PLEASE HELP ME I BEG YOU

A box contains 10 cards numbered 1 through 10. Which events are mutually exclusive? A. A card is drawn from the box and the numb

coldgirl [10]

The answer is A. 5 is not an even number, so both events can't take place together.

6 0

3 years ago

Evaluate the function for the given values of x.

galben [10]

Answer:

g(3) = 11

g(-3) = 16

g(-1) =3

Step-by-step explanation:

For g(3)

g(x)= x² +2

g(3) = 3² + 2

g(3) = 9+2

g(3) = 11

For g(-3)

g(x) = -3x + 7

g(-3) = -3(-3) + 7

g(-3) = 9 + 7

g(-3) = 16

For g(-1)

g(x) = x² + 2

g(-1) = (-1)² +2

g(-1) = 1+2

g(-1) =3

6 0

2 years ago

Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$

Then

$\dfrac{V_{\text{new flask}}}{V_{\text{flask}}} =\dfrac{8}{1}=8.$

Answer 2: correct choice is D.

8 0

3 years ago

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A gumball machine has just been filled with 50 black, 150 white, 100 red and 100 yellow gum balls that have been thoroughly mixe

Andrew [12]

<span>Exactly 33/532, or about 6.2% This is a conditional probability, So what we're looking for is the probability of 2 gumballs being selected both being red. So let's pick the first gumball. There is a total of 50+150+100+100 = 400 gumballs in the machine. Of them, 100 of the gumballs are red. So there's a 100/400 = 1/4 probability of the 1st gumball selected being red. Now there's only 399 gumballs in the machine and the probability of selecting another red one is 99/399 = 33/133. So the combined probability of both of the 1st 2 gumballs being red is 1/4 * 33/133 = 33/532, or about 0.062030075 = 6.2%</span>

6 0

3 years ago

Order least from greatest -50 44 -3 1.5

sp2606 [1]

Answer:

-50, -31.5, 44 I think

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers