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3 years ago

2,550 gallons in 30 days

2 answers:

8 0

If you would like to know how many gallons do you need for 1 day, you can calculate this using the following steps:

2550 gallons ... 30 days
x gallons = ? ... 1 day

2550 * 1 = 30 * x
2550 = 30 * x /30
x = 2550/30
x = 85 gallons

The correct result would be 85 gallons.

serg [7]3 years ago

8 0

Find the unit rate. Solve:-

2550 ÷ 30 = 85
85 gallons per day

Check our work:-

85 × 30 = 2550
Correct!

So, 85 gallons per day.

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A lawn service owner is testing new weed killers. He discovers that a particular weed killer is effective 89% of the time. Suppo

marissa [1.9K]

Answer:

The 90% confidence interval for p is (0.8236, 0.9564). The upper confidence limit for p is 0.9564.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

He discovers that a particular weed killer is effective 89% of the time. Suppose that this estimate was based on a random sample of 60 applications.

This means that $\pi = 0.89, n = 60$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.89 - 1.645\sqrt{\frac{0.89*0.11}{60}} = 0.8236$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.89 + 1.645\sqrt{\frac{0.89*0.11}{60}} = 0.9564$

The 90% confidence interval for p is (0.8236, 0.9564). The upper confidence limit for p is 0.9564.

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3 years ago

Read 2 more answers

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Minchanka [31]

Remember that the y-intercept is where the function has a value of $x = 0$ and the x-intercept is where the function has a value of $y = 0$ . Thus, to find the x-intercept and y-intercept, we can substitute these values into the equation and solve for the other variable.