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2 years ago

Complete the table by squaring each positive x-value listed 2, 3,4,5,6,7,8,9,10,11

Mathematics

1 answer:

elena55 [62]2 years ago

4 0

2²=4 2×2

3²=9 3×3

4²=16 4×4

5²=25 5×5

6²=36 6×6

7²=49 7×7

8²=64 8×8

9²=81 9×9

10²=100 10×10

11²=121 11×11

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Determine whether the underlined number describes a population parameter or a sample statistic. Explain your reasoning. Sixty-fi

Levart [38]

Answer: A. The number is a sample statistic because it is a numerical description of all of the passengers that survived.

Step-by-step explanation:

A population is simply similar items or events that a researcher or an experimenter is interested and wants to carry out an experiment on.

A statistic is simply referred to as the piece of information gotten from the population while a sample statistic is the piece of statistical information which a researcher will be able to get from the statistic.

In this scenario, the number is a sample statistic because it is a numerical description of all of the passengers that survived.

8 0

3 years ago

angel's dad bought a television for her completing all of her digital learning day lesson. he want to warp the gift. the box mea

atroni [7]

Answer: 13 sheets of paper

Step-by-step explanation:

We are given the dimensions of the box and the wrap paper:

Box:

$(56 in)(8 in)(36 in)$

Warp paper:

$(26 in)(17 in)$

Now we need to find the surface area of the box and the area of the wrap paper:

Box:

$surface_{1}=(56 in)(8 in)(2)=896 in^{2}$

$surface_{2}=(8 in)(36 in)(2)=576 in^{2}$

$surface_{3}=(56 in)(36 in)(2)=4032 in^{2}$

$surface-area=896 in^{2}+576 in^{2}+4032 in^{2}=5504 in^{2}$

Warp paper:

$area=(26 in)(17 in)=442 in^{2}$

Dividing the area of the box by the area of the paper:

$\frac{surface-area}{area}=\frac{5504 in^{2}}{442 in^{2}}=12.45 \approx 13$

This means Angel's dad needs to purchase 13 sheets of wrapping paper.

4 0

3 years ago

You have 2 nickels, 11 dimes and 5 quarters in your pocket. What is the ratio of nickels to quarters?

daser333 [38]

Answer:

2:5

Step-by-step explanation:

Since there are 2 nickles and 5 quarters there are 2 nickles for every 5 quarters so answer is 2:5

4 0

2 years ago

What are the solutions to the system of equations graphed below?

Morgarella [4.7K]

Answer:

looking at the straight line, it touches the axis at (-2,0) and (6,0)

Step-by-step explanation:

6 0

2 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

2 years ago