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3 years ago

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A bag contains 9 marbles: 2 are green, 4 are red, and 3 are blue. Mary chooses a marble at random, and without putting it back,

chooses another one at random. What is the probability that both marbles she chooses are blue? Write your answer as a fraction in simplest form.

1 answer:

amm18123 years ago

7 0

There is a 2/9 chance that both marbles drawn will be blue.

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Someone please help and show work

aleksklad [387]

Answer:

a) 4 $x^{6}$

b) 2 $w^{4}$

Step-by-step explanation:

a) multiply the coefficients and add the exponents

(4)(1) · $x^{3+3}$ = 4 $x^{6}$

b) divide the coefficients and subtract the exponents

(14 ÷ 7) · $w^{6-2}$ = 2 $w^{4}$

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3 years ago

I need help with this homework to practice for a quiz may you please help me ?

laiz [17]

Answer:

set the equations equal to each other, the equations are equal. then find value of x

Step-by-step explanation:

5x+17=3x-4

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3 years ago

Which is a factor of 54xy+45x+18y+15?<br> a) x-5<br> b)6y+5<br> c)6y+1<br> d)3y+5 ...?

tia_tia [17]

If you will simplify this expression you will get this (9x + 3)(6y + 5)
and the factor, which is <span>b)6y+5. </span>

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3 years ago

Read 2 more answers

Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$

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3 years ago

Please help and answer all the questions

raketka [301]

Answer:

least = minimum

no more than = maximum

3 0

3 years ago