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3 months ago

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A rocket leaves the surface of Earth at time t=0 and travels straight up from the surface. The height, in feet, of the rocket ab

ove the surface of Earth is given by y(t), where t is measured in seconds for 0≤t≤600. Values of y(t) for selected values of t are given in the table above. Of the following values of t, at which value would the speed of the rocket most likely be greatest based on the data in the table?
(1) t=100

(2) t=200

(3)t=300

(4) t=400

1 answer:

Inessa [12.5K]3 months ago

3 0

Answer:

Keep in mind:

Speed equals distance divided by time.

Next, we can calculate the average speed for different intervals.

Let's define a model in which the average speed at a specific time t = t0 can be calculated as:

AS(t0) = (y(b) - y(a))/(b - a)

Here, b represents the subsequent t0 value, and a is the earlier one. The mid-point in this segment is t0.

Therefore:

If t0 is set at 100 seconds:

AS(100s) = (400ft - 0ft)/(200s - 0s) = 2ft/s.

If t0 is at 200 seconds:

AS(200s) = (1360ft - 50ft)/(300s - 100s) = 6.55 ft/s.

If t0 is at 300 seconds:

AS(300s) = (3200ft - 400ft)/(400s - 200s) = 14ft/s.

If t0 is at 400 seconds:

AS(400s) = (6250s - 1360s)/(500s - 300s) = 24.45 ft/s.

This implies for one of the options, t = 400s has the highest speed.

This makes sense, as while the timing differences remain uniform at 100 seconds, the variances in the height values are increasing.

Thus, we can determine that the rocket is ascending with greater acceleration, leading to a higher average speed as time progresses.

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2 months ago

A random sample of 20 individuals who graduated from college five years ago were asked to report the total amount of debt (in $)

AnnZ [12381]

Response:

a. As student debt rises, current investment diminishes.

b. Y= 68778.2406 - 1.9112X

For each dollar increase in college debt, the average current investments decrease by 1.9112 dollars.

c. A substantial linear correlation exists between college debt and current investment as the P-value falls below 0.1.

d. Y= $59222.2406

e. R²= 0.9818

Step-by-step breakdown:

Hello!

Data has been gathered on a random sample of 20 individuals who completed their college education five years ago. The variables under consideration are:

Y: Current investment by an individual who graduated from college five years prior.

X: Total debt of an individual upon graduating five years ago.

a)

To explore the relationship between debt and investment, creating a scatterplot with the sample data is ideal.

The scatterplot demonstrates a negative correlation, indicating that as these individuals' debt increases, their current investments decrease.

Therefore, the statement that accurately describes this is: As college debt rises, current investment decreases.

b)

The population regression equation is Y= α + βX +Ei

To develop this equation, estimates for alpha and beta are required:

a= Y[bar] -bX[bar]

a= 44248.55 - (-1.91)*12829.70

a= 68778.2406

b= $\frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }$

b= $\frac{9014653088-\frac{(256594)(884971)}{20} }{4515520748-\frac{(256594)^2}{20} }$

b= -1.9112

∑X= 256594

∑X²= 4515520748

∑Y= 884971

∑Y²= 43710429303

∑XY= 9014653088

n= 20

Averages:

Y[bar]= ∑Y/n= 884971/20= 44248.55

X[bar]= ∑X/n= 256594/20= 12829.70

The estimated regression equation becomes:

Y= 68778.2406 - 1.9112X

For every dollar increase in college debt, the average current investments drop by 1.9112 dollars.

c)

To evaluate if there's a linear regression between these variables, the following null hypotheses are formulated:

H₀: β = 0

H₁: β ≠ 0

α: 0.01

Testing can be performed utilizing either a Student t-test or Snedecor's F (ANOVA)

Using t= b - β = -1.91 - 0 = -31.83

Sb 0.06

The critical area and P-value for this test is two-tailed. The P-value equals: 0.0001

Since this P-value is underneath the significance level, we reject the null hypothesis.

In the case of ANOVA, the rejection area is also one-tailed to the right, corresponding to the P-value.

The P-value remains: 0.0001

Using this method, we similarly reject the null hypothesis. $F= \frac{MSTr}{MSEr}= \frac{4472537017.96}{4400485.72} =1016.37$

In conclusion, at a significance level of 1%, there exists a linear relationship linking current investment to college debt.

The accurate statement is:

There exists a significant linear association between college debt and current investment since the P-value is less than 0.1.

d)

To forecast the value of Y when X is set, it is essential to substitute X in the estimated regression equation.

Y/$5000

Y= 68778.2406 - 1.9112*5000

Y= $59222.2406

The anticipated investment for someone with a college debt of $5000 is $59222.2406.

e)

To determine the proportion of variation in the dependent variable that the independent variable accounts for, the coefficient of determination R² must be calculated.

R²= 0.9818

$R^2= \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{sumY^2-\frac{(sumY)^2}{n} }$

$R^2= \frac{-1.9112^2[4515520748-\frac{(256594)^2}{20} ]}{43710429303-\frac{(884971)^2}{20} }$

This indicates that 98.18% of the variability in current investments relates to college graduation debt within the projected regression model: Y= 68778.2406 - 1.9112X

I trust this is beneficial!

5 0

1 month ago