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3 years ago

Factorise the following

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

7 0

Answer:

Below.

Step-by-step explanation:

a) 5x-x² = x(5 - x).

b) x²-6x-27 = (x - 9)(x + 3).

c) x²+x-56 = (x - 7)(x + 8).

d) x³+8 = (x + 2)(x^2 - 2x + 4)

e) 3x⁴-54t = 3(x^4 - 18t)

f) x²+3x-10 = (x - 2)(x + 5)

g) 2(x⁴+x³-12x²) = 2x^2(x + 4)(x - 3)

h) 20(x³-1) = 20(x - 1)(x^2 + x + 1)

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Step-by-step explanation:

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Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

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And the critical values with $\alpha/2=0.005$ on each tail are:

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Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

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