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Studentka2010 [4]
4 years ago
11

Find the 64th term of the following arithmetic sequence. 17, 26, 35, 44, ...

Mathematics
1 answer:
goblinko [34]4 years ago
8 0

take the difference

26-17=9

35-26=9

the first term is 17

and the nth term is 64

use the formula

tn=a+(n-1) d

let a be 17

let n be 64

let d be 9

then you substitute

t64=17+(64-1)9

=17+(63)9

=17+567

=584

so the 64th term is 584.

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In a metal fabrication​ process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are n
Leto [7]

Answer:

95% Confidence interval: (14.4537 ,15.1463)

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 feet

Sample mean, \bar{x} = 14.8 feet

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, σ = 0.65 feet

Degree of freedom = n - 1 = 15

95% Confidence interval:

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 15 and}~\alpha_{0.05} = \pm 2.1314  

14.8 \pm 2.1314(\dfrac{0.65}{\sqrt{16}} ) \\\\= 14.8 \pm 0.3463 = (14.4537 ,15.1463)  

is the required confidence interval for the true mean length of rods.

3 0
3 years ago
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65% of what number is 39? A. 25.35 B. 60 C. 166.7 D. 2535
aleksandrvk [35]
B: 60. You solve it by dividing 39 by 0.65
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4 years ago
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Which has a measure that is equal to the sum of the measures of the interior angles of a triangle?
sergij07 [2.7K]
The answer is a straight line
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Suppose you know the length of a confidence interval of a population mean is 8.4 and the sample mean (x bar) is 10. Find the ​ma
SOVA2 [1]

Answer:

ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The lenght of the interval correspond to:

8.4 = 2ME

ME= \frac{8.4}{2}= 4.2

And since we know the margin of error we can find the limits for the confidence interval:

Lower = 10 -4.2=5.8

Upper = 10 +4.2=14.2

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

\bar X=10 represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n represent the sample size  

Solution to the problem

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The sample mean \bar X is distributed on this way:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})  

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

The margin of error is given by:

ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The lenght of the interval correspond to:

8.4 = 2ME

ME= \frac{8.4}{2}= 4.2

And since we know the margin of error we can find the limits for the confidence interval:

Lower = 10 -4.2=5.8

Upper = 10 +4.2=14.2

4 0
3 years ago
What is 24x128725(7x782)/199+4+23
ryzh [129]

Answer:

\bold{\frac{8455687800}{113}}

Step-by-step explanation:

\frac{24\times \:128725\left(7\times \:782\right)}{199+4+23}

\mathrm{Add\:the\:numbers:}\:199+4+23=226

=\frac{24\times \:128725\times \:7\times \:782}{226}

\mathrm{Factor\:the\:number:\:}\:24=2\times \:12

=\frac{2\times \:12\times \:128725\times \:7\times \:782}{226}

\mathrm{Factor\:the\:number:\:}\:226=2\times \:113

=\frac{2\times \:12\times \:128725\times \:7\times \:782}{2\times \:113}

\mathrm{Cancel\:the\:common\:factor:}\:2

=\frac{12\times \:128725\times \:7\times \:782}{113}

\mathrm{Multiply\:the\:numbers:}\:12\times \:128725\times \:7\times \:782=8455687800

\bold{=\frac{8455687800}{113}}

\mathrm{Decimal:74829095.57522}

5 0
3 years ago
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