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3 years ago

1. A ball's position, in meters, as it travels every second is represented by the position function s(t) = 4.9t2+ 450.

1 answer:

7 0

1) The ball's position is described by:
s(t) = 4.9t² + 450
We want to find the velocity, which is the 1st-order derivative of the displacement function (I assume this is an introductory calculus class)
s'(t) = v(t) = 9.8t
We get this by multiplying 4.9 x 2 and reducing the exponent by 1. Now we simply plug 5 in for t.
v(5) = 9.8* 5
v(5) = 49m/s

2) Our cost function is C(x) = x² - 10,000
To find the average rate of change between these units, we use this formula:
( C(101) - C(100) ) ÷1 .
We find the change in C, and divide by the change in x, which is just one.
C(101) = 101² - 10,000
C(101) = 201

C(100) = 100² - 10,000
C(100= 0

C(101) - C(100) = 201
Average rate of change in cost is 201 dollars/ unit between the two points.

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What is the standard deviation of a normal distribution, whose mean is 35, in which an x-value of 23 has a z-score of -1.63?

Dvinal [7]

Answer:

7.36

Step-by-step explanation:

z-score is:

z = (x − μ) / σ

Given that z = -1.63, x = 23, and μ = 35:

-1.63 = (23 − 35) / σ

σ ≈ 7.36

6 0

3 years ago

0.3km ≈ how many miles

Elena L [17]

<span>.3 Kilometers = 0.18641135767 Miles</span>

3 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago

Angle RZT=110<br> Angle RZS=3s<br> Angle TZS=8s<br> What are angles RZS and TZS?

nekit [7.7K]

Angles RZT, RZS, and TZS form triangle RZT, so they sum to 180. Thus, we have:

110+3s+8s=180

Simplifying, we have:

11s=70

Dividing by 11, we see that:

s=70/11

Therefore, RZS=3s=210/11, and TZS=8s=560/11.

6 0

3 years ago

Will and tommie bought a giant candy bar composed of 24 pieces. Will ate 1/4 of the bar and Tommie ate 2/3 of the bar. How many

hichkok12 [17]

Answer:

3/24 or 1/8

Step-by-step explanation:

1. convert 1/4 and 2/3 into fractions with common denominators of 24 (since there are 24 pieces.

2. find the multiples of 4 and 3

4, 8, 12, 16, 20, 24 = 5

3, 6, 9, 12, 15, 18, 21, 24 = 8

3. convert the fractions

1/4 * 5/5 = 5/24

2/3*8/8 = 16/24

4. add the fractions.

5/24 + 16/24 = 21/24

5. subtract 21/24 from 24/24 = 3/24

6. reduce by dividing both terms by 3 = 1/8

6 0

3 years ago