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3 years ago

Please help WILL MARK BRAINLIEST! (random answers will be reported)

Mathematics

2 answers:

bixtya [17]3 years ago

6 0

Answer:

Step-by-step explanation:

The formula for volume of a cylinder is V = π $r^{2}$ h. We replace r with the radius (to find it just divide the diameter by 2), which is 2.5. Hope this helped!! :]

Pachacha [2.7K]3 years ago

6 0

The answer is C I am positive

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Based on a sample of 100 employees a 95% confidence interval is calculated for the mean age of all employees at a large firm. Th

babymother [125]

Answer:

$\= x = 40.85$

$E = 5.85$

$t_c = 2.08$

$t_c = 1.282$

Explanation:

From the question we are told that

The sample size is n = 100

The upper limit of the 95% confidence interval is b = 47.2 years

The lower limit of the 95% confidence interval is a = 34.5 years

Generally the sample mean is mathematically represented as

$\= x = \frac{a + b }{2}$

=> $\= x = \frac{47.2 + 34.5 }{2}$

=> $\= x = 40.85$

Generally the margin of error is mathematically represented as

$E = \frac{b- a }{ 2}$

=> $E = \frac{47.2- 34.5 }{ 2}$

=> $E = 5.85$

Considering question C a

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

The sample size is n = 22

Given that the sample size is not sufficient enough i.e $n < 30$ we will make use of the student t distribution table

Generally the degree of freedom is mathematically represented as

$df = n- 1$

=> $df = 22 - 1$

=> $df = 21$

Generally from the student t distribution table the critical value of $\frac{\alpha }{2}$ at a degree of freedom of 21 is

$t_c =t_{\frac{\alpha }{2} , 21 } = 2.08$

Considering question C b

From the question we are told the confidence level is 80% , hence the level of significance is

$\alpha = (100 - 80 ) \%$

=> $\alpha = 0.20$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$t_c =Z_{\frac{\alpha }{2} } = 1.282$

4 0

3 years ago

Find the area of the region enclosed by f(x) and the x axis for the given function over the specified u reevaluate f(x)=x^2+3x+4

lord [1]

Answer:

A = 68 unit^2

Step-by-step explanation:

Given:-

The piece-wise function f(x) is defined over an interval as follows:

f(x) = { x^2+3x+4 , x < 3

f(x) = { x^2+3x+4 , x≥3

Domain : [ -3 , 4 ]

Find:-

Find the area of the region enclosed by f(x) and the x axis

Solution:-

- The best way to tackle problems relating to piece-wise functions is to solve for each part individually and then combine the results.

- The first portion of function is valid over the interval [ -3 , 3 ]:

$f(x) = x^2+3x+4$

- The area "A1" bounded by f(x) is given as:

$A1 = \int\limits^a_b {f(x)} \, dx$

Where, The interval of the function { -3 , 3 ] = [ a , b ]:

$A1 = \int\limits^a_b {x^2+3x+4} \, dx\\\\A1 = \frac{x^3}{3} + \frac{3x^2}{2} + 4x |\limits_-_3^3 \\\\A1 = \frac{3^3}{3} + \frac{3*3^2}{2} + 4*3 - \frac{-3^3}{3} - \frac{3(-3)^2}{2} - 4(-3)\\\\A1 = 9 + 13.5 +12 + 9-13.5+12\\\\A1 =42 unit^2$

- Similarly for the other portion of piece-wise function covering the interval [3 , 4] :

$f(x) = 4x+10$

- The area "A2" bounded by f(x) is given as:

$A2 = \int\limits^a_b {f(x)} \, dx$

Where, The interval of the function { 3 , 4 ] = [ a , b ]:

$A2 = \int\limits^a_b {4x+10} \, dx\\\\A2 = 2x^2 + 10x |\limits_3^4 \\\\A2 = 2*(4)^2 + 10*4 - 2*(3)^2 - 10*3\\\\A2 = 32 + 40 - 18-30\\\\A2 =26 unit^2$