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3 years ago

The y values of the t-table for the function y=3

Mathematics

1 answer:

ikadub [295]3 years ago

8 0

For the function y = 3, all y values will be 3, no matter the value of x.

This function is simply a straight line, and so it has a slope of 0.

However, it is a legitimate function.

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In the first quadrant, you start at (8, 11)

Anestetic [448]

Answer:

(11,2)

Step-by-step explanation:

(8+3,11-9)

(11,2)

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3 years ago

Which of the following statements best describes the graph of -5x+2y=1

ahrayia [7]

The answer is the third option.
x and y are both to the first degree so we know it makes a straight line.
Next check the ordered pairs to see which set are true in the equation.

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4 years ago

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Question 1 of 10

Jet001 [13]

The answer should be true

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3 years ago

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beni is shopping for a new cell phone plan. company a charges$72.25for a new phone and $85.50 per month.company b charges $151.2

tigry1 [53]

Missing questions and subsequent solutions:
(a) Write an equation for company A for cost, C, number of months, n, that Beni will pay for the phone.

Solution:
For company A:
C = 72.25 + 85.50n

(b) Write an eqyation for company B for cost, C, and number of months, n, that Bei will pay for the phone.

Solution:
For company B:
C = 151.25 + 65.75n

(c) Write an inequality when the cost from company A is better than cost from company B.

Solution:
72.25 + 85.50n ≤ 151.25 + 65.75n
(85.50-65.75)n ≤ (151.25 - 72.25)
19.75 n ≤ 79
n ≤ 4

(d) Value of n for which cost from the two companies will be the same.

Solution:
If cost for companies A and B are the same, then
72.25 + 85.50n = 151.25 + 65.75n
(85.5 - 65.75)n = 151.25 - 72.25
19.75n = 79
n = 79/19.75 = 4 months

After 4 months,
C = 72.25 + 85.5*4 = $414.25

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4 years ago

An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors

GarryVolchara [31]

Answer:

a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b) A sample of 408 is required.

c) A sample of 20465 is required.

Step-by-step explanation:

Question a:

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 $1 - \frac{\alpha}{2}$ .

Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.

This means that $n = 1600, \pi = \frac{8}{1600} = 0.005$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095$

The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?

The margin of error is of:

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

A sample of n is required, and n is found for M = 0.009. So

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

$0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}$

$0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}$

$\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2$

$n = 407.3$

Rounding up:

A sample of 408 is required.

c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

Since we have no estimate, we use $\pi = 0.5$