For three minutes the temperature of a feverish person has had negative first derivative and positive second derivative.

Mathematics

1 answer:

nydimaria [60]2 years ago

7 0

Answer:

A.) The temperature fell in the last minute, but less than it fell in the minute before.

Step-by-step explanation:

Given that for three minutes the temperature of a feverish person has had negative first derivative and positive second derivative.

i.e. it temperature is represented by T, temperature is variable with

first derivative T' <0 and second derivative T">0

i.e. rate of change of temperature is negative or temperature is falling down in 3 minutes.

But the rate of rate of change of temperature was positive i.e. the rate of change of temperature is increasing as time increases.

So correct option would be option A

A.) The temperature fell in the last minute, but less than it fell in the minute before.

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lora16 [44]

Answer:

a) We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:

Null hypothesis: $\mu \geq 5.5$

Alternative hypothesis: $\mu < 5.5$

b) For this case since we want to determine if the true mean is lower than an specified value is a one tailed test

c) For this case the significance level is given $\alpha=0.01$ and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:

$z_{cric}= -2.326$

And the rejection zone would be: $(\infty ,-2.326)$

And the region is on the figure attached

Step-by-step explanation:

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:

Null hypothesis: $\mu \geq 5.5$

Alternative hypothesis: $\mu < 5.5$

Part b

For this case since we want to determine if the true mean is lower than an specified value is a one tailed test

Part c

For this case the significance level is given $\alpha=0.01$ and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:

$z_{cric}= -2.326$

And the rejection zone would be: $(\infty ,-2.326)$

And the region is on the figure attached