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3 years ago

14

So im doing online schooling right now and I have to take geometry, and i dont understand this question. Find the image of the p

oint (-5, 4) under the reflection across the y-axis.

1 answer:

lana66690 [7]3 years ago

5 0

I am guessing it’s
(5,4)

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(1) In Section 1.4 we used the Axiom of Completeness (AoC) to prove the Archimedean Property of R (Theorem 1.4.2). Show that the

serious [3.7K]

Answer:

Both Answers are in the following attachment (a) and (b)

Step-by-step explanation:

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3 years ago

When a distribution is mound-shaped symmetrical, what is the general relationship among the values of the mean, median, and mode

yuradex [85]

Answer:

The mean, median, and mode are approximately equal.

Step-by-step explanation:

The mean, median, and mode are <em>central tendency measures</em> in a distribution. That is, they are measures that correspond to a value that represents, roughly speaking, "the center" of the data distribution.

In the case of a <em>normal distribution</em>, these measures are located at the same point (i.e., mean = median = mode) and the values for this type of distribution are symmetrically distributed above and below the mean (mean = median = mode).

When a <em>distribution is not symmetrical</em>, we say it is <em>skewed</em>. The skewness is a measure of the <em>asymmetry</em> of the distribution. In this case, <em>the mean, median and mode are not the same</em>, and we have different possibilities as the mentioned in the question: the mean is less than the median and the mode (<em>negative skew</em>), or greater than them (<em>positive skew</em>), or approximately equal than the median but much greater than the mode (a variation of a <em>positive skew</em> case).

In the case of the normal distribution, the skewness is 0 (zero).

Therefore, in the case of a <em>mound-shaped symmetrical distribution</em>, it resembles the <em>normal distribution</em> and, as a result, it has similar characteristics for the mean, the median, and the mode, that is, <em>they are all approximately equal</em>. So, <em>the </em><em>general</em><em> relationship among the values for these central tendency measures is that they are all approximately equal for mound-shaped symmetrical distributions, </em>considering they have similar characteristics of the <em>normal distribution</em>, which is also a mound-shaped symmetrical distribution (as well as the t-student distribution).

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2 years ago

A right prism has height 7½ and triangular bases with sides of length 5, 12, and 13 What is the: Total Surface Area of the Prism

sdas [7]

Given:

A right prism has height 7½ and triangular bases with sides of length 5, 12, and 13.

To find:

The total surface area of the prism.

Solution:

We have,

Height of prism = 7½ = 7.5

Sides of triangular base are 5, 12, 13. These sides of Pythagorean triplets because

$5^2+12^2=13^2$

$25+144=169$

$169=169$

So, the base of the prism is a right triangle.

Area of a triangle is

$Area=\dfrac{1}{2}\times base \times height$

$A_1=\dfrac{1}{2}\times 5\times 12$

$A_1=30$

The area of the base is equal to the area of the top, i.e., $A_2=30$ sq units.

Perimeter of the base is

$P=5+12+13$

$P=30$

The curved surface area of the prism is

$CSA=\text{Perimeter of the base}\times \text{Height of the prism}$

$CSA=30\times 7.5$

$CSA=225$

Now, the total area of the prism is

$A=A_1+A_2+CSA$

$A=30+30+225$

$A=285$

Therefore, the total surface area of the triangular prism is 285 square units.

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2 years ago

Ethan opens a new savings account.

Musya8 [376]

Answer:

$2070

Step-by-step explanation:

$= 2000 \times (1 + \frac{3.5}{100})^{2}$

$= 2070$

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2 years ago

Solve by quadratic formula:<br> 3x^2 - 10x - 20= 0

Arlecino [84]

this is your answer, i hope this helps you

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3 years ago