Mathematics College

The measured height, in feet, of an airplane at certain times, in minutes, after takeoff can be modeled using the regression equation y = StartFraction 29,864 Over 1 + 9.381 e Superscript negative 0.9948 x Baseline EndFraction. To the nearest hundred feet, what is the predicted height of the airplane after 18 minutes?

Answers

Answer 1

Answer:

29900 ft

Step-by-step explanation:

just plug in 18 for x

Answer 2

A function assigns the values. The predicted height of the airplane after 18 minutes of takeoff is 29,864 ft.

What is a Function?

A function assigns the value of each element of one set to the other specific element of another set.

The measure of the height of an airplane at a certain time is given by the function,

[tex]y = \dfrac{29,864 }{1+9.381 e^{-0.9948x}}[/tex]

where x is the time in minutes and y is the height of the airplane.

Now, the height of the plane after 18 minutes post takeoff can be calculated by substituting the value of time, x as 18 in the function.

[tex]y = \dfrac{29,864 }{1+9.381 e^{-0.9948x}}\\\\y = \dfrac{29,864 }{1+9.381 e^{-0.9948 \times 18}}\\\\y = \dfrac{29,864 }{1+9.381 e^{-7.9064}}[/tex]

[tex]y = \dfrac{29,864 }{1+9.381 e^{-7.9064}}\\\\y = \dfrac{29,864 }{1+(9.381 \times 1.6724 \times 10^8)}\\\\y = \dfrac{29,864 }{1+(1.5689 \times 10^8)}\\\\y = \dfrac{29,864 }{1.000000157}[/tex]

y = 29864 ft

Hence, the predicted height of the airplane after 18 minutes of takeoff is 29,864 ft.

Learn more about Function here:

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Related Questions

A food-protection agency counts the number of insect heads found per 100-gram batch of wheat flour. The researchers have 500 batches, and they want to know whether the frequency of insect heads in batches follows a distribution called a Poisson distribution. To generate the expected frequencies of batches with different numbers of insect heads under a Poisson distribution, they had to estimate the mean number of insect heads per batch from the data. The 500 batches included at least 5 batches having 0, 1, 2, 3, or 4 insect heads. No batches had more than four heads. Given this information, there are_____classes (k).

Answers

Answer:

The correct answer is k = 5.

Step-by-step explanation:

The Chi-square goodness of fit test would be used to determine whether the frequency of insect heads in batches follows a distribution called a Poisson distribution.

The hypothesis for the test can be defined as follows:

H₀: The frequency of insect heads in batches does not follows a Poisson distribution.

Hₐ: The frequency of insect heads in batches follows a Poisson distribution.

Assume that the significance level of the test is, α = 0.05.

The Chi-square test statistic is given by:

[tex]\chi^{2}=\sum\limits^{k}_{i=1}\frac{(O_{i}-E_{i})^{2}}{E_{i}}[/tex]

It is provided that the 500 batches included at least 5 batches having 0, 1, 2, 3, or 4 insect heads. No batches had more than four heads.

So, the number of classes or categories are, k = 5.

Thus, the correct answer is k = 5.

How do you solve this? Without going into anything to complicated as this should be Year 10 maths

Answers

Given : ABCD is a square with each side 5cm .

To Find : The area of the shaded region .

Solution : On observing the figure we can see two quadrants , quadrant ADC & quadrant ABC .

If we join A to C , then it will be common for triangles ADC & ABC . And they will be congruent by SSS congruence condition.

Therefore the area of both quadrants will also be equal . Now we can find area of quadrant as ;

[tex]\large\boxed{\red{\bf Area_{quadrant}=\dfrac{\pi r^2}{4}}}[/tex]

Here radius will be equal to 5cm .

⇒ Area = πr² / 4 .

⇒ Area = π (5cm)² / 4 .

⇒ Area = 22/7 × 25 × 4 cm².

⇒ Area = 19.64 cm² .

So , total area of both quadrants = 39.28 cm² .

Also , area of square will be :

[tex]\large\boxed{\bf{\red{Area_{square}=(side)^2}}}[/tex]

⇒ Area = 5cm × 5cm .

⇒ Area = 25 cm².

Now , subtract area of one quadrant from the area of square = 25cm² - 19.64 cm² = 5.36 cm².

Similarly area of other white region = 5.36cm² .

And the areas sum will be = 5.36cm² × 2 = 10.72cm² .

Now , from the figure it's clear that ,

⇒ Area of unshaded region + Area of shaded region = 25cm².

⇒ 10.72cm² + ar( Shaded region ) = 25cm².

⇒ ar ( Shaded region ) = 25cm² - 10.72cm².

⇒ ar ( Shaded region ) = 14.28 cm².

Hence the area of shaded region is 14.28 cm².

[tex]\large\boxed{\red{\bf Answer = 14.28cm^2}}[/tex]

A survey of 120 college students is conducted and the results displayed below show that 80 students have a smartphone, 65 have a tablet, and 10 have neither. Fill in the empty table cells in the two-way frequency table.

Smartphone No Smartphone Total
Tablet 65
No Tablet 10
Total 80 120
What is the probability (rounded to the nearest whole percent) that a randomly selected college student has a tablet but not a smartphone? (4 points)

a
25%

b
30%

c
54%

d
38%

Answers

Answer:

the answer is 54 %

Step-by-step explanation:

divide 120/65=0.54, then since you're looking for the percent you multiply it by 100, and the answer would be 54%

A florist must make 5 identical
bridesmaid bouquets for a wedding. The budget is
$160, and each bouquet must have 12 flowers. Roses
cost $2.50 each, lilies cost $4 each, and irises cost
$2 each. The florist wants twice as many roses as the
other two types of flowers combined.

Answers

Answer:

A. 40x + 10y + 10z = $160

B. 8 Roses, 2 lilies and 2 irises

1. 20x + 5y + 5z = $80

2. 4x + y + z = $16

3. 8x + 2y + 2z = $32

Step-by-step explanation:

Cost for each flower = $160/5 = $32

So we have $32 for each bouquet consisting of 12 flowers each.

Roses = x = $2.50 each

lilies = y = $4 each

irises = z = $2 each

8x + 2y + 2z = $32

8($2.50) + 2($4) + 2($2) = $32

$20 + $8 + $4 = $32

$32 = $32

a. Maximum budget is $160

40x + 10y + 10z = $160

40($2.50) + 10($4) + 10($2) = $160

$100 + $40 + $20 = $160

$160 = $160

b. From above

8x + 2y + 2z = $32

8 Roses, 2 lilies and 2 irises

c. No. There are other solutions If total cost is not limited

1. 20x + 5y + 5z

20($2.50) + 5($4) + 5($2)

$50 + $20 + $10

= $80

2. 4x + y + z

4($2.50) + $4 + $2

$10 + $4 + $2

= $16

3. 8x + 2y + 2z

8($2.50) + 2($4) + 2($2)

$20 + $8 + $4

= $32

A process is normally distributed and in control, with known mean and variance, and the usual three-sigma limits are used on the control chart, so that the probability of a single point plotting outside the control limits when the process is in control is 0.0027. Suppose that this chart is being used in phase I and the averages from a set of m samples or subgroups from this process are plotted on this chart. What is the probability that at least one of the averages will plot outside the control limits when m = 5? Repeat these calculations for the cases where m = 10, m = 20, m = 30, and m = 50.

Required:
Discuss the results that you have obtained.

Answers

Answer:

The solution to the issue is outlined in the following portion of the summary.

Step-by-step explanation:

The given value is:

p = 0.0027

When m = 5,