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2 years ago

in survival at 40 the author includes a map and explained cause and effect relationships to the reader?

Mathematics

1 answer:

Studentka2010 [4]2 years ago

4 0

Answer:

To inform

Explanation:

It sounds like they're writing something informational and including a map. If what they're writing is very factual in nature then the purpose is to inform. If it's more telling a story then it's to entertain. If it's debating on a topic then it's to persuade. I hope this helps you.

You might be interested in

Simplify. Assume that no denominator equals zero.

Ann [662]

$\frac{x^2-25}{x^2-2x}\times \frac{x-2}{x^2+5x}$

Let's factor all four of these before we multiply.

x² - 25
We want two numbers that add to -25 and multiply to 0.
= (x+5)(x-5)

x² - 2x
Both terms are divisible by x.
= x(x-2)

x - 2
unfactorable

x² + 5x
Both terms are divisible by x.
= x(x+5)

Now we have this:

$\frac{(x+5)(x-5)}{x(x-2)}\times \frac{x-2}{x(x+5)}$

Let's go ahead and multiply across.

$\frac{(x+5)(x-5)(x-2)}{x(x-2)x(x+5)}$

Cancel out the (x+5) and (x-2) from the top and the bottom.

$\boxed{\frac{x-5}{x^2}}$

6 0

3 years ago

Read 2 more answers

Explain to me how to do this please

nataly862011 [7]

You must first attach the problems with your question.

6 0

3 years ago

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago

if (2x minus 1) is exactly divisible by 6 X square + ax - 4 and b x square - 11x + 3, then the value of b square minus a square

alexdok [17]

Step-by-step explanation:

Given that:-

$(2x - 1)$

is exactly divisible by

$6 {x}^{2} + ax - 4$

and

$b {x}^{2} - 11x + 3$

We need to simply, place the value of x.

$2x - 1 = 0$

$2x = 1$

$x = \frac{1}{2}$

Now,

$6 {( \frac{1}{2} )}^{2} + \frac{a}{2} - 4 = 0$

$\frac{3 + a}{2} = 4$

$3 + a = 4 \times 2 = 8$

$a = 8 - 3$

$a = 5$

Now, to find b.

$b {( \frac{1}{2}) }^{2} - \frac{11}{2} + 3 = 0$

$\frac{b}{4} - \frac{11}{2} = - 3$

$\frac{b - 22}{4} = - 3$

$b - 22 = - 12$

$b = 10$

is the answer.

Now, we need

${b}^{2} - {a}^{2} + a - b$

$100 - 25 + 100 + 5$

$= 205 - 25$

$= 180$

is the answer.

Hope it helps :D

3 0

3 years ago

Rothamsted Experimental Station (England) has studied wheat production since 1852. Each year many small plots of equal size but

podryga [215]

Answer:

(a) 0.653

(b) 0.0198

(c) Yes, after testing we conclude that the population variance of annual wheat production for the first plot is larger than that for the second plot.

Step-by-step explanation:

(a) We are give the sample of one year annual production of wheat (in pounds) ;

4.46, 4.21, 4.40, 4.81, 2.81, 2.90, 4.93, 3.54, 4.16, 4.48, 3.26, 4.74, 4.97, 4.02, 4.91, 2.59

For calculating sample variance, firstly we will calculate mean of the above data;

Mean of above data, $X_1bar$ = Sum of all values ÷ n (no. of values)

= $\frac{4.46 + 4.21+4.40+.......+4.91+2.59}{16}$ = 4.074

Sample Variance, $s_1^{2}$ = $\frac{\sum (X-X_1bar)^{2} }{n-1}$ = $\frac{(4.46-4.074)^{2}+(4.21-4.074)^{2}+.........+(4.91-4.074)^{2}+(2.59-4.074)^{2} }{16-1}$ = 0.653

(b) Another sample for annual wheat production (in pounds);

3.89, 3.81, 3.95, 4.07, 4.01, 3.73, 4.02, 3.78, 3.72, 3.96, 3.62, 3.76, 4.02, 3.73, 3.94, 4.03

Mean of above data, $X_2bar$ = Sum of all values ÷ n (no. of values)

= $\frac{3.89+3.81+3.95.......+3.94+4.03}{16}$ = 3.88

Sample Variance, $s_2^{2}$ = $\frac{\sum (X-X_2bar)^{2} }{n-1}$ = $\frac{(3.89-3.88)^{2}+(3.81-3.88)^{2}+.........+(3.94-3.88)^{2}+(4.03-3.88)^{2} }{16-1}$ = 0.0198

(c) Now, we have to test the claim that population variance of annual wheat production for the first plot is larger than that for the second plot i.e.;

Null Hypothesis, $H_0$ : $\sigma_1^{2} = \sigma_2^{2}$ or $H_0$ : $\frac{\sigma_1^{2} }{\sigma_2^{2} } = 1$

Alternate Hypothesis, $H_0$ : $\sigma_1^{2} > \sigma_2^{2}$ or $H_0$ : $\frac{\sigma_1^{2} }{\sigma_2^{2} } > 1$

The test statistics used here is;

$\frac{s_1^{2} }{s_2^{2} }* \frac{\sigma_1^{2} }{\sigma_2^{2}}$ ~ $F_n__1-1_,n_2-1$ where, $n_1 = 16$ and $n_2 =16$

Test Statistics = $\frac{0.653 }{ 0.0198}* 1$ ~ $F_1_5_,_1_5$

= 32.98

Since, we are not provided with any significance level so we assume it to be 5% and at this level, the F table gives critical value of 2.4282.

<em>Since our test statistics is higher than the critical value and it falls in the rejection region so we have sufficient evidence to reject null hypothesis and conclude that the population variance of annual wheat production for the first plot is larger than that for the second plot.</em>

8 0

3 years ago