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solong [7]
3 years ago
13

In an arithmetic sequence, a1=5 and a6=-5. Determine an equation for an, the nth term of this sequence.

Mathematics
1 answer:
GalinKa [24]3 years ago
6 0

Answer: 7 - 2n

Step-by-step explanation:

a1 = 5

a6 = -5 = a + (n - 1)d = a + (6 - 1)d

a + 5d = -5

Since a1 = 5

5 + 5d = -5

5d = -5 - 5

5d = -10

d = -10/5

d = -2

Therefore, the equation to calculate the nth term of this sequence will be:

= a + (n-1)d

= a + (n - 1)-2

= 5 -2n +2

= 7 - 2n

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LUCKY_DIMON [66]

Answer: x= -2, y= 3

Step-by-step explanation:

3x+5y=9

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Multiply bottom equation by 5

3x+5y=9

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Plug -2 back into equation of choice to find y

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3 years ago
Do positive or negative messages have a greater effect on behavior? Forty-two subjects were randomly assigned to one of two trea
Alex_Xolod [135]

Answer:

We conclude that a negative message results in a lower mean score than positive message.

Step-by-step explanation:

We are given that Forty-two subjects were randomly assigned to one of two treatment groups, 21 per group.

The 21 subjects receiving the negative message had a mean score of 9.64 with standard deviation 3.43; the 21 subjects receiving the positive message had a mean score of 15.84 with standard deviation 8.65.

<em>Let </em>\mu_1<em> = population mean score for negative message</em>

<em />\mu_2<em> = population mean score for positive message</em>

SO, Null Hypothesis, H_0 : \mu_1-\mu_2\geq0  or  \mu_1\geq \mu_2    {means that a negative message results in a higher or equal mean score than positive message}

Alternate Hypothesis, H_A : \mu_1-\mu_2  or  \mu_1   {means that a negative message results in a lower mean score than positive message}

The test statistics that will be used here is <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

                     T.S.  = \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t__n__1+_n__2-2

where, \bar X_1 = sample mean score for negative message = 9.64

\bar X_2 = sample mean score for positive message = 15.84

s_1 = sample standard deviation for negative message = 3.43

s_2 = sample standard deviation for positive message = 8.65

n_1 = sample of subjects receiving the negative message = 21

n_2 = sample of subjects receiving the positive message = 21

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} }  =  \sqrt{\frac{(21-1)\times 3.43^{2}+(21-1)\times 8.65^{2}  }{21+21-2} }  = 6.58

So, <u><em>the test statistics</em></u>  =  \frac{(9.64-15.84)-(0)}{6.58 \times \sqrt{\frac{1}{21}+\frac{1}{21}  } }  ~  t_4_0

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<em>Now at 0.05 significance level, the t table gives critical value of -1.684 at 40 degree of freedom for left-tailed test. Since our test statistics is less than the critical value of t as -3.053 < -1.684, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.</em>

Therefore, we conclude that a negative message results in a lower mean score than positive message.

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