(4, 2); slope = 3 How do you do this problem and show work

Use the position equation given below, where s represents the height of the object (in feet), v0 represents the initial velocity

iris [78.8K]

Answer:

(a) 7.5 seconds

(b) The horizontal distance the package travel during its descent is 1737.8 ft

Step-by-step explanation:

(a) The given function for the height of the object is s = -16·t² + v₀·t + s₀

The initial height of the object s₀ = 900 feet

The initial vertical velocity of the object v₀ $_y$ = 0 m/s

The time it takes the package to strike the ground is found as follows;

0 = -16·t² + 0×t + 900

900 = 16·t²

t² = 900/16 = 62.25

t = √62.25 = 7.5 seconds

(b) Given that the horizontal velocity of the package is given as 158 miles/hour, we have

158 miles/hour = 231.7126 ft/s

The horizontal distance the package covers in the 7.5 second of vertical flight = 231.7126 ft/s × 7.5 s = 1737.8445 feet = 1737.8 ft to one decimal place

The horizontal distance the package travel during its descent = 1737.8 ft.

4 0

2 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

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

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

2 years ago

Sec (of theta)=25/24 find cot(of theta)

Len [333]

If secant of theta = 1.0416666667 then
theta = 16.26 Degrees
cotangent of theta = 3.4286

4 0

3 years ago

Read 2 more answers

What is 15x7 for the real answer?

Molodets [167]

Your answer would be 105. Hope this helps!

7 0