Given f(x), find g(x) and h(x) such that f(x) = g(h(x)) and neither g(x) nor h(x) is solely x.

Mathematics

1 answer:

aniked [119]2 years ago

6 0

Answer:

Please look at the different answers. I wasn't 100% sure what your expression was.

Step-by-step explanation:

If you mean 2/(5x-1) then g(x) will take the fraction into account with a constant 2 for the numerator and a variable for the denominator since that is where our variable is so g(x)=2/x.

Now h(x)=5x-1 since if you plug in 5x-1 into 2/x where x is you will get our original expression.

Now if you did mean (2/5)x-1 I would take notice of where the variable is which is in (2/5)x so g(x)=x-1 where h(x)=(2/5)x since if you plug (2/5)x in place of the x in x-1 you will get the original.

Please let me know if I didn't interpret your expression correctly.

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The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a-1) At α = .05 i

fenix001 [56]

Answer:

Null hypothesis: $\mu \geq 250$

Alternative hypothesis: $\mu < 250$

$p_v =P(t_{15}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=510$ represent the mean for the sample

$s=50$ represent the standard deviation for the sample

$n=16$ sample size

$\mu_o =530$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:

Null hypothesis: $\mu \geq 530$

Alternative hypothesis: $\mu < 530$

We don't know the population deviation and the sample size is lees than 30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{510-530}{\frac{50}{\sqrt{16}}}=-1.60$