Arrow is launched upward at 256 feet per second from top of 85 foot building. What is maximum height of arrow?

Answers

Answer 1

Answer: As arrow rises to its maximum height, its velocity decreases from 256 ft/s to 0 ft/s at the rate of 32 ft/s each second. Use the following equation to determine the distance it rises.

vf^2 = vi^2 + 2 * a * d, vf = 0, a = -32
0 = 256^2 + 2 * -32 * d
64 * d = 65,536
d = 65,536 ÷ 64 = 1024 ft
Maximum height = 85 + 1024 = 1109 ft

Answer 2

Answer:

The arrow's maximum height above the ground can be calculated using maximum height. The maximum height reached from the launch point is 1024 feet. Adding the height of the building, the overall maximum height of the arrow is 1,109 feet.

To find the maximum height of the arrow, you can use the following formula:

Maximum Height = Initial Height + (InitialVelocity^2 / (2 * Acceleration due to gravity))

In this case:

Initial Height = 85 feet

Initial Velocity = 256 feet per second

Acceleration due to gravity (approximated as 32 feet per second squared)

Maximum Height = 85 + (256^2 / (2 * 32))

Maximum Height = 85 + (65536 / 64)

Maximum Height = 85 + 1024

Maximum Height = 1109 feet

So, the maximum height the arrow reaches above the ground is 1,109 feet.

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Find the value of x and y in each triangle

Answers

The value of variables are,

⇒ x = 13.85

⇒ y = 16

What is mean by Triangle?

A triangle is a three sided polygon, which has three vertices and three angles which has the sum 180 degrees.

We have to given that;

Base of triangle = 8

Now, We get;

⇒ cos 60° = 8 / y

⇒ 1/2 = 8/y

⇒ y = 8 × 2

⇒ y = 16

And, By Pythagoras theorem we get;

⇒ x² + 8² = y²

⇒ x² + 8² = 16²

⇒ x² + 64 = 256

⇒ x² = 256 - 64

⇒ x² = 192

⇒ x = √192

⇒ x = 13.85

Thus, The value of variables are,

⇒ x = 13.85

⇒ y = 16

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y = 8/cos60 = 16
x = sqr(16^2 + 8^2) = 8sqr(3) = 13.86 approx

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Answers

Answer:

10/9

Step-by-step explanation:

1/3+7/9=10/9

Step-by-step explanation:

=1+7

3 9

= 9+21

=28

27ans

Jack buys a bag of 5 apples, eachequal in size. He eats of 1/2 of one apple.
What fraction of the bag of
apples did he eat?

Answers

Answer:

4 1/2

Step-by-step explanation:

5 apples - 1/2 apple =

4 1/2 apple

9/2

Find the probability for choosing a letter at random from the word: Probability P(not P) A.) 1/11 B.) 2/11 C.) 5/11 D.) 10/11

Answers

Answer:

D.) 10/11

Step-by-step explanation:

Here the given word Probability has 11 letters in it. And we have to calculate the Probability of not selecting a letter P from the above word.

So the formula for calculating any probability is Total Favorable outcomes / Total number of outcomes.

Here total number of outcomes are 11 as word Probability has 11 letters.

So the probability of selecting letter P from word Probability = $(1)/(11)$

Now the P(not P) = 1 - P(selecting letter P)

= 1 - $(1)/(11)$ = $(10)/(11)$

The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, Southwest Air had the best reate with 80 % of its flights arriving on time. A test is conducted by randomly selecting 10 Southwest flights and observing whether they arrive on time. (a) Find the probability that at least 3 flights arrive late.

Answers

Answer:

There is a 32.22% probability that at least 3 flights arrive late.

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it arrives on time, or it arrives late. This means that we can solve this problem using binomial probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_(n,x).\pi^(x).(1-\pi)^(n-x)$

In which $C_(n,x)$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

And $\pi$ is the probability of X happening.

In this problem, we have that:

There are 10 flights, so $n = 10$ .

A success in this case is a flight being late. 80% of its flights arriving on time, so 100%-80% = 20% arrive late. This means that $\pi = 0.2$ .

(a) Find the probability that at least 3 flights arrive late.

Either less than 3 flights arrive late, or at least 3 arrive late. The sum of these probabilities is decimal 1. This means that:

$P(X < 3) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_(n,x).\pi^(x).(1-\pi)^(n-x)$

$P(X = 0) = C_(10,0).(0.2)^(0).(0.8)^(10) = 0.1074$

$P(X = 1) = C_(10,1).(0.2)^(1).(0.8)^(9) = 0.2684$

$P(X = 2) = C_(10,2).(0.2)^(2).(0.8)^(8) = 0.3020$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1074 + 0.2684 + 0.3020 = 0.6778$

Finally

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.6778 = 0.3222$

There is a 32.22% probability that at least 3 flights arrive late.

Final answer:

The problem is solved by calculating the probability of the complementary event (0,1,2 flights arriving late) using the binomial distribution, then subtracting this from 1 to find the probability of at least 3 flights arriving late.

Explanation:

This problem is typically solved by using a binomial probability formula, which is used when there are exactly two mutually exclusive outcomes of a trial, often referred to as 'success' and 'failure'.
Here, our 'success' is a flight arriving late. The probability of success, denoted as p, is thus 20% or 0.2 (since 80% arrive on time, then 100%-80% = 20% arrive late). The number of trials, denoted as n, is 10 (the number of randomly selected flights).
We want to find the probability that at least 3 flights arrive late, in other words, 3,4,...,10 flights arrive late. The problem can be solved easier by considering the complementary event: 0,1,2 flights arrive late. Then subtract the sum of these probabilities from 1.

The binomial probability of exactly k successes in n trials is given by:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
Where C(n, k) is the binomial coefficient, meaning choosing k successes from n trials.
We calculate like so:
P(X=0) = C(10, 0) * (0.2)^0 * (0.8)^10
P(X=1) = C(10, 1) * (0.2)^1 * (0.8)^9
P(X=2) = C(10, 2) * (0.2)^2 * (0.8)^8
Sum these up and subtract from 1 to get the probability that at least 3 flights arrive late. This gives the solution to the question.