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2 years ago

Use the approach in Gauss's Problem to find the following sums of arithmetic sequences.

Mathematics

1 answer:

Svet_ta [14]2 years ago

6 0

The total value of the sequence is mathematically given as

498501

<h3>The sum of the sequence is..?</h3>

Generally, the equation for Gauss's Problem is mathematically given as

The sum of an arithmetic series;

1+2+3+...+n= n(n+1)/2

Given an arithmetic sequence,

1+2+3+...+998,

Here,

n = 998

1+2+3+...+n=n(n+1)/2

1+2+3+...+998=98(998 + 1)/2

998 x 999 1+2+3+...+998 =2

1+2+3+...+998 = 498501

In conclusion, 498501 is the total value of the sequence.

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Which of the following are solutions to the system graphed below.

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Weird. I think you just need to look if the point falls on the shaded area. But only (-5,5) does ...

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3 years ago

From an industrial area 70 companies were selected at random and 45 of them were panning for expansion next year. Find 95% confi

Lana71 [14]

Answer:

Confidence limit = [52.8%, 75.2%]

Step-by-step explanation:

$P=\frac{45}{70}= 0.64$

$(1-P)=1-0.64=0.36$

$n= 70$

$P$ ± $z$ $\sqrt{\frac{P(1-P)}{n} }$

where the value $z$ will be taken from the z-table for 95% confidence interval

1-0.95= 0.05/2= 0.025

0.95+0.025= 0.0975

From the z-table the value of $z$ corresponding to 0.0975 is 1.96

$0.64$ ± $1.96$ $\sqrt{\frac{0.64*0.36}{70} }$

$0.64$ ± $1.96$ $(0.057)$

$0.64$ ± $0.112$

$64%$ % ± $11.2$ %

so the confidence interval is

$64+11.2=75.2$ %

$64-11.2=52.8$ %

$[52.8, 75.2]$

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3 years ago

To test the effectiveness of a business school preparation course, 8 students took a general business test before and after the

Lisa [10]

Solution :

Group Before After

Mean 693.75 743.75

Sd 155.37 143.92

SEM 54.93 50.88

n 8 8

Null hypothesis : The preparation course not effective.

$H_0: \mu_d = 0$

Alternative hypothesis : The preparation course is effective in improving the exam scores.

$H_a : \mu_d>0$ (after - before)

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2 years ago

To test if the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken and the value of t

ivolga24 [154]

Answer:

$p_v =P(t_{(29)}>1.42)=0.0831$

For this case the p value calculated is higher than the significance level used of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

a) do not reject the null hypothesis and conclude that the mean IQ is not greater than 100

Step-by-step explanation:

Information given

We want to verify if he mean IQ of employees in an organization is greater than 100 , the system of hypothesis would be:

Null hypothesis: $\mu \leq 100$

Alternative hypothesis: $\mu > 100$

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The statistic calculated for this case $t_{calc}= 1.42$

The degrees of freedom are given by:

$df=n-1=30-1=29$

Now we can find the p value using tha laternative hypothesis and we got:

$p_v =P(t_{(29)}>1.42)=0.0831$

a) do not reject the null hypothesis and conclude that the mean IQ is not greater than 100

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3 years ago

Uanita and Nina are bowling together. The probability of Juanita getting a strike next game is 24\%24%24, percent. The probabili

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Answer:

Juanita gets a strike next game.

Step-by-step explanation:

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3 years ago

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