How to write 12430000

What is the exact volume of the cylinder with a radius of 13 and a height of 30?

torisob [31]

The volume would be 15927.87
so
V ≈ 15927.87to be sure with my answer.

8 0

3 years ago

mardi received an inheritance of 70,000. She invested part at 12% and deposited the remainder in tax-free bonds at 9%. Her total

cluponka [151]

Here ya go (: hope this helps.

4 0

3 years ago

Round 1.49882 nearest to the nearest given place thousandths

I am Lyosha [343]

Okie doke. So, we are rounding this number to the nearest thousandths place, which is three digits behind the decimal. The rules for rounding are if the number is 5 or more in the digit behind it, the number goes up. If it is 4 or less, the number goes back. In other words, we depend on the digit right of the digit we are rounding to in order to see what we do. The number we are rounding is 1.49882. The 8 is in the thousandths place and the 8 is to the right of that, which is the ten thousandths place. Because 8 is greater than 5, the number rounds up. So the number rounded to the nearest thousandth is 1.500.

7 0

3 years ago

An angle is equal to one-third of its complement. Find the angle's measure.

artcher [175]

Let $x^\circ$ be the measure of the angle in question. Its complement has measure $(90-x)^\circ$ . We're told that this angle is congruent to 1/3 of its complement, so that

$x^\circ=\dfrac{(90-x)^\circ}3\implies3x^\circ=(90-x)^\circ\implies4x^\circ=90^\circ\implies x=22.5^\circ$

8 0

3 years ago

The principal at Crest Middle School, which enrolls only sixth-grade students and seventh-grade students, is interested in deter

AlekseyPX

Answer:

a) [ -27.208 , -12.192 ]

b) New procedure is not recommended

Step-by-step explanation:

Solution:-

- It is much more common for a statistical analyst to be interested in the difference between means than in the specific values of the means themselves.

- The principal at Crest Middle School collects data on how much time students at that school spend on homework each night.

- He/She takes a " random " sample of n = 20 from a sixth and seventh grades students from the school population to conduct a statistical analysis.

- The summary of sample mean ( x1 & x2 ) and sample standard deviation ( s1 & s2 ) of the amount of time spent on homework each night (in minutes) for each grade of students is given below:

Mean ( xi ) Standard deviation ( si )

Sixth grade students 27.3 10.8

Seventh grade students 47.0 12.4

- We will first check the normality of sample distributions.

We see that sample are "randomly" selected.
The mean times are independent for each group
The groups are selected independent " sixth " and " seventh" grades.

The means of both groups are conforms to 10% condition of normality.

Hence, we will assume that the samples are normally distributed.

- We are to construct a 95% confidence interval for the difference in means ( u1 and u2 ).

- Under the assumption of normality we have the following assumptions for difference in mean of independent populations:

Population mean of 6th grade ( u1 ) ≈ sample mean of 6th grade ( x1 )

Population mean of 7th grade ( u2 ) ≈ sample mean of 6th grade ( x2 )

Therefore, the difference in population mean has the following mean ( u ^ ):

u^ = u1 - u2 = x1 - x2

u^ = 27.3 - 47.0

u^ = -19.7

- Similarly, we will estimate the standard deviation (Standard Error) for a population ( σ^ ) represented by difference in mean. The appropriate relation for point estimation of standard deviation of difference in means is given below:

σ^ = √ [ ( σ1 ^2 / n1 ) + ( σ2 ^2 / n2 ) ]

Where,

σ1 ^2 : The population variance for sixth grade student.

σ2 ^2 : The population variance for sixth grade student.

n1 = n2 = n : The sample size taken from both populations.

Therefore,

σ^ = √ [ ( 2*σ1 ^2 / n )].

- Here we will assume equal population variances : σ1 ≈ σ2 ≈ σ is "unknown". We can reasonably assume the variation in students in general for the different grade remains somewhat constant owing to other reasons and the same pattern is observed across.

- The estimated standard deviation ( σ^ ) of difference in means is given by:

σ^ =

$s_p*\sqrt{\frac{1}{n_1} + \frac{1}{n_2} } = s_p*\sqrt{\frac{1}{n} + \frac{1}{n} } = s_p*\sqrt{\frac{2}{n}}\\\\\\s_p = \sqrt{\frac{(n_1 - 1 )*s_1^2 + (n_2 - 1 )*s_2^2}{n_1+n_2-2} } = \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{n+n-2} } = \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{2n-2} } \\\\s_p = \sqrt{\frac{(20 - 1 )*s_1^2 + (20 - 1 )*s_2^2}{2(20)-2} } \\\\s_p = \sqrt{\frac{19*10.8^2 + 19*12.4^2}{38} } = \sqrt{135.2} \\\\s_p = 11.62755$

σ^ = 11.62755*√2/20

σ^ = 3.67695

- Now we will determine the critical value associated with Confidence interval ( CI ) which is defined by the standard probability of significance level ( α ). Such that:

Significance Level ( α ) = 1 - CI = 1 - 0.95 = 0.05

- The reasonable distribution ( T or Z ) would be determined on the basis of following conditions:

The population variances ( σ1 ≈ σ2 ≈ σ ) are unknown.
The sample sizes ( n1 & n2 ) are < 30.

Hence, the above two conditions specify the use of T distribution critical value. The degree of freedom ( v ) for the given statistics is given by:

v = n1 + n2 - 2 = 2n - 2 = 2*20 - 2

v = 38 degrees of freedom

- The t-critical value is defined by the half of significance level ( α / 2 ) and degree of freedom ( v ) as follows:

t-critical = t_α / 2, v = t_0.025,38 = 2.024

- Then construct the interval for 95% confidence as follows:

[ u^ - t-critical*σ^ , u^ + t-critical*σ^ ]

[ -19.7 - 2.042*3.67695 , -19.7 + 2.042*3.67695 ]

[ -19.7 - 7.5083319 , -19.7 + 7.5083319 ]

[ -27.208 , -12.192 ]

- The principal should be 95% confident that the difference in mean times spent of homework for ALL 6th and 7th grade students in this school (population) lies between: [ -27.208 , -12.192 ]

- The procedure that the matched-pairs confidence interval for the mean difference in time spent on homework prescribes the integration of time across different sample groups.

- If we integrate the times of students of different grades we would have to make further assumptions like:

The intelligence levels of different grade students are same
The aptitude of students from different grades are the same
The efficiency of different grades are the same.

- We have to see that both samples are inherently different and must be treated as separate independent groups. Therefore, the above added assumptions are not justified to be used for the given statistics. The procedure would be more bias; hence, not recommended.

8 0

3 years ago