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3 years ago

For items 5 - 7 consider the score distribution of 15

Mathematics

2 answers:

Vika [28.1K]3 years ago

7 0

The median is 79. Hope this helps!

jeka57 [31]3 years ago

7 0

The answer is C I hoped this helped

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Please help me with this :) i give 30 points .

Tatiana [17]

Answer:

2 B 4/75

3 B 1/7

Step-by-step explanation:

4 red, 3 blue, 2 yellow, 6 orange

total = 15

2. P(red) = red/total = 4/15

replace it

so we still have 4 red, 3 blue, 2 yellow, 6 orange

total = 15

P(blue) = blue/total = 3/15 = 1/5

P(red, replace, blue) = 4/15 * 1/5 = 4/75

3 P(orange) = orange/total = 6/15 = 2/5

not replacing it

so we have 4 red, 3 blue, 2 yellow, 5 orange

total = 14

P(orange) = orange/total = 5/14

P(orange, no replace, orange) = 2/5*5/14 = 2/14 = 1/7

6 0

3 years ago

Read 2 more answers

which of the following expresses the coordinates of the foci of the conic section shown below? (x-2)^2/4+(y+5)^2/9

elena55 [62]

Step-by-step explanation:

$\frac{(x - 2) {}^{2} }{4} + \frac{(y + 5) {}^{2} }{9} = 1$

This is the equation of the ellipse. Since the denominator is greater for the y values, we have a vertical ellipse. Remember a>b, so a

The formula for the foci of the vertical ellipse is

(h,k+c) and (h,k-c).

where c is

Our center (h,k) is (2, -5)

${c}^{2} = {a}^{2} - {b}^{2}$

Here a^2 is 9, b^2 is 4.

${c}^{2} = 9 - 4$

${c}^{2} = 5$

$c = \sqrt{5}$

So our foci is

$(2, - 5 + \sqrt{5} )$

and

$(2, - 5 - \sqrt{5} )$

6 0

1 year ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Mkey [24]

For this case, we have to:

By definition, we know:

The domain of $f (x) = \sqrt [3] {x}$ is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

$y = \sqrt [3] {x-2}$ with $x = 0$ : $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x+2}$ with $x = 0:\ y = \sqrt [3] {2}$ is also defined.

$f (x) = \sqrt {x}$ has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .