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sertanlavr [38]

3 years ago

6

A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectil

e is modeled using the equation h(t) = –16t2 + 48t + 190.
What is the maximum height of the projectile?
A. 82 feet
B. 190 feet
C. 226 feet
D. 250 feet

2 answers:

IceJOKER [234]3 years ago

7 0

The answer is C on edge

Nostrana [21]3 years ago

6 0

Answer:

Option c is correct

226 feet is the maximum height of the projectile.

Step-by-step explanation:

A quadratic equation $y=ax^2+bx+c$ , ....[1]

then the axis of symmetry is given by:

$x = -\frac{b}{2a}$

As per the statement:

The path of the projectile is modeled using the equation :

$h(t) = -16t^2+48t+190$ ....[2]

where, h(t) is the height after t time.

On comparing with [1] we have;

a = -16 and b = 48

then;

$t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5$ sec

Substitute this in [2] we have;

$h(1.5) = -16(1.5)^2+48(1.5)+190$

⇒ $h(1.5) = -36+72+190$

Simplify:

$h(1.5) = 226 ft$

Therefore, the maximum height of the projectile at 1.5 sec is, 226 feet.

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a) $52.8-2.02\frac{3.9}{\sqrt{40}}=51.554$

b) $52.8+2.02\frac{3.9}{\sqrt{40}}=54.046$

c) $52.8-2.71\frac{3.9}{\sqrt{40}}=51.129$

d) $52.8+2.71\frac{3.9}{\sqrt{40}}=54.471$

e) Yes, depends of the confidence level.

At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X =52.8$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=3.9 represent the sample standard deviation

n=40 represent the sample size

a) What is the lower limit of the 95% interval? Give your answer to three decimal places

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=40-1=39$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,39)".And we see that $t_{\alpha/2}=2.02$

Now we have everything in order to replace into formula (1):

$52.8-2.02\frac{3.9}{\sqrt{40}}=51.554$

b) What is the upper limit of the 95% interval? Give your answer to three decimal places

$52.8+2.02\frac{3.9}{\sqrt{40}}=54.046$

So on this case the 95% confidence interval would be given by (51.554;54.046)

c) What is the lower limit of the 99% interval? Give your answer to three decimal places

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,39)".And we see that $t_{\alpha/2}=2.71$

Now we have everything in order to replace into formula (1):

$52.8-2.71\frac{3.9}{\sqrt{40}}=51.129$

d) What is the upper limit of the 99% interval? Give your answer to three decimal places

$52.8+2.71\frac{3.9}{\sqrt{40}}=54.471$

So on this case the 99% confidence interval would be given by (51.129;54.471)

e) Consider the claim that the mean amount of garbage per bin is 54.1985 pounds. Is the following statement true or false? The decision about the claim would depend on whether we use a 95% or 99% confidence interval: True/False

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At 95 % of confidence the value 54.1985 pounds is not included on the interval. At 5% of significance the statement is FALSE.

At 99 % of confidence the value 54.1985 pounds is included on the interval. So at 1% of significance the statement is TRUE.

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