In ΔUVW, the measure of ∠W=90°, the measure of ∠V=71°, and VW = 55 feet. Find the length of UV to the nearest tenth of a foot.

In bungee jumping, a bungee jump is the amount the cord will stretch at the bottom of the fall. The stiffness of the cord is rel

aleksandr82 [10.1K]

Answer:

6.7ft

Step-by-step explanation:

190÷28=6.7 is correct

6 0

3 years ago

Milan makes dollars for each hour of work. Write an equation to represent his total pay after working hours.

swat32

Answer:

p=9h

Step-by-step explanation:

7 0

4 years ago

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the n

Ivanshal [37]

Answer:

a) The probability that exactly 4 flights are on time is equal to 0.0313

b) The probability that at most 3 flights are on time is equal to 0.0293

c) The probability that at least 8 flights are on time is equal to 0.00586

Step-by-step explanation:

The question posted is incomplete. This is the complete question:

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures.

a) The probability that exactly 4 flights are on time is =

b) The probability that at most 3 flights are on time is =

c)The probability that at least 8 flights are on time is =

<h2>Solution to the problem</h2>

a) Probability that exactly 4 flights are on time

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

$P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}$

Where $_nC_x$ is the number of different combinations of x success in n trials.

$_nC_x=\frac{x!}{n!(n-x)!}$

Hence,

$P(X=4)=_9C_4\cdot (0.5)^4\cdot (0.5)^{5}$

$_9C_4=\frac{4!}{9!(9-4)!}=126$

$P(X=4)=126\cdot (0.5)^4\cdot (0.5)^{5}=0.03125$

b) Probability that at most 3 flights are on time

The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=0)=(0.5)^9=0.00195313$ . . . (the probability that all are not on time)

$P(X=1)=_9C_1(0.5)^1(0.5)^8=9(0.5)^1(0.5)^8=0.00390625$

$P(X=2)=_9C_2(0.5)^2(0.5)^7=36(0.5)^2(0.5)^7=0.0078125$

$P(X=3)= _9C_3(0.5)^3(0.5)^6=84(0.5)^3(0.5)^6=0.015625$

$P(X\leq 3)=0.00195313+0.00390625+0.0078125+0.015625=0.02929688\\\\ P(X\leq 3) \approx 0.0293$

c) Probability that at least 8 flights are on time 

That at least 8 flights are on time is the same that at most 1 is not on time.

That is, 1 or 0 flights are not on time.

Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.

$P(Y=0)=_0C_9(0.5)^0(0.5)^9=0.00195313\\ \\ P(Y=1)=_1C_9(0.5)^1(0.5)^8=0.0039065\\ \\ P(Y=0)+P(Y=1)=0.00585938\approx 0.00586$

6 0

4 years ago

Solve 3(x-3) = 0???????

kenny6666 [7]

Using distributive property, we get:

3x-9=0
3x=9
x=3

So, x=3. Now that was easy, huh? :)

5 0

3 years ago

Read 2 more answers

how many pounds of nuts selling for $9 per lb and raisins selling for $3 per lb should a person combine to obtain 60 lb of a tra

marta [7]

Let n be the number of lbs of nuts.
Let r be the number of lbs of raisins.
9n + 3r = 60 * 5 = 300 .............(1)
n + r = 60
n = 60 - r ...............(2)
Plugging the expression for n in (2) into (1) we get:
9(60 - r) + 3r = 300
which expands to:
540 - 9r + 3r = 300 .............(3)
Simplifying and rearranging (3) we get:
-6r = -240
Therefore r, the weight of raisins = 40 lbs
and the weight of nuts is 60 - 40 = 20 lbs

8 0

3 years ago