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The kittens, Annie and Josie, are pushing a ball, Annie with a force magnitude of 80 N in a direction of 133 degrees, and Josie

kvv77 [185]

Answer: Then the magnitude of the force is 37.86N, and the direction is 54.35°

Step-by-step explanation:

We can write the forces as vectors.

We know that Annie pushes with a magnitude of 80N in a direction of 133° (Remember that the angles are always measured from the x-axis)

The components of this force, (Ax, Ay), are then:

Ax = 80N*cos(133°)

Ay = 80N*sin(133°)

And we know that Josie pushes with a magnitude of 95N in direction of 290°

The components of this force, (Jx, Jy), are:

Jx = 95N*cos(290°)

Jy = 95N*sin(290°)

When we add these forces, the total force acting on the ball is:

F = (80N*cos(133°) , 80N*sin(133°)) + (95N*cos(290°), 95N*sin(290°))

F = (80N*cos(133°) + 95N*cos(290°), 80N*sin(133°) + 95N*sin(290°))

Now, the third kitten wants to do a force K, in a direction θ, such that the net force acting on the ball is zero.

Then we must have that, each component of the force of the third cat (K*cos(θ) on the x-axis and k*sin(θ) on the y-axis), is such that:

K*cos(θ) + 80N*cos(133°) + 95N*cos(290°) = 0

k*sin(θ) + 80N*sin(133°) + 95N*sin(290°) = 0

Now we need to solve that system for k and θ

if we simplify the equations we get:

k*cos(θ) - 22.07N = 0

k*sin(θ) -30.76N = 0

Now we can rewrite them as:

k*cos(θ) = 22.07N

k*sin(θ) = 30.76N

Now we can take the quotiet between both equations to get:

(k*sin(θ))/(k*cos(θ)) = 30.76N/22.07N

Tan(θ) = 1.394

θ = Atan(1.394) = 54.35°

Now that we know the angle, we can find the value of the magnitude k, by using one of the two equations of the system:

k*cos(54.35°) = 22.07N

k = 22.07N/cos(54.35°) = 37.86N

Then the magnitude of this force is 37.86N, and the direction is 54.35°

7 0

2 years ago

An unfair coin with p(h) = 0.6 is flipped 3 times and the result of each toss is noted. what is the probability that there are a

mr_godi [17]

Total possible outcomes: (2)³=8 (heads or tails possibility on each toss)

Draw out a Tree diagram to illustrate the given example, labelling the probability on each branch.

Adding all the possible situations with P(t)≥2, we get 0.352 as the probability.

Hope I helped :)

4 0

3 years ago

Determine the total numbers of zero<br> f(x) = 23 – 3x2 – 2x + 6

AleksandrR [38]

Answer:

f(x) = 29

Step-by-step explanation:

f(x)=23-(3x0)-(2X0)+6

f(x)=23+6

f(x)=29

4 0

3 years ago

Exercise 11.21) Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad ba

stiv31 [10]

Answer:

Step-by-step explanation:

Hello!

The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.

Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.

Be:

X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)

n₁= 12

X[bar]₁= 20.50 hs

S₁²= 199.76hs² (S₁= 14.13hs)

X₂: Battery duration between charges when the primary usage of the phone is internet applications.

n₂= 10

X[bar]₂= 8.50

S₂²= 33.29hs² (S₂= 5.77hs)

Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)

The parameters of interest are σ₁² and σ₂²

a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1

The hypotheses are:

H₀: σ₁²/σ₂² ≤ 1

H₁: σ₁²/σ₂² > 1

There is no level of significance listed so I've chosen α: 0.05

b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:

S²= $\frac{1}{n-1}$ *[∑X²-(∑X)²/n]

For the first sample:

n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36

S₁²= $\frac{1}{11}$ *[7240.36-(246)²/12]= 199.76hs²

S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs

For the second sample:

n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12

S₂²= $\frac{1}{9}$ *[1022.12-(85)²/10]= 33.2911hs²

S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs

c)

For this hypothesis test, the statistic to use is a Snedecors F:

$F= \frac{S_1^2}{S_2^2} *\frac{Sigma_1^2}{Sigma_2^2} ~~F_{n_1-1;n_2-1}$

This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:

$F_{n_1-1;n_2-1; 1-\alpha }= F_{11;9;0.95}= 3.10$

The rejection region is then F ≥ 3.10

$F_{H_0}= \frac{199.76}{33.29} * 1= 6.0006$

p-value: 0.006

Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.

Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.

I hope this helps!

8 0

3 years ago

A video club wants to reward its best members with a discount based on the member’s number of movie rentals and the number of ne

ivanzaharov [21]

Answer:

public class MovieRental

{

public static void main(String[] args)

{

Scanner in = new Scanner(System.in);

System.out.print("Enter the number of movie rentals: ");

int movieRentals = in.nextInt();

System.out.print("Enter the number of members referred to the video club: ");

int memberReferral = in.nextInt();

in.close();

double discountVal = Math.min(movieRentals + memberReferral, 75);

System.out.println("The discount is equal to: " + discountVal);

}

8 0

3 years ago