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3 years ago

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Mariana tried to drink a slushy as fast as she could. She drank the slushy at a rate of 4.5 milliliters per second. After 17 sec

onds, 148.5 milliliters of slushy remained

2 answers:

Fynjy0 [20]3 years ago

8 0

This is what i did to get the answer.

225/ 4.5 = 50 seconds to finish the slushy.

erma4kov [3.2K]3 years ago

3 0

Answer:

(a) 225 mm

(b)50 seconds

Step-by-step explanation:

Let the equation that represents this situation is y = mx+ b

Here, y represents the amount left in milliliters and x represents the time in seconds.

It has been given that she drank the slushy at a rate of 4.5 milliliters per second.

So, m = -4.5 (negative because the amount decreases)

And after 17 seconds, 148.5 milliliters of slushy remained

So, when x = 17, y = 148.5

Substituting these values in the above model y = mx +b

148.5 = -4.5(17) + b

148.5 = -76.5 +b

b = 225

Therefore, the model is y = -4.5x + 225

(a)

When x = 0, we have to find y

y = -4.5 (0) + 225

y =225

Thus, 225 mm of slushy was originally in the cup.

(b)

Now, we have to find x for y =0

0= -4.5x + 225

4.5 x = 225

x = 50

So, Mariana took 50 seconds to drink all the slushy.

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Which point would be a solution to the system of linear inequalities ?

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Answer:

(12,-6)

Step-by-step explanation:

we have

$y\leq \frac{4}{3}x+5$ ----> inequality A

$y\geq -\frac{5}{2}x+5$ ---> inequality B

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

<u><em>Verify each point</em></u>

Substitute the value of x and the value of y of each ordered pair in the inequality A and in the inequality B

case 1) (0,-1)

Inequality A

$-1\leq \frac{4}{3}(0)+5$

$-1\leq5$ ----> is true

Inequality B

$-1\geq -\frac{5}{2}(0)+5$

$-1\geq 5$ ----> is not true

therefore

The ordered pair is not a solution of the system

case 2) (0,3)

Inequality A

$3\leq \frac{4}{3}(0)+5$

$3\leq5$ ----> is true

Inequality B

$3\geq -\frac{5}{2}(0)+5$

$3\geq 5$ ----> is not true

therefore

The ordered pair is not a solution of the system

case 3) (-6,-6)

Inequality A

$-6\leq \frac{4}{3}(-6)+5$

$-6\leq-3$ ----> is true

Inequality B

$-6\geq -\frac{5}{2}(-6)+5$

$-6\geq 20$ ----> is not true

therefore

The ordered pair is not a solution of the system

case 4) (12,-6)

Inequality A

$-6\leq \frac{4}{3}(12)+5$

$-6\leq21$ ----> is true

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$-6\geq -\frac{5}{2}(12)+5$

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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 $\pi = 0.5$

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

$0.009 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.009\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.009}$

$(\sqrt{n})^2 = (\frac{2.575*0.5}{0.009})^2$

$n = 20464.9$

Rounding up:

A sample of 20465 is required.

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