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olga2289 [7]
3 years ago
7

Let log base aU=X and log base aV=Y, then a to the x power =? and a to y power =?

Mathematics
1 answer:
Andrews [41]3 years ago
3 0

Answer:

{ \bf{ log_{a}(U)  = x}} \\ { \boxed{ \tt{ {a}^{x}  = U}}} \\  \\ { \bf{ log_{a}(V)  = y}} \\ { \boxed{ \tt{ {a}^{y} = V }}}

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Round 35295 to the nearest thousand
Alex_Xolod [135]

35000, rounding down from five.

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3 years ago
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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=<u> (n₁-1)S₁² + -(n₂-1)S₂² + (n₃-1)S₃²</u>

                        n₁+n₂+n₃-k

Se²=<u> 19*14.3 + 19*17.2 + 19*16.7 </u>= <u>  915.8   </u>  = 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
I will mark you as brainlinest for correct answer!!!!!pleaseeeee help!!!!!!
mel-nik [20]
A^2 - b^2 = (a+b)(a-b)

answer :

t^2 - 1
n^2 - 225 
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3 years ago
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10 POINTS !! :( ANSWER BOTH PLS !! DUE BEFORE 11:59 PM
fenix001 [56]

Answer:

Easy,

1) To find the area you need to multiply. Therefore 10 x 4 = 40 square centimeters.

2) 20, why? A calculator online for area of a triangle, it really helps! :) just look up triangle area in the future.

Step-by-step explanation:

8 0
2 years ago
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.6 Mbp
olga55 [171]

Answer:

a) 59.98

b) 2.99

c) 2.99

d) Significantly High

Step-by-step explanation:

Part a)

Highest speed measured = x = 75.6 Mbps

Average/Mean speed = \overline{x} = 15.62 Mbps

Standard Deviation = s = 20.03 Mbps

We need to find the difference between carrier's highest data speed and the mean of all 50 data​ speeds i.e. x - \overline{x}

x - \overline{x} = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

\frac{59.98}{20.03}=2.99

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

z=\frac{x-\overline{x}}{s}

Using the given values, we get:

z=\frac{75.6-15.62}{20.03}=2.99

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly​ high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the  carrier's highest data speed​ is significantly higher.

3 0
3 years ago
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