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3 years ago

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Mathematics

1 answer:

salantis [7]3 years ago

6 0

Answer:

C. The difference of the medians is 4 times the interquartile range.

Step-by-step explanation:

The diagram show two box plots.

Mahoney's box plot:

Median $=33$

Interquartile range $=34-32=2$

Martin's box plot:

Median $=41$

Interquartile range $=42-40=2$

The interquartile ranges are the same.

The difference of the medians $=41-33=8$

Hence, the difference of the medians is 4 times the interquartile range.

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The price of a visit to the dentist is \$50$50dollar sign, 50. If the dentist fills any cavities, an additional charge of \$100$

Goshia [24]

Answer:

The expense of the visit to dental specialist is equivalent to (50+100n)

Step-by-step explanation:

7 0

2 years ago

What is the distance between (−5,−4) and (-6,4)?

Helga [31]

The distance formula is: d = sqrt( (x2 - x1)2 + (y2 - y1)2 )

For this problem, let (-5, -4) be the "first" point, so x1 = -5 and y2 = -4

and let (-6, 4) be the "second" point, so x2 = -6 and y2 = 4.

Then: d = sqrt( (-6 - -5)2 + (4 - -4)2 ) = sqrt( (-1)2 + (8)2 ) = sqrt( 1 + 64 ) = sqrt( 65)

The distance formula is just the Pythagorean Theorem applied to an x-y graph.

You would get the same final answer if you let (-5, -4) be the second point and (-6, 4) be the first point.

6 0

3 years ago

Need help asap

harina [27]

Answer:

2 hours, 9 minutes (and 19 seconds if u wanna be exact)

Step-by-step explanation:

4 0

3 years ago

Solve the system by using a matrix equation. --4x - 5y = -5 -6x - 8y = -2

evablogger [386]

Answer:

Solution : (15, - 11)

Step-by-step explanation:

We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

$\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}$

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )

Row Echelon Form :

$\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}$

Step # 1 : Swap the first and second matrix rows,

$\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}$

Step # 2 : Cancel leading coefficient in row 2 through $R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1$ ,

$\begin{pmatrix}-6&-8&-2\\ 0&\frac{1}{3}&-\frac{11}{3}\end{pmatrix}$

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

$\begin{bmatrix}1&0&|&15\\ 0&1&|&-11\end{bmatrix}$

As you can see our solution is x = 15, y = - 11 or (15, - 11).

4 0

3 years ago

Once again co sines . Please help me and include explanation with a clear answer

vladimir1956 [14]

Using the law os cosines formula b^2 = a^2 + c^2 - 2*a*c*cos(B)

a = 17, b = 8, c = 16

8^2 = 17^2 + 16^2 - 2*17*16* cos(B)

64 = 289 + 256 - 544 * cos(B)

544*cos(B) = 289 + 256 - 64

544 * cos(B) = 481

cos (B) = 481/544

B = arccos(481/544)

B = 27.8 degrees

8 0

3 years ago