The function W(t) = 40e^–0.35t models the healing of a wound after t hours, where W(t) represents the wound area in square milli
1 answer:
You might be interested in
Answer:
a) P(x)=14x-224,000 and b) P(21,000)=70,000.
Step-by-step explanation:
a.) Use the formulas shown to write the company's profit function, P, from producing and selling x radios.
The company's profit is given by its revenue minus the production cost. Then, the profit function P(x) will be given by
b.) Find the company's profit if 21,000 radios are produced and sold
To obtain the company's profit we just need to substitute the value given (x=21,000) in the formula 1). Therefore,
.
Answer:
We conclude that the mean Ohio score is below the national average.
Step-by-step explanation:
We are given that a random survey of 1000 students nationwide showed a mean ACT score of 21.1. Ohio was not used.
A survey of 500 randomly selected Ohio scores showed a mean of 20.8. The population standard deviation is 3.
<u><em>Let </em></u><u><em> = mean Ohio scores.</em></u>
So, Null Hypothesis, :
21.1 {means that the mean Ohio score is above or equal the national average}
Alternate Hypothesis, :
< 21.1 {means that the mean Ohio score is below the national average}
The test statistics that would be used here <u>One-sample z test statistics</u> as we know about population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean Ohio score = 20.8
= population standard deviation = 3
n = sample of Ohio = 500
So, <u><em>test statistics</em></u> =
= -2.24
The value of z test statistics is -2.24.
<em>Now, at 0.1 significance level the z table gives critical value of -1.2816 for left-tailed test.</em><em> Since our test statistics is less than the critical values of z as -2.24 < 1.2816, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>
Therefore, we conclude that the mean Ohio score is below the national average.