This doesn't makes sense can anyone explain? Test on Friday!!

From base camp, a hiker walks 3.5 miles west and 1.5 miles north. Another hiker walks 2 miles east and 0.5 miles south. To the n

levacccp [35]

Answer:

The hikers are 5.9 miles apart.

Step-by-step explanation:

Let O represents the base camp,

Suppose after walking 3.5 miles west, first hiker's position is A, then after going 1.5 miles north from A his final position is B,

Similarly, after walking 2 miles east, second hiker's position is C then going towards 0.5 miles south his final position is D.

By making the diagram of this situation,

Let D' is the point in the line AB,

Such that, AD' = CD

In triangle BD'D,

BD' = AB + AD' = 1.5 + 0.5 = 2 miles,

DD' = AC = AO + OC = 3.5 + 2 = 5.5 miles,

By Pythagoras theorem,

$BD^2 = BD'^2 + DD'^2$

$BD = \sqrt{2^2 + 5.5^2}=\sqrt{4+30.25}=\sqrt{34.25}\approx 5.9$

Hence, the hikers are 5.9 miles apart.

8 0

3 years ago

Which of the following is the graph of y= start root -x -3 endroot?

prisoha [69]

Answer:

Step-by-step explanation:

8 0

3 years ago

Sheila eats 3/4 of a bag of carrots each week, how many bags of baby carrots does she eat in 6 weeks? Write in simplest form

Alla [95]

3/4 x 6 = 18/4 = 4 1/2

she eats 4 1/2 bags of carrots in 6 weeks

8 0

3 years ago

Read 2 more answers

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

3 years ago

Hunter-Best [27]

Answer:

a.The ratio of the heights is the square roots of the both of them √64 and √144

which is 8 and 12

b.To find the height of the larger prism you use the ratio of the heights,in a ratio and proportion

8 - 49

12-x

8x/8=588/8

x=73.5

I hope this helps

4 0

3 years ago