What is the approximate ratio of the difference in means to the MADs of the data sets? 4 6 7 31

Answer:

-0.2 + 0.5m + 9

Step-by-step explanation: simplify

Answer:

2) 9/4

Step-by-step explanation:

12/5 in decimal form is 2.4. As a mixed number, 12/5 is 2 2/5. And, 12/5 is just a reduced form of 36/15. So, the only answer it can be is answer 2: 9/4

Answer:

Standard Deviation and Below are, in fact, the right answer.

Step-by-step explanation:

Edge. :)

Answer:

The person above is right, they deserve brainliest.

-standard deviations

-below

ED2021

Answer:

Option (4) and Option (6)

Step-by-step explanation:

By the zero product property,

If (x - a)(x - b) = 0

Then, (x - a) = 0 ⇒ x = a

(x - b) = 0 ⇒ x = b

By following the zero product property,

From the given equation,

3(x - 4)(2x - 3) = 0

x - 4 = 0 ⇒ x = 4

2x - 3 = 0 ⇒ x = [tex]\frac{3}{2}[/tex]

Therefore, options (4) and (6) will be the correct options.

Answer:

4 and 6 :)

Step-by-step explanation:

edge

the last one seems to be the best

Answer:

D

Step-by-step explanation:

Answer:

81°

Step-by-step explanation:

x° and 24° are interior angles of a triangle that are opposite to an exterior angle of measure 105°. According to the exterior angle theorem of a triangle, thus:

[tex] x + 24 = 105 [/tex]

Subtract 24 from both sides

[tex] x + 24 - 24 = 105 - 24 [/tex]

[tex] x = 81 [/tex]

Value of x = 81°

Because of the 45 45 90 theorem since the hypotenuse is 9√2 x and y are both 9

Answer:

Option C. 2(x + 8)

Step-by-step explanation:

To know which option is the correct answer, let us simplify each expression given in the options to know which will not result to 2x² – 16. This is illustrated below:

For Option A:

2x² + 4x – 4x – 16

= 2x² – 16

For Option B:

2(x² – 8)

Clear bracket

2x² – 16

For Option C:

2(x + 8)

Clear bracket

2x + 8

For Option D:

(4x² – 32) /2

4x²/2 – 32/2

2x² – 16

From the above illustration, option C did not result to 2x² – 16.

Therefore, option C is the answer to the question.

Answer:

Answer is 79 and 28

Step-by-step explanation:

Let two numbers be x and y.

x + y = 107. --> 1

x - y = 51 -->2

solving above equations,

2x = 158

x =79

y = 28

Answer:

3/2 x-5

Step-by-step explanation:

Answer:

y=3/2x-5

Step-by-step explanation:

m=y²-y¹/x²-x¹

m=1-4/4-6

m=-3/-2

m3/2

y-y=m(x-x1)

y-1=3/2(x-4)

y-1=3/2x-6

+1 +1

y=3/2x-5

Step-by-step explanation:

18x+30y=-7

18x+12y=0

42y=-7

y=-0.17

3x+5(-0.17)=-7

3x-0.85=-7

3x=-6.15

x=-2.05

Answer:

The National Woman Suffrage Association advocated for a range of reforms to make women equal members of society, such as the right to own property, etc. The American Woman Suffrage Association lobbied state governments to enact laws granting or expanding women's right to vote in the United States.

Answer:

C: The NWSA wanted to address the voting rights of both women and formerly enslaved people.

Step-by-step explanation:

The NWSA wanted to address and justify the rights of both former slaves and women, while the AWSA only wanted to either help women or former slaves, not simultaneously.

Answer:

Mrs Barnes has the highest

Answer:

a. Yes

Step-by-step explanation:

Answer:

3

Step-by-step explanation:

cause its a pretty simple question

Answer:

2.94 in

Step-by-step explanation:

Answer:

8.no answer A. x=8

Step-by-step explanation:

10.no answer B. x= -3

Answer:

8. (A) x = 8 , 10. (C.) x = 3

Step-by-step explanation:

Number 8 you could do 75 - 35 = 40

so what can you multiply 5 by (x) to get 40? you would multiply 5 by 8 or (x) and get 40

then add 35 + 40 to get 75 degrees which is the parallel angle to the other 75 degrees.

Number 10 you just want to get them to the same number.

so you would do ( 97 - 3 ) to get the ANSWER 87.

so x = 3

Answer:

diameter ÷ 2 = 8 ÷ 2 = 4

Step-by-step explanation:

Answer:

p=2 (l+w)

Step-by-step explanation:

Answer:

4,099 and 5,011

Step-by-step explanation:

Answer:

See explanation

Step-by-step explanation:

Solving (a):

[tex]a_n = 6a_{n-1}[/tex] where [tex]a_0 = 2[/tex]

n = 1

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_1 = 6a_{1-1}[/tex]

[tex]a_1 = 6a_{0}[/tex]

Substitute 2 for [tex]a_0[/tex]

[tex]a_1= 6 * 2[/tex]

[tex]a_1= 12[/tex]

n = 2

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_2 = 6a_{2-1}[/tex]

[tex]a_2 = 6a_{1}[/tex]

Substitute 12 for [tex]a_1[/tex]

[tex]a_2= 6 * 12[/tex]

[tex]a_2= 72[/tex]

n = 3

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_3 = 6a_{3-1}[/tex]

[tex]a_3 = 6a_2[/tex]

Substitute 72 for [tex]a_2[/tex]

[tex]a_3= 6 * 72[/tex]

[tex]a_3= 432[/tex]

n = 4

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_4 = 6a_{4-1}[/tex]

[tex]a_4 = 6a_{3}[/tex]

Substitute 432 for [tex]a_3[/tex]

[tex]a_4 = 6 * 432[/tex]

[tex]a_4 = 2592[/tex]

n = 5

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_5 = 6a_{5-1}[/tex]

[tex]a_5 = 6a_4[/tex]

Substitute 2592 for [tex]a_4[/tex]

[tex]a_5 = 6 * 2592[/tex]

[tex]a_5 = 15552[/tex]

n = 6

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_6 = 6a_{6-1}[/tex]

[tex]a_6 = 6a_{5}[/tex]

Substitute 15552 for [tex]a_5[/tex]

[tex]a_6 = 6 * 15552[/tex]

[tex]a_6 = 93312[/tex]

[tex]a_1= 12[/tex]   [tex]a_2= 72[/tex]   [tex]a_3= 432[/tex]   [tex]a_4 = 2592[/tex]   [tex]a_5 = 15552[/tex]   [tex]a_6 = 93312[/tex]

Solving (b):

[tex]a_n = a^2_{n - 1[/tex] where [tex]a_1 = 2[/tex]

We have

[tex]a_1 = 2[/tex] which serves as the first term

n =2

[tex]a_n = a^2_{n - 1[/tex]

[tex]a_2 = a^2_{2-1}[/tex]

[tex]a_2 = a^2_{1}[/tex]

Substitute 2 for [tex]a_1[/tex]

[tex]a_2 = 2^2[/tex]

[tex]a_2 = 4[/tex]

n = 3

[tex]a_3 = a^2_{3-1}[/tex]

[tex]a_3 = a^2_{2}[/tex]

[tex]a_3 = 4^2[/tex]

[tex]a_3 = 16[/tex]

n = 4

[tex]a_4 = a^2_{4-1}[/tex]

[tex]a_4 = a^2_3[/tex]

[tex]a_4 = 16^2[/tex]

[tex]a_4 = 256[/tex]

n =5

[tex]a_5 = a^2_{5-1[/tex]

[tex]a_5 = a^2_4[/tex]

[tex]a_5 = 256^2[/tex]

[tex]a_5 = 65536[/tex]

n = 6

[tex]a_6 = a^2_{6-1[/tex]

[tex]a_6 = a^2_{5[/tex]

[tex]a_6 = 65536^2[/tex]

[tex]a_6 = 4294967296[/tex]

[tex]a_1 = 2[/tex]   [tex]a_2 = 4[/tex]   [tex]a_3 = 16[/tex]   [tex]a_4 = 256[/tex]   [tex]a_5 = 65536[/tex]   [tex]a_6 = 4294967296[/tex]

Solving (c):

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] [tex]a_0=1[/tex]  ; ;  [tex]a_1=2[/tex]

[tex]a_1=2[/tex] ---- First term

n = 2

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_2=a_{2-1}+3a_{2-2}[/tex]

[tex]a_2=a_1+3a_0[/tex]

Substitute values for a1 and a0

[tex]a_2=2+3 * 1[/tex]

[tex]a_2=2+3[/tex]

[tex]a_2=5[/tex]

n = 3

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_3=a_{3-1}+3a_{3-2}[/tex]

[tex]a_3=a_{2}+3a_{1}[/tex]

[tex]a_3=5+3 * 2[/tex]

[tex]a_3=5+6[/tex]

[tex]a_3=11[/tex]

n = 4

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_4=a_{4-1}+3a_{4-2}[/tex]

[tex]a_4=a_{3}+3a_{2}[/tex]

[tex]a_4=11+3 * 5[/tex]

[tex]a_4=11+15[/tex]

[tex]a_4=26[/tex]

n = 5

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_5=a_{5-1}+3a_{5-2};[/tex]

[tex]a_5=a_{4}+3a_3[/tex]

[tex]a_5=26+3 * 11[/tex]

[tex]a_5=26+33[/tex]

[tex]a_5=59[/tex]

n = 6

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_6=a_{6-1}+3a_{6-2}[/tex]

[tex]a_6=a_{5}+3a_4[/tex]

[tex]a_6=59+3*26[/tex]

[tex]a_6=59+78[/tex]

[tex]a_6=137[/tex]

[tex]a_1=2[/tex]   [tex]a_2=5[/tex]   [tex]a_3=11[/tex]   [tex]a_4=26[/tex]   [tex]a_5=59[/tex]   [tex]a_6=137[/tex]

Solving (d):

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex];     [tex]a_0=1[/tex]; [tex]a_1=1[/tex]

[tex]a_1=1[/tex] --- First term

n = 2

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_2=2 * a_{2-1}+2^2a_{2-2}[/tex]

[tex]a_2=2 * a_1+4*a_0[/tex]

[tex]a_2=2 * 1+4*1[/tex]

[tex]a_2=2 +4[/tex]

[tex]a_2=6[/tex]

n = 3

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_3=3 * a_{3-1}+3^2 * a_{3-2}[/tex]

[tex]a_3=3 * a_{2}+9 * a_{1}[/tex]

[tex]a_3=3 * 6+9 * 1[/tex]

[tex]a_3=18+9[/tex]

[tex]a_3=27[/tex]

n = 4

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_4=4*a_{4-1}+4^2*a_{4-2}[/tex]

[tex]a_4=4*a_{3}+16*a_{2}[/tex]

[tex]a_4=4*27+16*6[/tex]

[tex]a_4=204[/tex]

n = 5

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_5=5 * a_{5-1}+5^2 * a_{5-2}[/tex]

[tex]a_5=5 * a_{4}+25 * a_{3}[/tex]

[tex]a_5=5 * 204+25 *27[/tex]

[tex]a_5=1695[/tex]

n = 6

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_6=6 * a_{6-1}+6^2*a_{6-2}[/tex]

[tex]a_6=6 * a_{5}+36*a_{4}[/tex]

[tex]a_6=6 * 1695+36*204[/tex]

[tex]a_6=17514[/tex]

[tex]a_1=1[/tex]   [tex]a_2=6[/tex]   [tex]a_3=27[/tex]   [tex]a_4=204[/tex]   [tex]a_5=1695[/tex]   [tex]a_6=17514[/tex]

Solving (e):

[tex]a_n= a_{n-1}+a_{n-3};\ a_0=1; a_1=2; a_2=0[/tex]

First term: [tex]a_1=2[/tex]

Second Term: [tex]a_2=0[/tex]

n = 3

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_3= a_{3-1}+a_{3-3}[/tex]

[tex]a_3= a_{2}+a_0[/tex]

[tex]a_3= 0+1[/tex]

[tex]a_3= 1[/tex]

n = 4

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_4= a_{4-1}+a_{4-3}[/tex]

[tex]a_4= a_{3}+a_{1}[/tex]

[tex]a_4= 1+2[/tex]

[tex]a_4=3[/tex]

n = 5

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_5= a_{5-1}+a_{5-3}[/tex]

[tex]a_5= a_{4}+a_{2}[/tex]

[tex]a_5= 3 + 0[/tex]

[tex]a_5= 3[/tex]

n = 6

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_6= a_{6-1}+a_{6-3}[/tex]

[tex]a_6= a_{5}+a_{3}[/tex]

[tex]a_6= 3 + 1[/tex]

[tex]a_6= 4[/tex]

[tex]a_1=2[/tex]   [tex]a_2=0[/tex]   [tex]a_3= 1[/tex]   [tex]a_4=3[/tex]   [tex]a_5= 3[/tex]   [tex]a_6= 4[/tex]

Answer:

61.7

Step-by-step explanation:

Answer:

What's the question?

Step-by-step explanation:

Answer:

It is $15.00

Step-by-step explanation:

It is B because A is not even money, C is fifteen cents which is not correct because the question is asking for 75% of $20.00, and finally D is incorrect because it is more than the amount of money asked in the question.

The x intercept is (9/2,0) or (4.5,0). The y intercept is (0,-1):)

Answer:

what's that ,we are all ears

Answer:

1/26

Step-by-step explanation:

There are two sets colors; red and black.

13 ranks, each rank has 2 black cards and 2 red

but there are only 2 black eights

2/52

Answer:

Real

Step-by-step explanation:

i^6-122-7i^2

(i^2)^3-122-7i^2

(-1)^3-122-7×(-1)

-1-122+7

-123+7

-116

So the final answer is -116

Answer:

Answers below

Step-by-step explanation:

Equation

Let's call:

p = number of pennies Janesa has in her pocket

n = number of nickels Janesa has in her pocket

She has 24 coins in total, thus:

[tex]p + n=24\qquad\qquad [1][/tex]

Since each penny is worth $0.01 and each nickel is worth $0.05, and Janesa has $1 in total:

[tex]0.01p+0.05n=1[/tex]

Multiplying by 100:

[tex]p+5n=100\qquad\qquad [2][/tex]

A. Completing the equation:

[tex]5n+p=100[/tex]

B.

From [1]:

[tex]p=24-n[/tex]

Substituting into [2]:

[tex]24-n+5n=100[/tex]

Simplifying:

[tex]24+4n=100[/tex]

Rearranging:

[tex]4n=100-24=76[/tex]

[tex]n=76 / 4= 19[/tex]

[tex]p=24-19=5[/tex]

B. There are 5 pennies in Janesa's pocket

C. Since each penny is worth $0.01, the total value of pennies is 5*$0.01=$0.05

The total value of the pennies is $0.05

D. The total value of the nickels is 19*$0.05 = $0.95

E. Jackson should say she has 19 nickels and 5 pennies

janesa has 24 coins in her pocket worth exactly $1.00. she tells her younger brother jackson that he can have the coins if he correctly guesses how many of each coin she has. she gives him the hint that there are only nickels and pennies. complete the equation that represents the total value of the nickels and p pennies 5( blank ) + p = 100 The type of coins that Janessa has = nickels and pennies.

Janessa has a total number of 24 coins

The sum of the nickels and pennies = $1.00

Let us represent the number of nickels as x

The number of pennies as y