A real estate purveyor purchases a 60{,}00060,00060, comma, 000 square foot \left(\text{ft}^2\right)(ft 2 )(, start text, f, t,
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Solution :
Let and
represents the proportions of the seeds which germinate among the seeds planted in the soil containing
and
mushroom compost by weight respectively.
To test the null hypothesis against the alternate hypothesis
.
Let denotes the respective sample proportions and the
represents the sample size respectively.
The test statistic can be written as :
which under follows the standard normal distribution.
We reject at
level of significance, if the P-value
or if
Now, the value of the test statistics = -1.368928
The critical value =
P-value =
= 0.171335
Since the p-value > 0.05 and , so we fail to reject
at
level of significance.
Hence we conclude that the two population proportion are not significantly different.
Conclusion :
There is not sufficient evidence to conclude that the of the seeds that
with the percent of the
in the soil.
Answer:
Part A
The polynomial expression for the profit from selling and making 'y' cars is 0.003·y² - 10·y - 1,000,000
Part B
The profit made from selling 30,000 cars is 1,670,000
Step-by-step explanation:
Part A
Taking the revenue, 'R', made by the car company for the sale of 'y' cars as R = 0.003·y² + 10·y
The cost to produce 'y' cars, C = 20·y + 1,000,00
The profit, 'P', from making and selling 'y' cars is given as follows;
P = R - C
∴ 0.003·y² + 10·y - (20·y + 1,000,00) = 0.003·y² - 10·y - 1,000,00
An expression of the polynomial for the profit 'P', from selling and making 'y' cars is therefore presented as follows;
0.003·y² - 10·y - 1,000,000
Part B
The profit made from selling 30,000 cars, (y = 30,000) is given as follows;
= 0.003·y² - 10·y - 1,000,000
= 0.003 × 30,000,000² - 10×3,000 - 1,000,000 = 1,670,000
The profit made from selling 30,000 cars = 1,670,000.