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yawa3891 [41]
3 years ago
13

Please help me on this

Mathematics
2 answers:
earnstyle [38]3 years ago
6 0
The answer is D because the square root of 3^15 is 2187 square root of 3. Now if you add 3^7 then that gives you 2187. So your answer would be D
Liula [17]3 years ago
4 0
<h2>Hello!</h2>

The answer is:

The correct option is D. \sqrt{3^{15} }=3^{7}* \sqrt{3}

<h2>Why?</h2>

To solve this problem, we must remember the following properties:

Product of powers with the same base:

a^{m}a^{n}=a^{m+n}

Power of a power:

(a^{m})^{n}=a^{m*n}

Also, we must remember that if we want to introduce a number into a root, in order to not change the expression, we need to introduce it with the same exponent:

a*\sqrt{b}=\sqrt{a^{2}*b }

So, solving the problem, we have:

\sqrt{3^{15} }=3^{7}*\sqrt{3}\\\\3^{7}*\sqrt{3}=\sqrt{(3^{7})^{2} *3}\\\\\sqrt{(3^{7})^{2} *3}=\sqrt{(3^{14} *3}\\\\\sqrt{3^{14} *3}=\sqrt{3^{14+1}

\sqrt{3^{14+1} } =\sqrt{3^{15}}

Hence,

\sqrt{3^{15} }=3^{7}* \sqrt{3}

So, the correct option is D. \sqrt{3^{15} }=3^{7}* \sqrt{3}

Have a nice day!

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What is the minimum number of degrees that a regular hexagon can be rotated before it carries onto itself?
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Two surfers and statistics students collected data on the number of days on which surfers surfed in the last month for 30 longbo
Alisiya [41]

Answer:

Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders

Step-by-step explanation:

The null hypothesis is that  the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders

H0:  μL > μs      against the claim Ha:  μL≤ μs

the alternate hypothesis is the mean days surfed for all long boarders isless or equal to  the mean days surfed for all short boarders (because long boards can go out in many different surfing conditions)

The test statistic is

t= x1- x2/  √s1/n1+ s2/n2

1) Calculations

Longboards

Mean

ˉx=∑x/n=4+8+9+4+9+7+9+6+6+11+15+13+16+12+10+12+18+20+15+10+15+19+21+9+22+19+23+13+12+10/30

=377/30

=12.5667

Longboard Variance S2=[∑dx²-(∑dx)²/n]/n-1

=[831-(-13)²/30]/29

=831-5.6333/29

=825.3667/29

=28.4609

Shortboard Mean

ˉx=∑x/n=6+4+6+6+7+7+7+10+4+6+7+5+8+9+4+15+13+9+12+11+12+13+9+11+13+15+9+19+20+11/30

=288/30

=9.6

Shortboard Variance S2=[∑x²-(∑x)²/n]/n-1

=[ 3270-(288)2/30]/29

=3270-2764.8/29

=505.2/29

=17.4207

2) Putting values in the test statistic

t=|x1-x2|/√S²1/n1+S²2/n2

t =|12.5667-9.6|/√28.4609/30+17.4207/30

t =|2.9667|/√0.9487+0.5807

t=|2.9667|/√1.5294

t=|2.9667|/1.2367

t=2.3989

3) Degree of freedom =n1+n2-2=30+30-2=58

4) The critical region is t ≤ t(0.05) (58) =1.6716

5) Since the calculated t= 2.4 does not fall in the critical region t(0.05) (58)  ≤ 1.6716 we do not reject H0.

The p-value is 0.008969. The result is significant at p <0 .05.

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3 years ago
What is the solution to the system of equations below?
Lilit [14]

Answer:

  x,y = -4, -15

Step-by-step explanation:

4 0
3 years ago
WILL GIVE BRAINLIEST!!!!
german
3/-7 is equivalent to -3/7z
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3 years ago
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