Find the product of 6x^2(x+4)(x-5)

The vectors <4,2> and <-4,-2> are opposite directions and are orthogonal to the vector u = <2,- 4>. That’s bec

Naddik [55]

This is true since the dot product is zero

5 0

3 years ago

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 44 o

Cerrena [4.2K]

Answer:

a) No 95% of values will fall between (24;64); 68,27% will fall between (34;54)

b)71,83 % will fall between 34 and 64 ounces

Step-by-step explanation:

Empirical rule establishes, for a normal distribution with mean μ and σ as standard deviation:

In interval μ ± σ or ( μ + σ ; μ - σ) we should find 68.27 % of all values of the population, and by simmetry 68.27/2 = 34,14 % should be over the mean and the other half would be values below the mean

Therefore in our case

μ + σ = 44 + 10 = 54

And

μ - σ = 44 - 10 = 34

a) Then 68,34 % of values will fall in this interval

We know now that value 34 is 1* σ below the mean, and is at the limit of 34,14 %

b) μ + 2*σ = 44 * 2*10 = 44 + 20 = 64

64 is the upper limit for the interval μ + 2*σ and we know that 95.45 % of all values will fall between ( μ - 2*σ ; μ + 2*σ ) and by simmetry just one side of this interval (the right side ) will have 95.45/2 = 47;73 %

Then in interval going from ( 34 ; 64 ) we shoud find 47.73 + 34,14

71,83 % of all values will fall between 34 and 64

7 0

3 years ago

Study the coordinate plane below. A certain polygon has its vertices at the following points: (1, 1), (1, 8), (4, 7), and (4, 3)

jarptica [38.1K]

Answer:

pentegon

Step-by-step explanation:

3 0

2 years ago

Solve for x. plz help

LiRa [457]

$x^2-2x+10=0\\\\ (x^2-2x+1)-1+10=0\\\\(x-1)^2+9=0\\\\ (x-1)^2=-9$

Answer D

7 0

3 years ago

Read 2 more answers

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

3 years ago