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Lady bird [3.3K]
3 years ago
8

Simplify. square root (108 x^5 y^6)

Mathematics
1 answer:
sveta [45]3 years ago
3 0
\bf \sqrt{108x^5y^6}\qquad 
\begin{cases}
108=2\cdot 2\cdot 3\cdot 3\cdot 3\\
\qquad 2^2\cdot 3^2\cdot 3\\
\qquad (2\cdot 3)^2\cdot 3\\
\qquad 6^2\cdot 3\\
x^5=x^{4+1}\\
\qquad x^4\cdot  x^1\\
\qquad x^{2\cdot 2}\cdot x\\
\qquad (x^2)^2\cdot x\\
y^6=y^{3\cdot 2}\\
\qquad (y^3)^2
\end{cases}\implies \sqrt{6^2\cdot 3\cdot (x^2)^2\cdot x\cdot (y^3)^2}
\\\\\\
6x^2y^3\sqrt{3x}
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A study was conducted to determine whether magnets were effective in treating pain. The values represent measurements of pain us
amm1812

Answer:

Given:

Sham: n= 20,     x=0.44,   s=1.24,

Magnet:n= 20,  x =0.49,   s= 0.95

For Sham:

Sample size, n = 20

Sample mean = 0.44

Standard deviation = 1.24

For Magnet:

Sample size = 20

Sample mean = 0.49

Standard deviation = 0.95

The null and alternative hypotheses:

H0: s1²=s2²

H1: s1² ≠ s2²

a) To find the test statistics, use the formula:

\frac{s1^2}{s2^2}

\frac{1.24^2}{0.95^2} = \frac{1.5376}{0.9025} = 1.7037

Test statistics = 1.7037

b) P-value:

Sham: degrees of freedom = n - 1 = 20 - 1 = 19

Magnet: degrees of freedom = n - 1 = 20 - 1 = 19

The critical values:

[Za/2, df1, df2)], [(1 - Za/2), df1, df2]

f[0.05/2, 19, 19], f[(1 - 0.05/2), 19, 19]

f[0.025, 19, 19], f[0.975, 19, 19]

(2.526, 0.3958)

The rejection region:

Reject H0, if  F < 0.3958 or if F > 2.526

c) Conclusion:

Since the critical values of test statistic is between (0.3958 < 1.7037 < 2.526), we fail to reject null hypothesis H0.

There is insufficient evidence to to support the claim that those given a sham treatment have reductions that vary more than those treated with magnets

3 0
3 years ago
In a fruit cocktail, for every 20 ml of orange juice you need 10 ml of apple juice and 15 ml of coconut milk. What proportion of
MariettaO [177]

Answer:

1/3

Step-by-step explanation:

20 ml + 10 ml + 15 ml = 45 ml total

15 ml coconut milk/45 ml total = 15/45 = 1/3 (as a fraction)

7 0
3 years ago
Q2) The following travel times were measured for vehicles traversing a 2,000 ft. segment of an arterial: Vehicle Travel Time (s)
adelina 88 [10]

Answer:

TMS = 45.78 m/s

SMS = 45.66 m/s

Step-by-step explanation:

Vehicle Travel Time (s)

1 40.5

2 44.2

3 41.7

4 47.3

5 46.5

6 41.9

7 43.0

8 47.0

9 42.6

10 43.3

Time mean speed and space mean speed are parameters used to express the speed of a group of vehicles in traffic flow.

Time mean speed is given as the average speed of all the cars

The speed for each vehicle is calculated by dividing the (2000 ft) distance by the time taken by each vehicle to complete that segment of the journey.

Let vehicle = V

Travel time = time

Speed = s

V | Time | speed (ft/s)

1 | 40.5 | 49.383

2 | 44.2 | 45.249

3 | 41.7 | 47.962

4 | 47.3 | 42.283

5 | 46.5 | 43.011

6 | 41.9 | 47.733

7 | 43.0 | 46.512

8 | 47.0 | 42.553

9 | 42.6 | 46.948

10| 43.3 | 46.189

TMS = (Σsᵢ)/N

Σsᵢ = (49.383+45.249+47.962+42.283+43.011+47.733+46.512+42.553+46.948+46.189)

Σsᵢ = 457.823

N = number of vehicles = 10

TMS = 457.823 ÷ 10 = 45.7823 = 45.782 m/s

b) Space mean speed is given as

SMS = N ÷ [Σ (1/sᵢ)]

V | Time | speed | (1/sᵢ)

1 | 40.5 | 49.383 | 0.02025

2 | 44.2 | 45.249 | 0.02210

3 | 41.7 | 47.962 | 0.02085

4 | 47.3 | 42.283 | 0.02365

5 | 46.5 | 43.011 | 0.02325

6 | 41.9 | 47.733 | 0.02095

7 | 43.0 | 46.512 | 0.02150

8 | 47.0 | 42.553 | 0.02350

9 | 42.6 | 46.948 | 0.02130

10| 43.3 | 46.189 | 0.02165

Σ (1/sᵢ) = (0.02025+0.02210+0.02085+0.02365+0.02325+0.02095+0.02150+0.02350+0.02130+0.02165)

Σ (1/sᵢ) = 0.219

SMS = 10 ÷ [Σ (1/sᵢ)] = 10 ÷ 0.219

SMS = 45.662 m/s

Hope this Helps!!!

4 0
3 years ago
If f(x)=-9x-2, find f(7)
Elza [17]

Answer:

x= -1/5

Let me know if you need an explanation :)

3 0
3 years ago
In ΔLMN, \overline{LN} LN is extended through point N to point O, m∠NLM = (x+1)^{\circ}(x+1) ∘ , m∠LMN = (x+15)^{\circ}(x+15) ∘
mixer [17]

Answer:

26 degree

Step-by-step explanation:

We are given that  in triangle

\angle NLM=(x+1)^{\circ}

\angke LMN=(x+15)^{\circ}

\angle MNO=(4x+6)^{\circ}

\angle MNO+\angle MNL=180^{\circ}

By using linear pair angles property

\angle MNL=180-\angle MNO=180-(4x+6)

\angle MNL+\angle NLM+\angle LMN=180^{\circ}

By using triangle angles sum property

180-(4x+6)+x+15+x+1=180

2x+16=180-180+4x+6

16-6=4x-2x=2x

2x=10

x=\frac{10}{2}=5

Substitute the value

\angle MNO=4(5)+6=20+6=26^{\circ}

3 0
3 years ago
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