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klasskru [66]
3 years ago
8

Given that a=3, b=-2 and c= -4, evaluate the following expression. 2c-a/3c+b - 5a+4c/c-a

Mathematics
1 answer:
andrew-mc [135]3 years ago
5 0

Given:

Consider the given expression is:

\dfrac{2c-a}{3c+b}-\dfrac{5a+4c}{c-a}

To find:

The value of the given expression for a=3, b=-2, c=-4.

Solution:

We have,

\dfrac{2c-a}{3c+b}-\dfrac{5a+4c}{c-a}

Substituting a=3, b=-2, c=-4, we get

\dfrac{2(-4)-(3)}{3(-4)+(-2)}-\dfrac{5(3)+4(-4)}{(-4)-(3)}

=\dfrac{-8-3}{-12-2}-\dfrac{15-16}{-4-3}

=\dfrac{-11}{-14}-\dfrac{-1}{-7}

=\dfrac{11}{14}-\dfrac{1}{7}

Taking LCM, we get

=\dfrac{11-2}{14}

=\dfrac{9}{14}

Therefore, the value of the given expression for a=3, b=-2, c=-4 is \dfrac{9}{14}.

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Read 2 more answers
A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute
GREYUIT [131]

Answer:

(a) Null Hypothesis, H_0 : \sigma = 10 beats per minute  

     Alternate Hypothesis, H_A : \sigma\neq 10 beats per minute  

(b) The value of chi-square test statistics is 35.704.

(c) P-value = 0.4360.

(d) We conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

Step-by-step explanation:

We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.

If the range rule of thumb is applied to that normal​ range, the result is a standard deviation of 10 beats per minute.

Let \sigma = <u><em>population standard deviation for the pulse rates of men</em></u>.

(a) So, Null Hypothesis, H_0 : \sigma = 10 beats per minute      {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}

Alternate Hypothesis, H_A : \sigma\neq 10 beats per minute      {means that the pulse rates of men have a standard deviation different from 10 beats per minute}

The test statistics that will be used here is <u>One-sample chi-square test</u> for standard deviation;

                             T.S.  =  \frac{(n-1)\times s^{2} }{\sigma^{2} }  ~  \chi^{2}__n_-_1  

where, s = sample standard deviation = 10.1 beats per minute

            n = sample of men = 36

So, <u><em>the test statistics</em></u> =  \frac{(36-1)\times 10.1^{2} }{10^{2} }  ~ \chi^{2}__3_5

                                   =  35.70  4

(b) The value of chi-square test statistics is 35.704.

(c) Also, the P-value of the test statistics is given by;

                    P-value = P(\chi^{2}__3_5 > 35.704) = <u>0.4360</u>

(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as the test statistics will not fall in the rejection region.

Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

3 0
4 years ago
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