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RSB [31]
3 years ago
9

On the first day of baseball tournament Jessie scored 2 runs. On the second day, 4 runs. On the third day, 6 runs. If this patte

rn continues, how many runs should Jessie score on the eighth day?
Mathematics
1 answer:
Allushta [10]3 years ago
8 0

Answer:

16 runs.

Step-by-step explanation:

We have been given that on the first day of baseball tournament Jessie scored 2 runs. On the second day, 4 runs. On the third day, 6 runs.

We can see that runs scored by Jessie form an arithmetic sequence, where each successive term is 2 more than the previous term.

Since we know that formula for nth term of an arithmetic sequence is: a_n=a_1+(n-1)d, where,

a_n=\text{nth term of the sequence},

a_1=\text{1st term of the sequence},

n=\text{Number of terms of the sequence},

d=\text{Common difference}.

Since on the first day Jessie scored 2 runs, so a_1=2 and difference between two consecutive terms is 2 (4-2=2), so d will be 2.

Upon substituting our values in arithmetic sequence formula we will get,

a_n=2+(n-1)*2

a_n=2+2n-2

a_n=2n

Therefore, formula for nth term of sequence representing number scored by Jessie on the baseball tournament is a_n=2n.

Let us find the 8th term of sequence by substituting n=8 in our sequence formula.

a_8=2*8

a_8=16

Therefore, Jessie should score 16 runs on the eighth day of baseball tournament.

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BlackZzzverrR [31]

Answer:

(1) The value of P (A) is 0.4286.

(2) The value of P (B) is 0.50.

(3) The value of P (A ∩ B) is 0.2143.

(4) The the value of P (B|A) is 0.50.

(5) The events <em>A</em> and <em>B</em> are independent.

Step-by-step explanation:

The events are defined as follows:

<em>A</em> = a student in the class has a sister

<em>B</em> = a student has a brother

The information provided is:

<em>N</em> = 210

n (A) = 90

n (B) = 105

n (A ∩ B) = 45

The probability of an event <em>E</em> is the ratio of the favorable number of outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

The conditional probability of an event <em>X</em> provided that another event <em>Y</em> has already occurred is:

P(X|Y)=\frac{P(A\cap Y)}{P(Y)}

If the events <em>X</em> and <em>Y</em> are independent then,

P(X|Y)=P(X)

(1)

Compute the probability of event <em>A</em> as follows:

P(A)=\frac{n(A)}{N}\\\\=\frac{90}{210}\\\\=0.4286

The value of P (A) is 0.4286.

(2)

Compute the probability of event <em>B</em> as follows:

P(B)=\frac{n(B)}{N}\\\\=\frac{105}{210}\\\\=0.50

The value of P (B) is 0.50.

(3)

Compute the probability of event <em>A</em> and <em>B</em> as follows:

P(A\cap B)=\frac{n(A\cap B)}{N}\\\\=\frac{45}{210}\\\\=0.2143

The value of P (A ∩ B) is 0.2143.

(4)

Compute the probability of <em>B</em> given <em>A</em> as follows:

P(B|A)=\frac{P(A\cap B)}{P(A)}\\\\=\frac{0.2143}{0.4286}\\\\=0.50

The the value of P (B|A) is 0.50.

(5)

The value of P (B|A) = 0.50 = P (B).

Thus, the events <em>A</em> and <em>B</em> are independent.

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