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3 years ago

A student wanted to find the sum of all the even numbers from 1 to 100. He said:

Mathematics

1 answer:

murzikaleks [220]3 years ago

3 0

Answer:

The statements are incorrect as: The sum of even numbers from 1 to 100(i.e. 2550) is not double\twice of the sum of odd numbers from 1 to 100(i.e. 2500).

Step-by-step explanation:

We know that sum of an Arithmetic Progression(A.P.) is given by:

where 'n' denotes the "number" of digits whose sum is to be determined, 'a' denotes the first digit of the series and '' denote last digit of the series.

Now the sum of even numbers i.e. 2+4+6+8+....+100 is given by the use of sum of the arithmetic progression since the series is an A.P. with a common difference of 2.

image with explanation

Hence, sum of even numbers from 1 to 100 is 2550.

Also the series of odd numbers is an A.P. with a common difference of 2.

sum of odd numbers from 1 to 100 is given by: 1+3+5+....+99

Hence, the sum of all the odd numbers from 1 to 100 is 2500.

Clearly the sum of even numbers from 1 to 100(i.e. 2550) is not double of the sum of odd numbers from 1 to 100(i.e. 2500).

Hence the statement is incorrect.

Step-by-step explanation:

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A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random

Yuliya22 [10]

Answer:

b) independent samples t-test

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$

Our notation on this case :

$n_1 =5$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

We can calculate the mean and deviation for eaach group with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =5.92$ represent the sample mean for the group 1

$\bar X_2 =5.74$ represent the sample mean for the group 2

$s_1=1.88$ represent the sample standard deviation for group 1

$s_2=1.81$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$

And now we can calculate the statistic:

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

The degrees of freedom are given by:

$df=5+5-2=8$

And now we can calculate the p value using the altenative hypothesis:

$p_v =2*P(t_{8}>0.154) =0.881$

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