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mixer [17]
2 years ago
10

One marble is selected at random from a bowl containing 15 blue marbles and 12 red marbles. What are the odds of selecting a blu

e marble
Mathematics
2 answers:
Softa [21]2 years ago
7 0

Answer:

5/9

Step-by-step explanation:

lana [24]2 years ago
4 0

Answer:

5/9

Step-by-step explanation:

You put the total amount of blue marbles over the total number of marbles and you get 15/27, which simplifies to 5/9

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What is the range of the given function? {(–2, 0), (–4, –3), (2, –9), (0, 5), (–5, 7)}
Natali5045456 [20]

Answer: R = {0, -3, -9, 5, 7}

Explanation: The range is the set of all y-coordinates for our ordered pairs.

So the range for this relation is {0, -3, -9, 5, 7}.

7 0
3 years ago
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110$ for 16ft find the unit rate ​
Neporo4naja [7]

Answer:

6.875 dollars per feet

Step-by-step explanation:

110/16

6.875

8 0
3 years ago
If their were thirty cows and twenty ate chickens how many didn't?
tresset_1 [31]

Answer:

10 didn’t eat chickens

Step-by-step explanation:

30-20=10

6 0
3 years ago
Read 2 more answers
I need helpp plsss:)
kakasveta [241]

Answer:

*read explanation for both questions*

Step-by-step explanation:

So the first one asks <em>”the first number in a pattern is 6, the pattern follows the rule ’</em><em>multiply by 4’ </em><em>“</em>

The question has the answer to it, so each number would be multiplied by 4. So we have the first one which is 6, next we multiply 6 x 4 = 24. Then we multiply 24 x 4 = 96. and finally you multiply 96 x 4 = 384.

That’s the pattern fo the first one

The second one asks <em>“The shapes have been sorted into two groups. explain what attribute was used to sort the shapes.”</em>

<em />

When we’re looking for attributes, look for what they have in common. For me, I see that Group 1 has the smaller versions of the same shapes in Group 2. So that could be the attribute the question is asking for.

3 0
3 years ago
A)A cuboid with a square x cm and height 2xcm². Given total surface area of the cuboid is 129.6cm² and x increased at 0.01cms-¹.
Nutka1998 [239]

Answer: (given assumed typo corrections)


(V ∘ X)'(t) = 0.06(0.01t+3.6)^2 cm^3/sec.


The rate of change of the volume of the cuboid in change of volume per change in seconds, after t seconds. Not a constant, for good reason.



Part B) y'(x+Δx/2)×Δx gives exactly the same as y(x+Δx)-y(x), 0.3808, since y is quadratic in x so y' is linear in x.


Step-by-step explanation:

This problem has typos. Assuming:

Cuboid has square [base with side] X cm and height 2X cm [not cm^2]. Total surface area of cuboid is 129.6 cm^2, and X [is] increas[ing] at rate 0.01 cm/sec.


129.6 cm^2 = 2(base cm^2) + 4(side cm^2)

= 2(X cm)^2 + 4(X cm)(2X cm)

= (2X^2 + 8X^2)cm^2

= 10X^2 cm^2

X^2 cm^2 = 129.6/10 = 12.96 cm^2

X cm = √12.96 cm = 3.6 cm


so X(t) = (0.01cm/sec)(t sec) + 3.6 cm, or, omitting units,

X(t) = 0.01t + 3.6

= the length parameter after t seconds, in cm.


V(X) = 2X^3 cm^3

= the volume when the length parameter is X.


dV(X(t))/dt = (dV(X)/dX)(X(t)) × dX(t)/dt

that is, (V ∘ X)'(t) = V'(X(t)) × X'(t) chain rule


V'(X) = 6X^2 cm^3/cm

= the rate of change of volume per change in length parameter when the length parameter is X, units cm^3/cm. Not a constant (why?).


X'(t) = 0.01 cm/sec

= the rate of change of length parameter per change in time parameter, after t seconds, units cm/sec.

V(X(t)) = (V ∘ X)(t) = 2(0.01t+3.6)^3 cm^3

= the volume after t seconds, in cm^3

V'(X(t)) = 6(0.01t+3.6)^2 cm^2

= the rate of change of volume per change in length parameter, after t seconds, in units cm^3/cm.

(V ∘ X)'(t) = ( 6(0.01t+3.6)^2 cm^3/cm )(0.01 cm/sec) = 0.06(0.01t+3.6)^2 cm^3/sec

= the rate of change of the volume per change in time, in cm^3/sec, after t seconds.


Problem to ponder: why is (V ∘ X)'(t) not a constant? Does the change in volume of a cube per change in side length depend on the side length?


Question part b)


Given y=2x²+3x, use differentiation to find small change in y when x increased from 4 to 4.02.


This is a little ambiguous, but "use differentiation" suggests that we want y'(4.02) yunit per xunit, rather than Δy/Δx = (y(4.02)-y(4))/(0.02).


Neither of those make much sense, so I think we are to estimate Δy given x and Δx, without evaluating y(x) at all.

Then we want y'(x+Δx/2)×Δx


y(x) = 2x^2 + 3x

y'(x) = 4x + 3


y(4) = 44

y(4.02) = 44.3808

Δy = 0.3808

Δy/Δx = (0.3808)/(0.02) = 19.04


y'(4) = 19

y'(4.01) = 19.04

y'(4.02) = 19.08


Estimate Δy = (y(x+Δx)-y(x)/Δx without evaluating y() at all, using only y'(x), given x = 4, Δx = 0.02.


y'(x+Δx/2)×Δx = y'(4.01)×0.02 = 19.04×0.02 = 0.3808.


In this case, where y is quadratic in x, this method gives Δy exactly.

6 0
3 years ago
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