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Lostsunrise [7]
2 years ago
8

PLS HELP

Mathematics
1 answer:
Anika [276]2 years ago
7 0

The chairs in the auditorium illustrates an arithmetic sequence

  • The recursive rule is:  an = an-1 + 8; a1 = 26
  • The explicit rule is:  an = 18 + 8n
  • There are 74 seats in the 7th row
  • 840 chairs can fit the auditorium

<h3>The recursive rule</h3>

From the question, we have the following sequence:

Seats: 18, 26, 34

Rewrite as:

26 = 18 + 8

34 = 26 + 8

The above means that:

The number of seats on a row is 8 added to the the number of seats on the previous row.

Hence, the recursive rule is:

an = an-1 + 8; a1 = 26

<h3>The explicit rule</h3>

In (a), we have:

a1 = 26 and d = 8

The explicit rule is calculated using:

an = a1 + (n -1) * d

This gives

an = 26 + (n - 1) * 8

Expand

an = 26 - 8 + 8n

Evaluate the difference

an = 18 + 8n

Hence, the explicit rule is:

an = 18 + 8n

<h3>The number of seats in the row 7</h3>

This means that n = 7.

So, we  have:

a7 = 18 + 8 * 7 = 74

Hence, there are 74 seats in the 7th row

<h3>The sigma notation</h3>

We have the maximum number of rows to be 12.

So, the sigma notation would be:

\sum\limits^{12}_{n=1} {18 + 8n}

The total number of seats is:

Sn = n/2(2a + (n -1) * d)

This gives

S12 = 12/2(2 * 26 + (12 -1) * 8)

S12 = 840

Hence, 840 chairs can fit the auditorium

Read more about arithmetic sequence at:

brainly.com/question/6561461

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A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In
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Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

                        P.Q. = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }  ~ N(0,1)

where, \hat p_1 = sample proportion of males having blood disorder= \frac{250}{1000} = 0.25

\hat p_2 = sample proportion of females having blood disorder = \frac{275}{1000} = 0.275

n_1 = sample of males = 1000

n_2 = sample of females = 1000

p_1 = population proportion of males having blood disorder

p_2 = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

= [ (0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }, (0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} } ]

 = [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

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