Guys pls help me!!!!!

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra

Anika [276]

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

$z=1$

$\frac{dx}{dt}=430$

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

$y=2$

According to Pythagoras

$hypotenuse^2=base^2+perpendicular^2$

$y^2=x^2+1^2$

$y^2=x^2+1$ .... (1)

Put z=1 and y=2, to find the value of x.

$2^2=x^2+1^2$

$4=x^2+1$

$4-1=x^2$

$3=x^2$

Taking square root both sides.

$\sqrt{3}=x$

Differentiate equation (1) with respect to t.

$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0$

Divide both sides by 2.

$y\frac{dy}{dt}=x\frac{dx}{dt}$

Put $x=\sqrt{3}$ , y=2, $\frac{dx}{dt}=430$ in the above equation.

$2\frac{dy}{dt}=\sqrt{3}(430)$

Divide both sides by 2.

$\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}$

$\frac{dy}{dt}=372.390923627$

$\frac{dy}{dt}\approx 372$

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

6 0

3 years ago

I NEED HELP PLEASE !!!

VLD [36.1K]

Dividing by coefficient other than +1

7 0

3 years ago

A certain company sends 40% of its overnight mail parcels by means of express mail service A1. Of these parcels, 4% arrive after

Harrizon [31]

Answer:

(a) The probability that a randomly selected parcel arrived late is 0.026.

(b) The probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c) The probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d) The probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

Step-by-step explanation:

Consider the tree diagram below.

(a)

The law of total probability sates that: $P(A)=P(A|B)P(B)+P(A|B')P(B')$

Use the law of total probability to determine the probability of a parcel being late.

$P(L)=P(L|A_{1})P(A_{1})+P(L|A_{2})P(A_{2})+P(L|A_{3})P(A_{3})\\=(0.04\times0.40)+(0.01\times0.50)+(0.05\times0.10)\\=0.026$

Thus, the probability that a randomly selected parcel arrived late is 0.026.

(b)

The conditional probability of an event A provided that another event B has already occurred is:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:

$P(A_{1}|L)=\frac{P(L|A_{1})P(A_{1})}{P(L)} \\=\frac{0.04\times 0.40}{0.026} \\=0.615$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{2}|L)=\frac{P(L|A_{2})P(A_{2})}{P(L)} \\=\frac{0.01\times 0.50}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{3}|L)=\frac{P(L|A_{3})P(A_{3})}{P(L)} \\=\frac{0.05\times 0.10}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

4 0

3 years ago

X^2+16=10x factor it NEED IT ASAP

Paraphin [41]

The solutions would be x=8 and x=2

5 0

3 years ago

Two airplanes are flying in the air at the same height: airplane a is flying east at 250 mi/h and airplane b is flying north at

Marta_Voda [28]

If the airport is located at the origin, for units of miles and hours, we can write the equations of position for airplanes "a" and "b" in rectangular coordinates as
a = (-30 +250t, 0)
b = (0, -40 +300t)

The distance between these (moving) points can be computed in the usual way using the Pythagorean theorem.
d = √((-30 +250t - 0)² +(0 - (-40 +300t))²)
d = √(2500 -39000t +152500t²)

Then the rate of change of d is the derivative of this.
d'(t) = (-19500 +152500t)/√(2500 -39000t +152500t²)

At the present time (t=0), the rate of change of distance between the planes is
d'(0) = -19500/√2500 = -390

The distance between the planes is decreasing at 390 mi/h.

3 0

3 years ago