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2 years ago

Graph Data ALG 2 Please Help!

Mathematics

1 answer:

Sergio039 [100]2 years ago

5 0

Answer:

1. Figure 3

2. 2881.4 cm³

Step-by-step explanation:

The graph in figure 3 has a curve that fits the data best because it captures most of the points unlike the other two curves.

The curve in figure 3 has the equation;

$y=-16.9x^2 +430x+523$

When using x=8, substitute the value of x in the equation with 8 to get

y = -16.9*64 + 430*8 + 523

y= 2881.4

= 2881.4 cm³

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2 years ago

4over5 n = 2over3. n = ____?<br>please helpp

olganol [36]

Answer:

n=0

Step-by-step explanation:

because 0 would be the only number that would make both sides equal

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2 years ago

What is the area PLEASE HELP!! :/

UkoKoshka [18]

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Step-by-step explanation:

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2 years ago

A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. A previous study indicates that the propo

never [62]

Answer:

A sample of 997 is needed.

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

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

A previous study indicates that the proportion of left-handed golfers is 8%.

This means that $\pi = 0.08$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a p-value of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

This is n for which M = 0.02. So

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

$0.02 = 2.327\sqrt{\frac{0.08*0.92}{n}}$

$0.02\sqrt{n} = 2.327\sqrt{0.08*0.92}$

$\sqrt{n} = \frac{2.327\sqrt{0.08*0.92}}{0.02}$

$(\sqrt{n})^2 = (\frac{2.327\sqrt{0.08*0.92}}{0.02})^2$

$n = 996.3$

Rounding up:

A sample of 997 is needed.

3 0

2 years ago

An engineer is going to redesign an ejection seat for an airplane. the seat was designed for pilots weighing between 130lb and 1

Vlad [161]

Answer: 0.4758

Step-by-step explanation:

Given : Mean : $\mu=137\text{ lb}$

Standard deviation : $\sigma =28.9\text{ lb}$

Also, the new population of pilots has normally distributed .

The formula to calculate the z-score :-

$z=\dfrac{x-\mu}{\sigma}$

For x=130 lb .

$z=\dfrac{130-137}{28.9}=-0.2422145\approx-0.24$

For x=171lb.

$z=\dfrac{171-137}{28.9}=1.1764705\approx1.18$

The p-value = $P(-0.24$

$=0.8809999-0.4051651=0.4758348\approx0.4758348\approx0.4758$

Hence, the required probability : 0.4758

6 0

2 years ago