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2 years ago

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One ring in the Stonehenge Monument was originally made of 30 Stones, each about six and a half feet wide, 3 ft thick, and 13 ft

tall.what is the approximate volume of the 17th stones that remain today? I'm desperate so please help me out XD

1 answer:

Temka [501]2 years ago

0 0

To get volume, multiply width by height by length/thickness. so multiply 6.5 by 3 by 13 to get your answer, and if you need to know it for all of them combined, multiply by the number of stones.

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Two different samples will be taken from the same population of test scores where the population mean and standard deviation are

Alenkinab [10]

Answer:

The sample consisting of 64 data values would give a greater precision.

Step-by-step explanation:

The width of a (1 - <em>α</em>)% confidence interval for population mean μ is:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}}$

So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (<em>n</em>).

That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.

Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.

The two sample sizes are:

<em>n</em>₁ = 25

<em>n</em>₂ = 64

The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.

Width for n = 25:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{25}}=\frac{1}{5}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]$

Width for n = 64:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]$

Thus, the sample consisting of 64 data values would give a greater precision

5 0

2 years ago

Read 2 more answers

There are 10 red counters and x blue counters in a bag. 2 counters are removed fron the bag. the probability that both counters

frutty [35]

Step-by-step explanation:

10 red 6 blue so 16 altogether.

apologies if I'm wrong

8 0

2 years ago

A golf ball is manufactured so that if it is dropped from A feet above the ground onto a hard surface, the maximum height of eac

Lunna [17]

Step-by-step explanation:

a.) To model this scenario

Let the height of ball = y

The height of 1st= 0.5y

2nd =0.5(0.5y)

3rd = 0.5*(0.5(0.5y))

Hence the height of nth bounce can be modeled as

Height of nth bounce =(0.5ⁿ-1)*y

The exponential equation is

hn= (0.5ⁿ-1)*y

b.) if the ball is dropped from 9ft above the ground

y= 9ft

On the 4th bounce

n=4

Substituting in the exponential equation we have

h4=(0.5^4-1)*9

h4=0.5³*9

h4= 0.125*9

h4= 1.125ft

On the 4th bounce, the ball will reach a height of 1.125ft

6 0

2 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

2 years ago

The time it takes to transmit a file always depends on the file size. Suppose you transmitted 30 files, with the average size of

Ostrovityanka [42]

Answer:

Y = 0.009042 + 0.0002457X

Y = 0.1073 seconds

Step-by-step explanation:

In the given problem we have two variables: the transmission time and average size of file.

Y = transmission time

X = average file size

The linear regression model is given by

Y = a + bX

The slope b is given by

b = correlation coefficient*(SDy/SDx)

Where SDy is the standard deviation of average transmittance time and SDx is the standard deviation of average file size.

b = 0.86(0.01/35)

b = 0.0002457

The y-intercept a is given by

a = y - bx

a = 0.04 - (0.0002457)126

a = 0.04 - 0.030958

a = 0.009042

Therefore, the linear regression model is

Y = 0.009042 + 0.0002457X

Predict the time it will take to transmit a 400 Kbyte file.

Substitute X = 400 in the regression model

Y = 0.009042 + 0.0002457(400)

Y = 0.009042 + 0.09828

Y = 0.1073 seconds

Therefore, the predicted time to transmit a 400 Kbyte file is 0.1073 seconds.

8 0

2 years ago