)[/tex]

[tex]C = (4, 8)[/tex]

[tex]D=(3, 5)[/tex]

Required

The true statement

Calculate slope (m) using

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Calculate distance using:

[tex]d= \sqrt{(x_2 - x_1)^2 + (y_2 -y_1)^2}[/tex]

Calculate slope and distance AB

[tex]m_{AB} = \frac{11 - 8}{3 - 2}[/tex]

[tex]m_{AB} = \frac{3}{1}[/tex]

[tex]m_{AB} = 3[/tex] -- slope

[tex]d_{AB}= \sqrt{(3 - 2)^2 + (11 -8)^2}[/tex]

[tex]d_{AB}= \sqrt{10}[/tex] -- distance

Calculate slope and distance BC

[tex]m_{BC} = \frac{8 - 11}{4 - 3}[/tex]

[tex]m_{BC} = \frac{- 3}{1}[/tex]

[tex]m_{BC} = -3[/tex] -- slope

[tex]d_{BC} = \sqrt{(4-3)^2+(8-11)^2[/tex]

[tex]d_{BC} = \sqrt{10}[/tex] --- distance

Calculate slope CD

[tex]m_{CD} = \frac{5 - 8}{3 - 4}[/tex]

[tex]m_{CD} = \frac{- 3}{- 1}[/tex]

[tex]m_{CD} = 3[/tex] -- slope

[tex]d_{CD} = \sqrt{(3-4)^2+(5-8)^2}[/tex]

[tex]d_{CD} = \sqrt{10}[/tex] -- distance

Calculate slope DA

[tex]m_{DA} = \frac{8 - 5}{2 - 3}[/tex]

[tex]m_{DA} = \frac{3}{- 1}[/tex]

[tex]m_{DA} = -3[/tex] -- slope

[tex]d_{DA} = \sqrt{(2-3)^2 + (8-5)^2}[/tex]

[tex]d_{DA} = \sqrt{10}[/tex]

From the computations above, we can see that all 4 sides are equal, i.e. [tex]\sqrt{10}[/tex]

And the slope of adjacent sides are negative reciprocal, i.e.

[tex]m_{AB} = 3[/tex]  and [tex]m_{CD} = -3[/tex]

[tex]m_{CD} = 3[/tex] and [tex]m_{DA} = -3[/tex]

The quadrilateral is a rhombus

Answer:

300%

Step-by-step explanation:

because 9/3×100=900/3=300 so it is 300%

Answer:

300%

Step-by-step explanation:

9/3 * 100%

900%/3 = 300%

Answer:

RM 1200 kalau ada gambar cuba insert

Answer:

1. D. 1

2. B. y=a³/x

3. A. y=1/x

Step-by-step explanation:

too long to give te explanations but they're there in the attachments

Answer:

Step-by-step explanation:

[tex]\frac{(5.27+x)}{2} =-4.51[/tex],[tex]\frac{8.21+y}{2} = 1.37[/tex]

(3.75,-5.47)

9514 1404 393

Answer:

  (8a -a²)/(a +2)

Step-by-step explanation:

Cancel common factors from numerator and denominator.

  [tex]\dfrac{-56+15a-a^2}{a^2+2a}\div\dfrac{a-7}{a^2}=-\dfrac{(a-7)(a-8)(a^2)}{a(a+2)(a-7)}\\\\=-\dfrac{a(a-8)}{a+2}=\boxed{\dfrac{8a-a^2}{a+2}}[/tex]

Answer:

3.82

Step-by-step explanation:

[tex]\sqrt{(9+\sqrt{32}) }[/tex] Do not confirm the answer unless your equation looks like that?

[tex]\sqrt{(9+\sqrt{32}) }[/tex] Start by the [tex]\sqrt{32}[/tex]

[tex]\sqrt{(9+5.65) }[/tex] Now add (9 + 5.65)

[tex]\sqrt{14.65}[/tex]         Finally Simplify

[tex]3.82[/tex]              Final answer

Some symbols and numbers are missing. I assume the system is supposed to read

2x - y + 2z + w = -3

x + 2y - 3z + w = 12

3x - y - z + 2w = 3

-2x + 3y + 2z - 3w = -3

In matrix form, this is

[tex]\begin{bmatrix}2&-1&2&1\\1&2&-3&1\\3&-1&-1&2\\-2&3&2&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix}-3\\12\\3\-3\end{bmatrix}[/tex]

which we can strip down to the augmented matrix,

[tex]\left[\begin{array}{cccc|c}2&-1&2&1&-3\\1&2&-3&1&12\\3&-1&-1&2&3\\-2&3&2&-3&-3\end{array}\right][/tex]

Now for the row operations:

• swap rows 1 and 2

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\2&-1&2&1&-3\\3&-1&-1&2&3\\-2&3&2&-3&-3\end{array}\right][/tex]

• add -2 (row 1) to row 2, -3 (row 1) to row 3, and 2 (row 1) to row 4

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&-7&8&-1&-33\\0&7&-4&-1&21\end{array}\right][/tex]

• add 7 (row 2) to -5 (row 3), and row 3 to row 4

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&16&-2&-24\\0&0&4&-2&-12\end{array}\right][/tex]

• multiply through rows 3 and 4 by 1/2

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&8&-1&-12\\0&0&2&-1&-6\end{array}\right][/tex]

• add -4 (row 4) to row 3

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&0&3&12\\0&0&2&-1&-6\end{array}\right][/tex]

• swap rows 3 and 4

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&-1&-6\\0&0&0&3&12\end{array}\right][/tex]

• multiply through row 4 by 1/3

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&-1&-6\\0&0&0&1&4\end{array}\right][/tex]

• add row 4 to row 3

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&0&-2\\0&0&0&1&4\end{array}\right][/tex]

• multiply through row 3 by 1/2

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]

• add -8 (row 3) and row 4 to row 2

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&-5&0&0&-15\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]

• multiply through row 2 by -1/5

[tex]\left[\begin{array}{cccc|c}1&2&-3&1&12\\0&1&0&0&3\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]

• add -2 (row 2) and 3 (row 3) and -1 (row 4) to row 1

[tex]\left[\begin{array}{cccc|c}1&0&0&0&-1\\0&1&0&0&3\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right][/tex]

Then the solution to the system is (x, y, z, w) = (-1, 3, -1, 4).

Answer:

The measure of the smallest angle is 30º

Step-by-step explanation:

Let the angles be:

[tex]x \to[/tex] the first angle (the smallest)

[tex]y \to[/tex] the second angle

[tex]z \to[/tex] the third angle

So, we have:

[tex]y = 2x[/tex]

[tex]z=x + 60[/tex]

Required

Find x

The angles in a triangle is:

[tex]x + y +z = 180[/tex]

Substitute values for y and z

[tex]x + 2x +x + 60 = 180[/tex]

[tex]4x + 60 = 180[/tex]

Collect like terms

[tex]4x = 180-60[/tex]

[tex]4x = 120[/tex]

Divide by 4

[tex]x = 30[/tex]

Answer:

157.8

Step-by-step explanation:

Add them all up to get 789 and divide them by 5 as there are five numbers to get the answer:)

Given: ????=????−2????+5????

and ????=2????+????+3????

To find: We need to find the value of ????×????

Solution: Here given,

????=????−2????+5????

and ????=2????+????+3????

Therefore, solving these two we have, ????=0

So,????×????=0

Answer:

12x2 – 75 = 3 (2x+5)(2x – 5)

Step-by-step explanation:

Answer:

You're correct

Step-by-step explanation:

Answer:

The solution to the equation  log3(x+2)=1 is given by x=1

Step-by-step explanation:

We are given that

[tex]log_3(x+2)=1[/tex]

We have to find the graph which shows the  solution to the equation log3(x+2)=1.

[tex]log_3(x+2)=1[/tex]

[tex]x+2=3^1[/tex]

Using the formula

[tex]lnx=y\implies x=e^y[/tex]

[tex]x+2=3[/tex]

[tex]x=3-2[/tex]

[tex]x=1[/tex]

Answer:

Cuboid = width*height*length

Cuboid = 24 cm^2

Answer:

The correct answer is =6.

Step-by-step explanation:

Solution,

Given;

11−17=49

or,11n-17=49

or,11−17+17=49+17

or,11=66

or,n=66/11

#n=6

HOPE IT HELPED♥︎

Answer:

A

Step-by-step explanation:

edge 2021

Answer:

0.8389 = 83.89% probability that the sample mean would be less than 133.5 liters.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean per capita consumption of milk per year is 131 liters with a variance of 841.

This means that [tex]\mu = 131, \sigma = \sqrt{841} = 29[/tex]

Sample of 132 people

This means that [tex]n = 132, s = \frac{29}{\sqrt{132}}[/tex]

What is the probability that the sample mean would be less than 133.5 liters?

This is the p-value of Z when X = 133.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{133.5 - 131}{\frac{29}{\sqrt{132}}}[/tex]

[tex]Z = 0.99[/tex]

[tex]Z = 0.99[/tex] has a p-value of 0.8389

0.8389 = 83.89% probability that the sample mean would be less than 133.5 liters.

Answer:

The maximum value of the objective function is 112 when x = 0 and y = 7.

Step-by-step explanation:

Given the constraints:

5x+3y≤37, 3x+5y≤35, x≥0, y≥0

Plotting the above constraints using geogebra online graphing tool, we get the solution to the constraints as:

A(0, 7), B(7.4, 0), C(5, 4) and D(0, 0)

The objective function is given as E =2x+16y, therefore:

At point A(0, 7):  E = 2(0) + 16(7) = 112

At point B(7.4, 0): E = 2(7.4) + 16(0) = 14.8

At point C(5, 4): E = 2(5) + 16(4) = 74

At point D(0, 0): E = 2(0) + 16(0) = 0

Therefore the maximum value of the objective function is at A(0, 7).

The maximum value of the objective function is 112 when x = 0 and y = 7.

Answer:

12

Step-by-step explanation:

First lets list all the factors of these numbers

72: 1,2 3,4,6,8,9,12,18,24,36,72

84: 1 , 2 , 3 ,4 , 6 , 7 , 12 , 14 ,  21 , 28 , 42 , 84

Now lets find the biggest number that is a factor of both 84 and 72  

as we can see the highest number that is the factor of both 84 and 72 is 12

12 is the hcf

Answer:

Should be 5 1/2 if thats on there

Step-by-step explanation:

u take 11/2 and take out the 1 u get 10/2 so u cut 10 in half get 5 then add the one and make it 5 1/2

Answer:

Median: 7.5

Mean: 8.6

Step-by-step explanation:

Median = the average of the 2 middle numbers of the set in ascending order, 6, 6, 6, 7, 8, 10, 12, 14

(7+8)/2 = 2

Mean = the sum of the numbers divided by the number of values

6 + 6+ 6+ 7 +8 +10 +12 +14/8

69/8

8.625

Hi there!  

»»————-　★　————-««

I believe your answer is:  

{7, -5, (2/3), 5.79}

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

[tex]\boxed{\text{\underline{\textbf{Some Examples of Rational Numbers Are...}}}}\\\\\rightarrow \text{Integers}\\\\\rightarrow \text{Perfect Squares}\\\\\rightarrow \text{Terminating Decimals}\\\\\rightarrow \text{Recurring Decimals}[/tex]

⸻⸻⸻⸻

⸻⸻⸻⸻

The rational numbers are {7, -5, (2/3), 5.79}.

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that [tex]\mu = 1262, \sigma = 118[/tex]

a) Determine the 26th percentile for the number of chocolate chips in a bag. ​

This is X when Z has a p-value of 0.26, so X when Z = -0.643.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.643 = \frac{X - 1262}{118}[/tex]

[tex]X - 1262 = -0.643*118[/tex]

[tex]X = 1186[/tex]

(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.

Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.

2.5th percentile:

X when Z has a p-value of 0.025, so X when Z = -1.96.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.96 = \frac{X - 1262}{118}[/tex]

[tex]X - 1262 = -1.96*118[/tex]

[tex]X = 1031[/tex]

97.5th percentile:

X when Z has a p-value of 0.975, so X when Z = 1.96.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.96 = \frac{X - 1262}{118}[/tex]

[tex]X - 1262 = 1.96*118[/tex]

[tex]X = 1493[/tex]

Between 1031 and 1493.

​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Difference between the 75th percentile and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.675 = \frac{X - 1262}{118}[/tex]

[tex]X - 1262 = -0.675*118[/tex]

[tex]X = 1182[/tex]

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.675 = \frac{X - 1262}{118}[/tex]

[tex]X - 1262 = 0.675*118[/tex]

[tex]X = 1342[/tex]

IQR:

1342 - 1182 = 160