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Kisachek [45]
3 years ago
14

<Q and <R are complementary angles. If m<Q = (31 - 3x)° and m<R = (19x - 5)°, find m<R​

Mathematics
2 answers:
Sergio [31]3 years ago
7 0

31-3x+19x-5=90

26+16x=90

-26 -26

--------------------

16x= 64

16x/16= 64/16

x=4

m <R=19 (4)-5= 71.

sergiy2304 [10]3 years ago
6 0

Answer:

Step-by-step explanation:

90=(31-3x)+(19x-5)

90=31-3x+19x-590=19x-3x+31-5

90=16x+26

90-26=17x

64=4x

x=4

m∠R=19(4)-5

m∠R=76-5

m∠R=71

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Step-by-step explanation:

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3 years ago
alpha and beta are the zeros of the polynomial x^2 -(k +6)x +2(2k -1). Find the value of k if alpha + beta = 1/2 alpha beta(ITS
serious [3.7K]

Answer:

k=\frac{-11}{2}.

Step-by-step explanation:

We are given \alpha and \beta are zeros of the polynomial x^2-(k+6)x+2(2k-1).

We want to find the value of k if \alpha+\beta=\frac{1}{2}.

Lets use veita's formula.

By that formula we have the following equations:

\alpha+\beta=\frac{-(-(k+6))}{1}  (-b/a where the quadratic is ax^2+bx+c)

\alpha \cdot \beta=\frac{2(2k-1)}{1} (c/a)

Let's simplify those equations:

\alpha+\beta=k+6

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If \alpha+\beta=k+6 and \alpha+\beta=\frac{1}{2}, then k+6=\frac{1}{2}.

Let's solve this for k:

Subtract 6 on both sides:

k=\frac{1}{2}-6

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k=\frac{1}{2}-\frac{12}{2}

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3 years ago
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Nostrana [21]

Kindly find complete question attached below

Answer:

Kindly check explanation

Step-by-step explanation:

Given a normal distribution with ;

Mean = 36

Standard deviation = 4

According to the empirical rule :

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That is ; mean ± 1(standard deviation)

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2.)

95% of the distribution is within 2 standard deviations of the mean ;

That is ; mean ± 2(standard deviation)

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Between 28 and 44

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99% is about 3 standard deviations of the mean :

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3 0
2 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=<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
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