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2 years ago

13

Lin runs 5 laps around a track in 6 minutes. How many laps per minute is that?

2 answers:

irga5000 [103]2 years ago

7 0

Answer:

divide 6 by 5 your answer is 1.2

lidiya [134]2 years ago

5 0

Answer: 5/6

Step-by-step explanation: Its very simple. All you have to do is list it out. If he gets 5 laps in 6 min, prematurely, you could assume that the laps and the minutes are linear. Then, if he runs 5 in 6min, you could divide both sides by 6, thus causing the answer of 5/6 laps in 1 min.

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Sedbober [7]

49 would be in the next space after 42 because this sequence is counting by 7s

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2 years ago

5 x 1 please?? im teaching my cousin right now

Pachacha [2.7K]

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2 years ago

Read 2 more answers

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ b(1)=−2 b(n)=b(n−1)−7 Find the 3 rd 3 rd 3, start superscript, start text, r, d, end text, end superscript ter

vitfil [10]

Answer:

-16

Step-by-step explanation:

Given the sequence:

b(n)=b(n−1)−7, where b(1)=−2

b(2)=b(2−1)−7

=b(1)−7

=-2-7

b(2)=-9

Therefore, the 3rd term of the sequence

b(3)=b(3−1)−7

=b(2)-7 (Recall b(2)=-9 from above)

=-9-7

b(3)=-16

The 3rd term of the sequence is -16.

5 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

What is point of view

melamori03 [73]

Answer:

"a particular attitude or way of considering a matter. "

hope this helps :3

4 0

3 years ago