All categories

3 years ago

Approximate the value of 14−−√ to the nearest tenths place. Plot the approximation on the number line.

Mathematics

2 answers:

gizmo_the_mogwai [7]3 years ago

8 0

Answer:

it's 3.7

Step-by-step explanation:

LiRa [457]3 years ago

3 0

Your answer would be 3.7
If you are in K12, just count the numbers after the 3
To get 3.7

You might be interested in

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

Write an equation to represent each situation.

mina [271]

Answer:

18 years

Step-by-step explanation:

2(8+x) = 34+x

16+2x = 34+x

-18 = -x

4 0

3 years ago

A cone has a height of 2.5 in. and a radius of 5 in. What is the volume of the cone? (Use 3.14 for pi. Round the answer to two d

Pie

Answer:

65.45

Step-by-step explanation:

you do 3.14*5^2*2.5/3

7 0

2 years ago

A rectangular area adjacent to a river is fenced in; no fence is needed on the river side. The enclosed area is 1500 square fee

ZanzabumX [31]

Answer:

a) C(x) = 15000/x + 6x +80

b) Domain of C(x) { R x>0 }

Step-by-step explanation:

We have:

Enclosed area = 1500 ft² = x*y from which y = 1500 / x (a) where x is perpendicular to the river

Cost = cost of sides of fenced area perpendicular to the river + cost of side parallel to river + cost of 4 post then

Cost = 10*y + 2*3*x + 4*20 and accoding to (a) y = 1500/x

Then

C(x) = 10* ( 1500/x ) + 6*x + 80

C(x) = 15000/x + 6x +80

Domain of C(x) { R x>0 }

5 0

2 years ago

63-) find the value of x can someone please help me

Helen [10]

Answer:

61° is the answers for the question

Step-by-step explanation:

please give me brainlest

6 0

2 years ago

Read 2 more answers