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3 years ago

Position to term rule of 2, 6, 10, 14, 18 multiply by _ and subtract by _

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

4 0

Answer:

Please check the explanation.

Step-by-step explanation:

Given the sequence

2, 6, 10, 14, 18

An arithmetic sequence has a constant difference and is defined as

$\:a_n=a_1+\left(n-1\right)d$

compute the differences of all the adjacent terms

$6-2=4,\:\quad \:10-6=4,\:\quad \:14-10=4,\:\quad \:18-14=4$

The difference between all the adjacent terms is the same.

Thus,

$d=4$

and

$a_1=2$

Therefore, the nth term is computed by:

$a_n=4\left(n-1\right)+2$

$a_n=4n-2$

Thus, position to term rule of 2, 6, 10, 14, 18 multiply by __4___ and subtract by __2__.

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In Major League Baseball, the American League (AL) allows a designated hitter (DH) to bat in place of the pitcher, but in the Na

hammer [34]

Answer:

Step-by-step explanation:

From the given information:

Let:

$\mu_1$ represent the population mean no. for DH group.

$\mu_2$ represent the population mean no. for no DH group

$n_1$ represent the sample sizes of DH

$n_2$ represent the sample sizes of no DH

Sample size: $n_1 = n_2 = 20$

For both groups, the population standard deviation of runs scored = 2.54

i.e

$\sigma^2_1 = \sigma_2^2 = 2.54$

The null & alternative hypothesis;

$H_o : \mu_1 \le \mu_2 \\ \\ H_a: \mu_1 > \mu_2$

Level of significance:

The test statistics is:

$z = \dfrac{\bar x_1 - \bar x_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma_2^2}{n_2} }}$

where;

$\bar x_1$ = sample mean for DH group

$\implies \dfrac{1}{n_1} \sum \limit ^{n_1}_{i=1} x_1 \\ \\ = \dfrac{1}{20}(0+6+8+2+2+4+7+7+6+5+1+1+5+4+4+5+7+11+10+0) \\ \\ = \dfrac{92}{20}$

= 4.6

$\bar x_2$ = sample mean for no DH group

$\implies \dfrac{1}{n_2} \sum \limit ^{n_1}_{i=1} x_1 \\ \\ = \dfrac{1}{20}(3+6+2+4+0+5+7+6+1+8+12+4+6+3+4+0+5+2+1+4) \\ \\ = \dfrac{83}{20}$

= 4.15

Now:

$z = \dfrac{4.60- 4.15}{\sqrt{\dfrac{2.54^2}{20} + \dfrac{2.54^2}{20} }} \\ \\ = \dfrac{0.45}{\sqrt{0.32258 +0.32258 }} \\ \\ = \dfrac{0.45}{0.803219} \\ \\$

= 0.5602

Since the test is one-tailed by looking at that $H_a:$

The P-value = $P(Z > z)$

$\implies 1 - P(Z \le z) \\ \\ = 1 - P(Z \le 0.5602)$

Using Excel Function " =normdist(z)"; we have":

$= 1- 0.7123$

P-value = 0.2877

Decision rule: To reject $H_o$ , if $p-value < \alpha \ at \ 0.10$

Conclusion: SInce P = 0.2877 which is $> \ \alpha \ at \ 0.10.$

We fail to reject the $H_o$ and conclude that there is insufficient evidence to support the given claim that: $\text{more runs are scored in games for which DH is used.}$