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Evgesh-ka [11]
3 years ago
10

In the warmer months, Ms.Selwa likes to run to school to get in her daily exercise. She can run 1/6 of a kilometer in a minute.

Michael Power Saint Joseph is 3/6 of a kilometer away from her home. At this speed, how long will it take Ms.Selwa to run to school every morning?
Mathematics
1 answer:
Lerok [7]3 years ago
8 0

Answer:

3 minutes

Step-by-step explanation:

If Ms. Selwa can run 1/6 of a kilometer in a minute, she can run 3/6 of a kilometer in 3 minutes.

This is because 1/6 times 3 equals 3/6, which is the distance of the school from her home.

So, it will take Ms. Selwa 3 minutes to run to school.

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Find the sum of the first 20 terms of an arithmetic sequence with an 18th term of 8.1 and a common difference of 0.25.
il63 [147K]

The sum of the first 20 terms of an arithmetic sequence with the 18th term of 8.1 and a common difference of 0.25 is 124.5

Given,

18th term of an arithmetic sequence = 8.1

Common difference = d = 0.25.

<h3>What is an arithmetic sequence?</h3>

The sequence in which the difference between the consecutive term is constant.

The nth term is denoted by:

a_n = a + ( n - 1 ) d

The sum of an arithmetic sequence:

S_n = n/2 [ 2a + ( n - 1 ) d ]

Find the 18th term of the sequence.

18th term = 8.1

d = 0.25

8.1 = a + ( 18 - 1 ) 0.25

8.1 = a + 17 x 0.25

8.1 = a + 4.25

a = 8.1 - 4.25

a = 3.85

Find the sum of 20 terms.

S_20 = 20 / 2 [ 2 x 3.85 + ( 20 - 1 ) 0.25 ]

         = 10 [ 7.7 + 19 x 0.25 ]

         = 10 [ 7.7 + 4.75 ]

         = 10 x 12.45

         = 124.5

Thus the sum of the first 20 terms of an arithmetic sequence with the 18th term of 8.1 and a common difference of 0.25 is 124.5

Learn more about arithmetic sequence here:

brainly.com/question/25749583

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8 0
1 year ago
One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R
marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

Urn\ 1 = \{B_1, R_1, R_2, R_3\}

Urn\ 2 = \{R_4, R_5, B_2, B_3\}

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

If urn 2 is selected, the following selection exists:

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

<em>See attachment for possibility tree</em>

Solving (b): The total number of outcome

<u>For urn 1</u>

There are 4 balls in urn 1

n = \{B_1,R_1,R_2,R_3\}

Each of the balls has 3 subsets. i.e.

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

So, the selection is:

Urn\ 1 = 4 * 3

Urn\ 1 = 12

<u>For urn 2</u>

There are 4 balls in urn 2

n = \{B_2,B_3,R_4,R_5\}

Each of the balls has 3 subsets. i.e.

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

So, the selection is:

Urn\ 2 = 4 * 3

Urn\ 2 = 12

Total number of outcomes is:

Total = Urn\ 1 + Urn\ 2

Total = 12 + 12

Total = 24

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Answer:

A≈61.94

Just plug it into a calculator

     

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If the radius of the circle is multiplied by a factor of 3, then the area of the circle is multiplied by a factor of
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3 years ago
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