All of the objects are falling from the same height. Which statement is true about their motion?

A peqch pie is made with 13 peaches, which are evenly distributed throughout the pie. If Joe ate one slice that contained exactl

bekas [8.4K]

Answer:

138.5°

Step-by-step explanation:

Since the peach pie is a circle, we have a total angle of 360°.

We need to determine what fraction of the total pie of 13 peaches is the 5 peaches contained in Joe's slice.

This is number of peaches in Joe's slice/total number of peaches in pie = 5/13. To find the angle Joe's slice measures, we multiply 5/13 by 360°.

So, 5/13 × 360°

= 0.3846 × 360°

= 138.46°

≅ 138.5° to the nearest tenth of a degree

3 0

3 years ago

The polynomial of degree 3, P(x), has a root of multiplicity 2 at x = 3 and a root of multiplicity 1 at x = -2. The y-intercept

Maru [420]

Answer:

$p(x) = - 0.7(x - 3)^{2} (x + 2)$

Step-by-step explanation:

We have that the polynomial, has a root of multiplicity 2 at x = 3 and a root of multiplicity 1 at x = -2.

This means the factored form of the polynomial will be.

$p(x) = a(x - 3)^{2} (x + 2)$

Also, it was given that, the y-intercept is y=-12.6.

This implies that:

$- 12.6= a(0 - 3)^{2} (0 + 2)$

$- 12.6= a(- 3)^{2} (2)$

$- 12.6= 18a$

Divide both sides by 18;

$a = \frac{ - 12.6}{18}$

$a= - 0.7$

Therefore the polynomial is

$p(x) = - 0.7(x - 3)^{2} (x + 2)$

7 0

4 years ago

What is the length of AC? A. 12cm B. 10cm C. 15 cm D. 20 cm

Anna35 [415]

D.20 cm hope this helps

7 0

3 years ago

Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p

Ymorist [56]

Answer:

$E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours$

$Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2$

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

$A\sim Uniform(1 ,7)$

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

$B\sim Exp(\lambda=\frac{1}{30 min})$

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

$E(A +B)$

Since the two random variables are assumed independent, then we have this

$E(A+B) = E(A)+E(B)$

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by $E(X) =\frac{a+b}{2}$ where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

$E(A)=\frac{1+7}{2}=4 hours$

The expected value for the exponential distirbution is given by :

$E(X)= \int_{0}^\infty x \lambda e^{-\lambda x} dx$

If we use the substitution $y=\lambda x$ we have this:

$E(X)=\frac{1}{\lambda} \int_{0}^\infty y e^{-\lambda y} dy =\frac{1}{\lambda}$

Where X represent the random variable and $\lambda$ the parameter. If we apply this formula to our case we got:

$E(B) =\frac{1}{\lambda}=\frac{1}{\frac{1}{30}}=30min$

We can convert this into hours and we got E(B) =0.5 hours, and then we can find:

$E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours$

And in order to find the variance for the random variable A+B we can find the individual variances:

$Var(A)= \frac{(b-a)^2}{12}=\frac{(7-1)^2}{12}=3 hours^2$

$Var(B) =\frac{1}{\lambda^2}=\frac{1}{(\frac{1}{30})^2}=900 min^2 x\frac{1hr^2}{3600 min^2}=0.25 hours^2$

We have the following property:

$Var(X+Y)= Var(X)+Var(Y) +2 Cov(X,Y)$

Since we have independnet variable the Cov(A,B)=0, so then:

$Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2$

3 0

4 years ago

Simplify: x+4-5x-2 A:-5x+2 B:-5x-2 C:-4x+2 D:-4x-2

mezya [45]

Answer:

C:-4x+2

Step-by-step Explanation:

$x + 4 - 5x - 2 \\ = x - 5x + 4 - 2 \\ = - 4x + 2 \\$

5 0

4 years ago