Can anyone help me with this??

An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each oth

sattari [20]

They are traveling at right angles to each other so we can say one is traveling north to south and the other west to east. Then we can say that there positions, y and x are:

y=150-600t x=200-800t

By using the Pythagorean Theorem we can find the distance between these two planes as a function of time:

d^2=y^2+x^2, using y and x from above

d^2=(150-600t)^2+(200-800t)^2

d^2=22500-180000t+360000t^2+40000-320000t+640000t^2

d^2=1000000t^2-500000t+62500

d=√(1000000t^2-500000t+6250)

So the rate of change is the derivative of d

dd/dt=(1/2)(2000000t-500000)/√(1000000t^2-500000t+6250)

dd/dt=(1000000t-250000)/√(1000000t^2-500000t+6250)

So the rate depends upon t and is not a constant, so for the instantaneous rate you would plug in a specific value of t...

...

To find how much time the controller has to change the airplanes flight path, we only need to solve for when d=0, or even d^2=0...

1000000t^2-500000t+62500=0

6250(16t^2-8t+1)=0

6250(16^2-4t-4t+1)=0

6250(4t(4t-1)-1(4t-1))=0

6250(4t-1)(4t-1)=0

6250(4t-1)^2=0

4t-1=0

4t=1

t=1/4 hr

Well technically, the controller has t<1/4 because at t=1/4 impact will occur :)

7 0

3 years ago

Sequence of number - 25,23,20,16<br>do with process

Nitella [24]

Answer:

what should we be finding?

8 0

3 years ago

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I am very confused plz help

prohojiy [21]

Answer:

a 26

b 8 2/3

Step-by-step explanation:

a) 4 1/3 * 6

Change the mixed number to an improper fraction

(3*4+1)/3 *6

13/3 *6

= 78/3

= 26

b) 2 3/5 * 3 1/3

Change the mixed numbers to improper fractions

(5*2+3)/5 * (3*3+1)/3

13/5 * 10/3

130/15

Top and bottom divide by 5

26/3

Change back to a mixed number

3 goes into 26 8 times (3*8 =24) with 2 left over

8 2/3

6 0

3 years ago

Can someone answer this question please answer it correctly if it’s corect I will mark you brainliest

Alina [70]

Answer:

26

Step-by-step explanation:

3 0

3 years ago

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An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago