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3 years ago

How many 3/4 teaspoons of salt are in 1/3 of a teaspoon of salt?

Mathematics

1 answer:

vlabodo [156]3 years ago

3 0

Answer:

4/9

Step-by-step explanation:

Multiply each answer choice by 3/4 or 1/3 and they will all be wrong except for 4/9.

3 x 4 = 1/3

so 4/9

You might be interested in

Zoe wants to make a driveway.

-Dominant- [34]

Well the volume of dirt she needs to remove is equal to length*width*height=6.2*4.9*0.225=6.8355

Now let's find how many bags she needs.

To solve for this, simply divide room in bags from total room.
6.8355/0.87=7.857

Thus Zoe has barely enough, but enough bags.

6 0

3 years ago

2a-6=4a Solve for a Check your work

algol [13]

Answer:

-3

Step-by-step explanation:

2a - 6 = 4a

subtract 2a giving you -6 = 2a

divide by 2 from both sides so that a will be by itself

giving you -6 ÷ 2 = a

-6 ÷ 2 = -3

so a = -3

5 0

3 years ago

Solve. 10 · w = 90 w = ______

Ratling [72]

Answer:

Step-by-step explanation:

10- w = 90

w = 90 -10

w = 80

5 0

3 years ago

Read 2 more answers

What is the slope of the line that passes through the points (6,7) and (4,1)?

Scorpion4ik [409]

Answer:

Step-by-step explanation:

m= y2 - y1/ x2 - x 1

(6, 7) (4, 1)

1-7= -6

4-6= -2

-6/-2 = 3

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