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3 years ago

(a2 − 6) and (a2 + 4) are the factors of

Mathematics

1 answer:

morpeh [17]3 years ago

6 0

Just multiply them together and get a4-2a2-24.
(a2-6)(a2+4)
F: a4
O:4a2
I:-6a2
L:-24
Add together
a4-2a2-24

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2 years ago

A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len

Gnom [1K]

Answer:

$t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723$

E. -0.723

$df=n-1=93-1=92$

$p_v =2*P(t_{(92)}$

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

$\bar X=5.97$ represent the sample mean for the length

$s=0.4$ represent the sample standard deviation

$n=93$ sample size

$\mu_o =6$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:

Null hypothesis: $\mu = 6$

Alternative hypothesis: $\mu \neq 6$

since we don't know the population deviation the statistic is:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing in formula (1) we got:

$t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723$

E. -0.723

P value

The degrees of freedom are given by:

$df=n-1=93-1=92$

Since is a two tailed test the p value would be:

$p_v =2*P(t_{(92)}$

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

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