Which equation demonstrates the additive identity property?

<img src="https://tex.z-dn.net/?f=%20%7B4%7D%5E%7B3%7D%20" id="TexFormula1" title=" {4}^{3} " alt=" {4}^{3} " align="absmiddle"

quester [9]

Answer:

64

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the area of this figure?

KATRIN_1 [288]

Answer:

10.5

Step-by-step explanation:

Area is length times width so, 3.5 times 3 equals 10.5. 10.5 is the area of the given figure.

PLZ MARK BRAINLIEST!!! HOPE THIS HELPS!!!

3 0

2 years ago

Help me please past due! Complete all 5 problems in simplest form and tell me what you did

Sonbull [250]

3/6= 1/2 because 3 is the half of 6 and 1 is the half of 2

5/20= 1/4 because 1 x 5= 5 and 4 x 5= 20

9/27= 1/3 because 1 x 9 =9 and 3 x 9=27

1/2 is the simplest form already

12/36= 1/3 because 1 x 12= 12 and 3 x 12= 36

6 0

2 years ago

Able, Ben and Cal each played a game able’s score was six times ben’s score, Cale score was a hird time of Able’s score write do

Feliz [49]

Answer:

6:1:2

Step-by-step explanation:

Let a = Able's score, b = Ben's score, and c = Cal's score.

Since

Able's score was 6 times Ben's score, that means a = 6b.

Cal's score was a third of Able's score, so that means c = a/3. And since a = 6b, that means c = 6b / 3 = 2b.

Thus, the ratio of Able's score to Ben's score to Cal's score, a:b:c, is 6:1:2, because c is twice as much as b and a is 6 times as much as b.

3 0

3 years ago

Imagine an experiment having three conditions and 20 subjects within each condition. The mean and variances of each condition ar

xxTIMURxx [149]

Answer:

1. Mean square B= 5.32

2. Mean square E= 16.067

3. F= 0.33

4. p-value: 0.28

Step-by-step explanation:

Hello!

You have the information of 3 groups of people.

Group 1

n₁= 20

X[bar]₁= 3.2

S₁²= 14.3

Group 2

n₂= 20

X[bar]₂= 4.2

S₂²= 17.2

Group 3

n₃= 20

X[bar]₃= 7.6

S₃²= 16.7

1. To manually calculate the mean square between the groups you have to calculate the sum of square between conditions and divide it by the degrees of freedom.

Df B= k-1 = 3-1= 2

Sum Square B is:

∑ni(Ÿi - Ÿ..)²

Ÿi= sample mean of sample i ∀ i= 1,2,3

Ÿ..= general mean is the mean that results of all the groups together.

General mean:

Ÿ..= (Ÿ₁ + Ÿ₂ + Ÿ₃)/ 3 = (3.2+4.2+7.6)/3 = 5

Sum Square B (Ÿ₁ - Ÿ..)² + (Ÿ₂ - Ÿ..)² + (Ÿ₃ - Ÿ..)²= (3.2 - 5)² + (4.2 - 5)² + (7.6 - 5)²= 10.64

Mean square B= Sum Square B/Df B= 10.64/2= 5.32

2. The mean square error (MSE) is the estimation of the variance error (σ $_{e}^2$ → $S_{e} ^{2}$ ), you have to use the following formula:

Se²= (n₁-1)S₁² + -(n₂-1)S₂² + (n₃-1)S₃²

n₁+n₂+n₃-k

Se²= 19*14.3 + 19*17.2 + 19*16.7 = 915.8 = 16.067

20+20+20-3 57

DfE= N-k = 60-3= 57

3. To calculate the value of the statistic you have to divide the MSB by MSE

$F= \frac{Mean square B}{Mean square E} = \frac{5.32}{16.067} = 0.33$

4. P(F $_{2; 57}$ ≤ F) = P(F $_{2; 57}$ ≤ 0.33) = 0.28

I hope you have a SUPER day!

3 0

2 years ago