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julia-pushkina [17]
3 years ago
8

You randomly draw a marble from a bag of marbles that contains 8 blue marbles, 5 green marbles, and 8 red

Mathematics
2 answers:
Gekata [30.6K]3 years ago
5 0

Answer:

8-8-21

Step-by-step explanation:

I think that is the answer

denis23 [38]3 years ago
4 0
16/21 is the probability
You might be interested in
if p(x) = x+ 7/ x-1 and q (x) = x^2 + x - 2, what is the product of p(3) and q(2)? a. 50 b. 45 c. 40 d. 20 e. 6
Amanda [17]

Answer:

d. 20

Step-by-step explanation:

To answer the question given, we will follow the steps below:

we need to first find p(3)

p(x) = x+ 7/ x-1

we will replace all x by 3 in the equation above

p(3) = 3+7 / 3-1

p(3) = 10/2

p(3) = 5

Similarly to find q(2)

q (x) = x^2 + x - 2,

we will replace x by 2 in the equation above

q (2) = 2^2 + 2 - 2

q (2) = 4 + 0

q (2) = 4

The product of p(3) and q(2)   =   5 × 4   = 20

6 0
3 years ago
The function f is defined by f(x) = x^2 + 2
jok3333 [9.3K]
It's simple,


f(x)=x^2+2
is the formula that you start with

you are then given f(x). f(4x)


you then plug in 4x

4x=(4x)^2+2

you then solve for x

4x=16x+2
-16x. -16x


-12x=2
÷-12. ÷-12

x=0.16 is your answer
7 0
3 years ago
ITS TIMED PLEASE HELP​
Lubov Fominskaja [6]

Answer:

The graph of the function f(x)=\frac{1}{2}x^{2}-4x+5 has a minimum located at (4,-3)

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

f(x)=a(x-h)^{2}+k

where

a is a coefficient

(h,k) is the vertex of the parabola

If a > 0 the parabola open upward and the vertex is a minimum

If a < 0 the parabola open downward and the vertex is a maximum

In this problem

The coefficient a must be positive, because we need to find a minimum

therefore

Check the option C and the option D

Option C

we have

f(x)=\frac{1}{2}x^{2}-4x+5

Convert to vertex form

f(x)-5=\frac{1}{2}x^{2}-4x

Factor the leading coefficient

f(x)-5=\frac{1}{2}(x^{2}-8x)

f(x)-5+8=\frac{1}{2}(x^{2}-8x+16)

f(x)+3=\frac{1}{2}(x^{2}-8x+16)

f(x)+3=\frac{1}{2}(x-4)^{2}

f(x)=\frac{1}{2}(x-4)^{2}-3

The vertex is the point (4,-3) ( is a minimum)

therefore

The graph of the function f(x)=\frac{1}{2}x^{2}-4x+5 has a minimum located at (4,-3)

5 0
4 years ago
If m∠WXY =160, what are m∠WXZ and m∠ZXY
Dominik [7]

Answer:

so it would be (7x-4)+(5x+8) = 160

x=13

4 0
3 years ago
Factor the following expression<br> <img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-7x-18" id="TexFormula1" title="x^{2} -7x-18
guapka [62]

Answer:

(x-9) (x=2)

Step-by-step explanation:

4 0
3 years ago
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