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ycow [4]
3 years ago
14

What is the sum of the first 14 terms of the sequence an = 6n -12

Mathematics
1 answer:
mina [271]3 years ago
6 0
When n = 1  first term  = -6 
n = 2  second term =  0
n = 3 third term = 6

n = 4 4th term = 12

so we have an Arithmetic sequence  first term = -6 and common difference = 6

Sum 14 terms  =  (14/2)[2*-6 + (14-1)*6]

 =  462  answer
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Caroline drove 350 miles to her grandmother's house. the trip took her 51/4 hours. what is the average speed to miles per hour?
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6 0
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Several students decide to start a T-shirt company after initial expenses of $280 they purchase each T-shirt wholesale for $3.99
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4 0
3 years ago
Read 2 more answers
HELP ME DEAR GOD-
romanna [79]

Answer:

The probability you will get a head at least once is 50%.

Step-by-step explanation:

Since the question is asking about the probability you will get, we can assume we’re answering based on theoretical probability. This type of probability is based on logic.

A coin always has two sides, one with head and the other with tails. So we can easily represent this as half and half. 1/2 as a fraction. 0.5 as a decimal. 50% as a percent. This means that P(H) will be equal to any one of these as they are all the same. The same can be said for the probability that a head does not appear, in other words, a tail appears. The reason being that the probability is split evenly between the two. This will again mean that P(T) will equal to any one of those.

So, A = 50% and B = 50%. The probability you will get a head at least once is 50%.

7 0
3 years ago
Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find
Papessa [141]

Answer:

1) \text{P(at least one boy and one girl)}=\frac{3}{4}

2) \text{P(at least one boy and one girl)}=\frac{3}{8}

3) \text{P(at least two girls)}=\frac{1}{2}

Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To  Find : The probability of each event.  

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)        

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

1) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, BGG, GBB, GBG, GGB=6

\text{P(at least one boy and one girl)}=\frac{6}{8}

\text{P(at least one boy and one girl)}=\frac{3}{4}

2) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, GBB=3

\text{P(at least one boy and one girl)}=\frac{3}{8}

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

\text{P(at least two girls)}=\frac{4}{8}

\text{P(at least two girls)}=\frac{1}{2}

4 0
4 years ago
Read 2 more answers
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