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Ganezh [65]
3 years ago
5

Find the 59th term of the arithmetic sequence 29,37,45,...

Mathematics
1 answer:
liberstina [14]3 years ago
4 0

Answer:

13,456

Step-by-step explanation:

Tn= a+(n-1)d

a is the first term, n is the given number and d is the common difference. d is given by subtracting the first term from the second term or the second from the third. Therefore the the 59th term of the sequence 29, 37, 45 is,

Tn= 29+(59-1)8

=29(58)8

=13,456

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Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean
LuckyWell [14K]

Answer:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

df = n_1 +n_2 -2 = 10+15-2= 23

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Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

\bar X_1 = 1085 sample mean for group 1

\bar X_2 = 1034 sample mean for group 2

n_1 = 10 sample size for group 1

n_2 = 15 sample size for group 2

s_1 = 52 sample deviation for group 1

s_2 = 61 sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: \mu_1= \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

And the statistic is given by:

t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}

And replacing we got:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 10+15-2= 23

And the p value would be:

p_v = 2*P(t_{23} >2.240) = 0.035

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

8 0
3 years ago
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saw5 [17]

Answer:

m∠1 =  81

m∠2 =  99

Step-by-step explanation:

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x = 27

m∠1 = 3x

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2 years ago
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Alex777 [14]

Answer:

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< HEF  also called <6 is a vertical angle to < 4

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3 years ago
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