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3 years ago

6

Consider the graph and equation, y = 3x, that represent Alonso’s walking speed. What relationship is represented by this equatio

n and graph? What would the points (3, 9) and (5, 15) represent?

2 answers:

svlad2 [7]3 years ago

7 0

Answer:

The equation y = 3x represents a proportional relationship, where y represents distance in miles, and x represents time in hours. The point (3, 9) represents that Alonso walks 9 miles in 3 hours. The point (5, 15) represents that Alonso walks 15 miles in 5 hours.

Ira Lisetskai [31]3 years ago

4 0

Answer:

Step-by-step explanation:

The x value represents the time and y value represents the distance

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Hi! I’m stuck on this problem on math! This is 5th grade math. Can anyone help me?

emmasim [6.3K]

Answer:

The answer is 4.2 kilograms of dog food

Step-by-step explanation:

it says that ray feeds his dog 0.12 kilogram of food each day

he needs for 1 month

1 month = 31 days or 30 days

since 1 day food is 0.12 kilograms you'll have to multiply 0.12 with 1 month

i.e; 0.12*31 or 0.12*30

0.12*31=3.72 and 0.12*30=3.6

since we don't have an option we'll have to choose something bigger than this

therefore, the answer is 4.2

7 0

3 years ago

HELP DUE IN 15 MINS!

Stells [14]

Answer:

n = 18

Step-by-step explanation:

<u>Sum of interior angles of a polygon:</u>

(n - 2)180 = 160n
180n - 360 = 160n
20n = 360
n = 18

8 0

3 years ago

Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

nadya68 [22]

I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

and so the standard deviation is $\sqrt{\dfrac{350}3}\approx\boxed{116.67}$

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

8 0

3 years ago

Can anyone help with this?

Len [333]

Answer:

226.2‬

Step-by-step explanation:

32.5+42.9=75.4

75.4*3=226.2‬

226.2 is the answer.

5 0

3 years ago

Find f of 1 and f of 4

Dafna1 [17]

You have to check f(1) in each one to see if it's right and for f(4)

4 0

3 years ago