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belka [17]
3 years ago
10

There below! Can somebody help me please?

Mathematics
2 answers:
Montano1993 [528]3 years ago
4 0

Answer: C. 24 units

Step-by-step explanation:

The circumference of a circle is 2πr.  Let's say x = 1/3r.  Thus, the area of circle A relative to circle B is 1:3.  Thus, circle B is 3 times larger than circle A.  So, simply divide 64π by 4 to get 16π.  Then multiply 16π*3=48π.  For the circumference of a circle to be 48, the radius would have to be 24.

Hope it helps <3

babymother [125]3 years ago
3 0

Answer:

The answer is C, 24 units.

Step-by-step explanation:

Let r_A be the radius is circle A and r_B be the radius of circle B.

Radius of circle A is 1/3 of the radius of circle B, or:

r_A=\frac{1}{3}r_B

3r_A=r_B

The sum of their circumferences is 64\pi. In other words:

2\pi r_A +2\pi r_B = 64\pi

Substitute r_B:

2\pi(r_A +r_B)=64\pi

2\pi (r_A +3r_A)=64\pi

4r_A=32, r_A=8

r_B=3r_A, r_B=3(8)=24

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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.

6 0
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For how many weeks would the class need to have car washes to earn $5000.
nikdorinn [45]
The correct answer would be for 29 weeks.
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Answer:

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SOH-CAH-TOA

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P = \frac{1}{14}

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