Which type of parent function is f(x) =1/2​

Answer:

(a):

Dannette                   Alphonso

[tex]\bar x_D = 4.33[/tex]                    [tex]\bar x_A = 5.17[/tex]

[tex]M_D = 2.5[/tex]                    [tex]M_A = 5[/tex]

[tex]\sigma_D = 3.350[/tex]                  [tex]\sigma_A = 1.951[/tex]

[tex]IQR_D = 7[/tex]                  [tex]IQR_A = 1.5[/tex]

(b):

Measure of center: Median

Measure of spread: Interquartile range

(c):

There are no outliers in Dannette's dataset

There are outliers in Alphonso's dataset

Step-by-step explanation:

Given

See attachment for the appropriate data presentation

Solving (a): Mean, Median, Standard deviation and IQR of each

From the attached plots, we have:

IQR_A = 1.5 ---- Dannette

[tex]A = \{3,4,4,4,4,5,5,5,5,6,6,11\}[/tex] ---- Alphonso

n = 12 --- number of dataset

Mean

The mean is calculated

[tex]\bar x = \frac{\sum x}{n}[/tex]

So, we have:

[tex]\bar x_D = \frac{1+1+1+1+2+2+3+7+8+8+9+9}{12}[/tex]

[tex]\bar x_D = \frac{52}{12}[/tex]

[tex]\bar x_D = 4.33[/tex] --- Dannette

[tex]\bar x_A = \frac{3+4+4+4+4+5+5+5+5+6+6+11}{12}[/tex]

[tex]\bar x_A = \frac{62}{12}[/tex]

[tex]\bar x_A = 5.17[/tex]  --- Alphonso

Median

The median is calculated as:

[tex]M = \frac{n + 1}{2}th[/tex]

[tex]M = \frac{12 + 1}{2}th[/tex]

[tex]M = \frac{13}{2}th[/tex]

[tex]M = 6.5th[/tex]

This implies that the median is the mean of the 6th and the 7th item.

So, we have:

[tex]M_D = \frac{2+3}{2}[/tex]

[tex]M_D = \frac{5}{2}[/tex]

[tex]M_D = 2.5[/tex] ---- Dannette

[tex]M_A = \frac{5+5}{2}[/tex]

[tex]M_A = \frac{10}{2}[/tex]

[tex]M_A = 5[/tex]  ---- Alphonso

Standard Deviation

This is calculated as:

[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}[/tex]

So, we have:

[tex]\sigma_D = \sqrt{\frac{(1 - 4.33)^2 +.............+(9- 4.33)^2}{12}}[/tex]

[tex]\sigma_D = \sqrt{\frac{134.6668}{12}}[/tex]

[tex]\sigma_D = 3.350[/tex] ---- Dannette

[tex]\sigma_A = \sqrt{\frac{(3-5.17)^2+............+(11-5.17)^2}{12}}[/tex]

[tex]\sigma_A = \sqrt{\frac{45.6668}{12}}[/tex]

[tex]\sigma_A = 1.951[/tex] --- Alphonso

The Interquartile Range (IQR)

This is calculated as:

[tex]IQR =Q_3 - Q_1[/tex]

Where

[tex]Q_3 \to[/tex] Upper Quartile       and        [tex]Q_1 \to[/tex] Lower Quartile

[tex]Q_3[/tex] is calculated as:

[tex]Q_3 = \frac{3}{4}*({n + 1})th[/tex]

[tex]Q_3 = \frac{3}{4}*(12 + 1})th[/tex]

[tex]Q_3 = \frac{3}{4}*13th[/tex]

[tex]Q_3 = 9.75th[/tex]

This means that [tex]Q_3[/tex] is the mean of the 9th and 7th item. So, we have:

[tex]Q_3 = \frac{1}{2} * (8+8) = \frac{1}{2} * 16[/tex]           [tex]Q_3 = \frac{1}{2} * (5+6) = \frac{1}{2} * 11[/tex]

[tex]Q_3 = 8[/tex] ---- Dannette                 [tex]Q_3 = 5.5[/tex] --- Alphonso

[tex]Q_1[/tex] is calculated as:

[tex]Q_1 = \frac{1}{4}*({n + 1})th[/tex]

[tex]Q_1 = \frac{1}{4}*({12 + 1})th[/tex]

[tex]Q_1 = \frac{1}{4}*13th[/tex]

[tex]Q_1 = 3.25th[/tex]

This means that [tex]Q_1[/tex] is the mean of the 3rd and 4th item. So, we have:

[tex]Q_1 = \frac{1}{2}(1+1) = \frac{1}{2} * 2[/tex]                  [tex]Q_1 = \frac{1}{2}(4+4) = \frac{1}{2} * 8[/tex]

[tex]Q_1 = 1[/tex] --- Dannette                   [tex]Q_1 = 4[/tex] ---- Alphonso

So, the IQR is:

[tex]IQR = Q_3 - Q_1[/tex]

[tex]IQR_D = 8 - 1[/tex]                                     [tex]IQR_A = 5.5 - 4[/tex]

[tex]IQR_D = 7[/tex] --- Dannette                      [tex]IQR_A = 1.5[/tex] --- Alphonso

Solving (b): The measures to compare

Measure of  center

By observation, we can see that there are outliers is the plot of Alphonso (because 11 is far from the other dataset) while there are no outliers in Dannette plot (as all data are close).

Since, the above is the case; we simply compare the median of both because it is not affected by outliers

Measure of  spread

Compare the interquartile range of both, as it is arguably the best measure of spread, because it is also not affected by outliers.

Solving (c): Check for outlier

To check for outlier, we make use of the following formulas:

[tex]Lower =Q_1 - 1.5 * IQR[/tex]

[tex]Upper =Q_3 + 1.5 * IQR[/tex]

For Dannette:

[tex]Lower = 1 - 1.5 * 7 = -9.5[/tex]

[tex]Upper = 8 + 1.5 * 7 = 18.5[/tex]

Since, the dataset are all positive, we change the lower outlier to 0.

So, the valid data range are:

[tex]Valid = 0 \to 18.5[/tex]

From the question, the range of Dannette's dataset is: 1 to 9. Hence, there are no outliers in Dannette's dataset

For Alphonso:

[tex]Lower = 4 - 1.5 * 1.5 =1.75[/tex]

[tex]Upper = 5.5 + 1.5 * 1.5 =7.75[/tex]  

So, the valid data range are:

[tex]Valid = 1.75\to 7.75[/tex]

From the question, the range of Alphonso's dataset is: 3 to 11. Hence, there are outliers in Alphonso's dataset

9514 1404 393

Answer:

  d, b, c, a, e

Step-by-step explanation:

The order of the letters in the congruence statement tells you ...

  ΔQOP ≅ ΔABC

  ∠Q ≅ ∠A

  ∠O ≅ ∠B ≅ 115°

  ∠P ≅ ∠C

  QO ≅ AB = 5 m

  OP ≅ BC = 8 m

  PQ ≅ CA

12z - 21 = 92

12z - 21 + 21 = 92 + 21

12z = 113

z = 113/12

Answer:

C. she is not beautiful and not clever.

Step-by-step explanation:

A. she is beautiful but not clever. B. she is beautiful and clever

C. she is not beautiful and not clever.

D. she is beautiful or not clever.

p: she is beautiful

q :she is clever

~p^ (~q) in verbal form

~p = she is not beautiful

~q = she is not clever

~p^ (~q) = she is not beautiful and not clever.

C. she is not beautiful and not clever.

Answer:

The lorry travels by 6 mph more than the expected speed.

Step-by-step explanation:

Velocity formula:

Velocity is distance divided by time, that is:

[tex]v = \frac{d}{t}[/tex]

Shortest route:

400 miles in 10 hours, which means that [tex]d = 400, v = 10[/tex]. So

[tex]v = \frac{d}{t} = \frac{400}{10} = 40[/tex]

In mph.

The lorry actually travels by a different route which increases the distance by 15%, but it still arrives in 10 hours.

Distance is multiplied by 100% + 15% = 115% = 1.15, so:

[tex]d = 1.15*400 = 460[/tex]

Then

[tex]v = \frac{d}{t} = \frac{460}{10} = 46[/tex]

46 mph

By how many more mph than the expected speed does the lorry travel?

46 - 40 = 6 mph

The lorry travels by 6 mph more than the expected speed.

Answer:

43.5 % decrease

Step-by-step explanation:

Take the original price and subtract the new price

850-480

370

Divide by the original price

370/850

.435294118

Change to percent form by multiplying by 100

43.5294118 % decrease

43.5 % decrease

Answer:

1/8 Of the animals are rabbits.

Step-by-step explanation:

3/8 Cats + 4/8 Dogs = 7/8

1/8 Is all you have room left for so 1/8 would be rabbits.

Answer:

1/8.

Step-by-step explanation:

Fraction of rabbits = 1 - (3/7 + 4/8)

= 1 - 7/8

= 8/8 - 7/8

= 1/8.

Answer:

73.0m

Step-by-step explanation:

see the attachment. Hope it helps.

Answer:

1. The first set of number is

1,2,3,4,5

Mean= 1+2+3+4+5/5

= 15/5

= 3

Median= 3

Mode= N/A

The next set of numbers is

2,3,4,5,6,6

Mean= 2+3+4+5+6+6/6

= 26/6

= 4.3

Median=4+5/2

= 9/2

= 4.5

Mode= 6

The next set of number is

6,7,5,4,5,6,2,5

Mean= 2+4+5+5+5+6+6+7/8

= 40/8

= 5

Median= 5+5/2

= 10/2

= 5

Mode= 5

Answer and Step-by-step explanation:

The answer is the second answer choice. y = 2x + 1

By looking at the graph, we see that there is a y-intercept at (0, 1), and it has a positive slope of 2.

#teamtrees #PAW (Plant and Water)

Step-by-step explanation:

-3,4 Is the answer Is it right or wrong if it is true plz mark me as brainliest

Answer:

Ano is correct

Step-by-step explanation:

-3,4is theoretically correct answer

Answer:

2.45.250- 94.750

= 92. 2975. this is the learners who got the admission

Answer:

1,50,500 students

Step-by-step explanation:

Hope this helps... vote as brainliest

ok check image file in the image is answers with color code

Answer:

A U B={1,2,3,4,5,6,7,8,9,10} represents A union B

(includes all the members of set Sand the members of set Bout together, do not repeat anyone that comes twice)

A n B={1,4} represents A interception B( this refers to the members that sets A and B have in common)

Answer:

A ∪ B = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 }

Step-by-step explanation:

A = { 1 , 3 , 4 , 5 , 7 , 9 }

B = { 1 , 2 , 4 , 6 , 8 , 10 }

A ∪ B = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 }

           It represents the whole numbers between 0 and 11

          [ A ∪ B is the elements of both A and B , without any repetition ]    

A ∩ B = { 1 , 4 }

            It represents the common numbers in both A and B

Answer:

[tex]x = 97[/tex]

Step-by-step explanation:

Given

[tex]t = 20[/tex] --- time (years)

[tex]A =1000[/tex] --- amount

[tex]r = 10\%[/tex] --- rate of interest

Required

The last 10 payments (x)

First, calculate the end of year 1 payment

[tex]y_1(end) = 10\% * 1000 * 150\%[/tex]

[tex]y_1(end) = 150[/tex]

Amount at end of year 1

[tex]A_1=A - y_1(end) - r * A[/tex]

[tex]A_1=1000 - (150 - 10\% * 1000)[/tex]

[tex]A_1 =1000 - (150- 100)[/tex]

[tex]A_1 =950[/tex]

Rewrite as:

[tex]A_1 = 0.95 * 1000^1[/tex]

Next, calculate the end of year 1 payment

[tex]y_2(end) = 10\% * 950 * 150\%[/tex]

[tex]y_2(end) = 142.5[/tex]

Amount at end of year 2

[tex]A_2=A_1 - (y_2(end) - r * A_1)[/tex]

[tex]A_2=950 - (142.5 - 10\%*950)[/tex]

[tex]A_2 = 902.5[/tex]

Rewrite as:

[tex]A_2 = 0.95 * 1000^2[/tex]

We have been able to create a pattern:

[tex]A_1 = 1000 * 0.95^1 = 950[/tex]

[tex]A_2 = 1000 * 0.95^2 = 902.5[/tex]

So, the payment till the end of the 10th year is:

[tex]A_{10} = 1000*0.95^{10}[/tex]

[tex]A_{10} = 598.74[/tex]

To calculate X (the last 10 payments), we make use of the following geometric series:

[tex]Amount = \sum\limits^{9}_{n=0} x * (1 + r)^n[/tex]

[tex]Amount = \sum\limits^{9}_{n=0} x * (1 + 10\%)^n[/tex]

[tex]Amount = \sum\limits^{9}_{n=0} x * (1 + 0.10)^n[/tex]

[tex]Amount = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]

The amount to be paid is:

[tex]Amount = A_{10}*(1 + r)^{10}[/tex] --- i.e. amount at the end of the 10th year * rate of 10 years

[tex]Amount = 1000 * 0.95^{10} * (1+r)^{10}[/tex]

So, we have:

[tex]Amount = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]

[tex]\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+r)^{10}[/tex]

[tex]\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+10\%)^{10}[/tex]

[tex]\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1+0.10)^{10}[/tex]

[tex]\sum\limits^{9}_{n=0} x * (1.10)^n = 1000 * 0.95^{10} * (1.10)^{10}[/tex]

The geometric sum can be rewritten using the following formula:

[tex]S_n = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]

[tex]S_n =\frac{a(r^n - 1)}{r -1}[/tex]

In this case:

[tex]a = x[/tex]

[tex]r = 1.10[/tex]

[tex]n =10[/tex]

So, we have:

[tex]\frac{x(r^{10} - 1)}{r -1} = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]

[tex]\frac{x((1.10)^{10} - 1)}{1.10 -1} = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]

[tex]\frac{x((1.10)^{10} - 1)}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]

[tex]x * \frac{1.10^{10} - 1}{0.10} = \sum\limits^{9}_{n=0} x * (1.10)^n[/tex]

So, the equation becomes:

[tex]x * \frac{1.10^{10} - 1}{0.10} = 1000 * 0.95^{10} * (1.10)^{10}[/tex]

Solve for x

[tex]x = \frac{1000 * 0.95^{10} * 1.10^{10} * 0.10}{1.10^{10} - 1}[/tex]

[tex]x = 97.44[/tex]

Approximate

[tex]x = 97[/tex]

Answer

19.013.153,42629466 giây

Step-by-step explanati

van toc ánh sán  =299.792.458 m/s

s= v*t

t=s/v

t= 5.699.999.999.999.999/299.792.458= 19.013.153,42629466 giây

Answer:

Step-by-step explanation:

question no . 5

ratio of angle L

take angle as reference angle

using cos rule

cos L = base / hypotenuse

= 4/5

ratio of angle N

take angle N as reference angle

using sin rule

sin N = opposite / hypotenuse

= 4/5

for measure of angle L

cos L = 4/5

cos L = 0.8

L = [tex]cos^{-1}(0.8)[/tex]

L = 36.9

for meanure of angle N

sin N = 4/5

sin N = 0.8

N = [tex]sin^{-1}(0.8)[/tex]

N = 53.1

Answer:

B.

Step-by-step explanation:

The equation of a circle with center at (h, k) and radius r is

[tex] (x - h)^2 + (y - k)^2 = r^2 [/tex]

We have center at (-5, 0). That makes h = -5, and k = 0.

The radius is 3, so r = 3.

[tex] (x - (-5))^2 + (y - 0)^2 = 3^2 [/tex]

[tex] (x + 5)^2 + y^2 = 9 [/tex]

Answer: B.

Answer:

B

Step-by-step explanation:

The equation of a circle has the form:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

Where (h, k) is the center of the circle and r is the radius.

From the graph, we can see that the center of the circle is at (-5, 0). So, (h, k) is (-5, 0), where h = -5 and k = 0.

And by counting, we can determine that the radius of the circle is three units. Hence, r = 3.

Substitute the information into the equation:

[tex](x-(-5))^2+(y-(0))^2=(3)^2[/tex]

Simplify. Therefore, our equation is:

[tex](x+5)^2+y^2=9[/tex]

Our answer is B.

Answer:

39 feet

Step-by-step explanation:

p = 15 + 8 + 10 + 6

p = 39

39 feet

Answer:

[tex] {x}^{2} + 26x = 11 \\ x = 0.4 \: and \: - 26.4[/tex]

Answer:

3:7

Step-by-step explanation:

We know that there are 4 girls and 3 boys and 4+3=7.

Answer:

1.tenisha

2.1\5+1\3=1\×

3.n-12=1\3(n+2)

4.500×=120

Answer: y = 4/3x -8

Step-by-step explanation:

(6,0) (0,-8)

0-(-8)/6-0 = 8/6 = 4/3

y intercept is given already

Answer:

33.51 cubic inches

Step-by-step explanation:

Formula for volume of a sphere is V= 4 /3 · π · r3

We know the diameter is 4, and the radius is half the diameter, so the radius is 2.

V = 4 /3 · π · [tex]2^{3}[/tex]

≈ 33. 51032

Rounded to the nearest hundreth is 33.51.

Answer:

The modulus of z is [tex]\sqrt{10}[/tex]

Step-by-step explanation:

Complex number:

A complex number, given by:

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

Can be represented also as a point (x,y).

The modulus is given by:

[tex]|z| = \sqrt{x^2 + y^2}[/tex]

Point z is at (negative 3, 1). What is the modulus of z?

[tex]x = -3, y = 1[/tex]. So

[tex]|z| = \sqrt{x^2 + y^2} = \sqrt{(-3)^2+1^2} = \sqrt{10}[/tex]

The modulus of z is [tex]\sqrt{10}[/tex]

Answer:

-2

Step-by-step explanation:

if this is the right problem it is -2

Answer:

48

Step-by-step explanation:

the answer is 48 I got this answer by multiplying 5 by 10 and subtracting is from 2 which gives me 50 - 2 which is 48

Answer:

48

Step-by-step explanation:

5×10-2=50-2

=48

hope it helps!!

Explanation:

The sample space is simply the list of possible outcomes. We list all the hair colors possible in this class.