Materials: Tall clear drinking glass or vase At least four of the following liquids: Fresh water Salt water Vegetable oil Rubbing alcohol Dish soap Honey Corn syrup Milk Maple syrup At least three different small items of your choice, such as: Ping pong ball Small screw, bolt, or nut Popcorn kernel Peanut Blueberry Grape Cherry tomato Instructions: Select four liquids and predict how you think they compare in density by ranking them from most dense to least dense in the data table below. Measure out ¼ cup volume of each liquid, and pour them one at a time into the clear glass or vase. Record your observations in the lab worksheet. Gently add the first small item to the liquids, and record your observation of where it settles. Repeat with the other small items. Clean up all lab materials (the liquids can be poured down the sink), and complete the lab worksheet. Data Table: Prediction: Rank the four liquids from lowest density (top) to highest density (bottom) Observation: Rank how the four liquids really compare, from lowest density (top) to highest density (bottom) Observations: What objects did you place in the liquid, and where did each settle? Object Layer where it settled Observations and Conclusions: Define density, and describe how this activity helps you compare the density of four different liquids without making mass measurements. How did the observations compare to your predictions? Did any of the results surprise you? How would the density of water change if you measured out ½ cup instead of ¼ cup? Explain your answer in complete sentences.

Answer:

Tangent

Step-by-step explanation:

Tangent intersects a circle at exactly one point

The function that relates the distance travelled d to the time t is d = f(t) = 140t.

What is Speed?

Speed is a scalar quantity which measures the rate of change of the position of an object without measuring on the direction.

In other words, it can be defined as the ratio of distance covered by an object to the time taken by the object to cover the distance.

The train leaves the station at time t = 0.

Distance travelled by the train = 280 kilometers

Time taken to travel the distance = 2 hours

Speed = Distance / Time

           = 280 / 2

           = 140 kilometers/ hour.

Given that train travels at a constant speed.

So for any distance 'd' and the time taken to travel the distance 't',

d / t = 140

d = 140t

d = f(t) = 140t

Hence the distance travelled by the train can be related to the time taken by the function, d = f(t) = 140t

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Answer:

d = 140t

Step-by-step explanation:

Answer:

538 books should be tested.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

How many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence?

n books should be tested.

n is found when

We have that

Rounding up

538 books should be tested.

Answer:

g(-1 )=-1 and g(2)+g(1)=7

Step-by-step explanation:

If g(x) = x^3+x^2-x-2 find g(-1)

if we find g(-1)

we substitute all the x's in the function with -1

-1^3+-1^2-(-1)-2

-1^3 = -1

-1^2 = 1

-1+1+1-2

(two minuses make a plus)

-1+1 = 0

0+1 = 1

1-2 = -1

if x=-1, g(-1) is -1

g(2)+g(1)

substitute the x's in the function with 2 and 1 and add your results

2^3+2^2-2-2

2^3 = 8, 2^2 = 4

8+4-2-2

8+4= 12, 12-2 = 10, 10-2 = 8

g(2)=8

g(1) now

1^3 + 1^2-1-2

1^3=1, 1^2 = 1

1+1-1-2

1+1 = 2, 2-1 = 1, 1-2 = -1

g(3) = -1

g(2) (which equals 8) + g(3) (which equals -1) =

8+(-1) = 7

g(2)+g(3)=7

Answer:

just do 68986

- 64997

______

3,989

Answer:

C(60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

Let x(t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is  

C(t) = 3*10⁻⁴*x(t).

The input rate is 6*(0.001/100) = 6*10⁻⁵.

The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)

The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.

The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.  

Remember, 1 h = 60 minutes. The initial value problem is  

dx/dt= 6*10⁻⁵ - 18*10⁻⁴x =  - 6* 10⁻⁴*(3x - 10⁻¹)               x(0) = 1.

The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.

The integration of both sides gives us  

Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C    or    |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).  

Therefore, 3x - 0.1 = C₁*e∧(-18*10⁻⁴t).

Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)

then  

C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)

If  t = 60

We have

C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵

3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵

t = 1870.72 s