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maxonik [38]
3 years ago
8

Sara can finish 5 laps around the track in 6 minutes. What is her average speed in miles per hour, if it takes 12 laps around th

e track to run a mile?
Mathematics
1 answer:
sveticcg [70]3 years ago
4 0

9514 1404 393

Answer:

  4 1/6 miles per hour

Step-by-step explanation:

6 minutes is 1/10 hour, so Sara's speed is ...

  speed = distance/time

  speed = (5/12 mi)/(1/10 h) = 50/12 mi/h

  speed = 4 1/6 mi/h

Sara's speed is 4 1/6 miles per hour.

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I need 2-4 please & show step by step explanation ( show the work ) please!
Likurg_2 [28]

Answer:

i don't know go on your browser

Step-by-step explanation:

3 0
3 years ago
H + 732 = -194 <br> Solve for H
lilavasa [31]
H + 732 = -194

* combine like terms

H = -194 - 732

H = -926
4 0
3 years ago
Are they unique more than one or no triangle
Tanya [424]
The measures of the angles of a triangle add to 180.

45 + 45 + 90 = 180
The measures of these three angles do add to 180, ,so there is at least one triangle with these angle measures.
Using AA triangle similarity, any triangle with the same angle measures will be similar.

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8 0
3 years ago
Find the first four terms of the sequence an = 4n-6
natulia [17]

Answer:

The first four terms of the sequence are-6, -2, 2 and 6

Step-by-step explanation:

The given sequence is a_n=4n-6

In order to find the first four term of the sequence we put n =0, 1, 2 and 3.

For n =0

a_0=4(0)-6=-6

For n =1

a_1=4(1)-6=-2

For n =2

a_2=4(2)-6=2

For n =3

a_3=4(3)-6=6

Therefore, the first four terms of the sequence are

-6, -2, 2 and 6

5 0
3 years ago
Daily low temperatures in Columbus, OH in January 2014 were approximately normally distributed with a mean of 15.45 and a standa
Alona [7]

Answer: 12.10

Step-by-step explanation:

Given : Mean : \mu = 15.45

Standard deviation : \sigma = 13.70

The formula to calculate the z-score :-

z=\dfrac{x-\mu}{\sigma}

For x= 5 degrees

z=\dfrac{5-15.45}{13.70}=-0.7627737226\approx-0.76

For x= 10 degrees

z=\dfrac{10-15.45}{13.70}=-0.397810218\approx-0.40

The P-value : P(-0.76

=0.3445783-0.2236273=0.120951\approx0.1210

In percent , 0.1210\times100=12.10\%

Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%

5 0
3 years ago
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