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3 years ago

Which expression is equivalent to (- 2c / d^2)^3

Mathematics

2 answers:

nikdorinn [45]3 years ago

4 0

-8c^3/d^6 is equivalent to the expression

Snezhnost [94]3 years ago

4 0

Answer:

-8c^3/d^6

You are correct!

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Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50% of all such batches contain no defect

ad-work [718]

Answer:

a) P (0 defective component) = 0.5

P( 1 defective component ) = 0.35

P( 2 defective component ) = 0.15

b) P( 0 ) = 0

p ( 1 ) = 1

p ( 2 ) = 0.33

Step-by-step explanation:

p( no defective ) = 0.5

p( 1 is defective ) = 0.35

p( 2 is defective ) = 0.15

Given that 2 components are selected at random

a) Given that neither component is defective

Probability of 0 defective component = 0.5

P( 1 defective component ) = 0.35

P( 2 defective component ) = 0.15

b) Given that one of the two tested component is defective

P( 0 defective ) = 0

P( 1 defective ) = P ( $\frac{x=1}{x\geq 1}$ ) = p( x = 1 ) / 1 - P ( x = 0 )

= ( 0.5 )^1 ( 0.5 )^0 / 1 - ( 0.5)^0 (0.5)^1

= 0.5 * 1 / 1 - 0.5 = 0.5 / 0.5 = 1

p ( 2 defective ) = p( x = 3 ) / 1 - P ( x = 0 )

= ( 0.5 )^2 ( 0.5 )^0 / 1 - ( 0.5)^0 (0.5)^2

= 0.25 / 0.75 = 0.33

5 0

2 years ago

A tank contains 10 liters of pure water. Saline solution with a variable concentration 5 grams of salt per liter is pumped into

algol [13]

Answer:

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

Step-by-step explanation:

The mass flow rate dQ(t)/dt = mass flowing in - mass flowing out

Since 5 g/L of salt is pumped in at a rate of 4 L/min, the mass flow in is thus 5 g/L × 4 L/min = 20 g/min.

Let Q(t) be the mass present at any time, t. The concentration at any time ,t is thus Q(t)/volume = Q(t)/10. Since water drains at a rate of 4 L/min, the mass flow out is thus, Q(t)/10 g/L × 4 L/min = 2Q(t)/5 g/min.

So, dQ(t)/dt = mass flowing in - mass flowing out

dQ(t)/dt = 20 g/min - 2Q(t)/5 g/min

Since the salt just begins to be pumped in, the initial mass of salt in the tank is zero. So Q(0) = 0

So, the initial value problem is thus

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

3 0

2 years ago

A line has a slope of -1/4 and passes through point (-5/4, 1) What is the equation of the line?

Vedmedyk [2.9K]

The line of the equation would be y = -1/4x + 11/16.

In order to find that, we'll put the ordered pair and the slope into slope intercept form. Then we can solve for the intercept.

y = mx + b
1 = -1/4(-5/4) + b
1 = 5/16 + b
11/16 = b

So then we put that in the equation with slope for the final equation.

7 0

2 years ago

X + 78 = 90 can someone solve this for me please?

kondaur [170]

Let's solve your equation step-by-step.

x+78=90

Step 1: Subtract 78 from both sides.

x+78−78=90−78

x=12

Please put brainliest!

8 0

3 years ago

48-5+2x3+6.

kkurt [141]

The parentheses must be placed here 48 - (5 + 2) * 3 + 6 in order for the equation to equal 33.

3 0

3 years ago