A rectangular tank 4 feet long, 3 feet wide, and 5 feet deep is full of oil with weight density 50 lb ft 3 lbft3 . Calculate the work required to pump all of the oil out over the top of the tank.

Answer:

See Below.

Step-by-step explanation:

We are given that:

[tex]\displaystyle I = I_0 e^{-kt}[/tex]

Where I₀ and k are constants.

And we want to prove that:

[tex]\displaystyle \frac{dI}{dt}+kI=0[/tex]

From the original equation, take the derivative of both sides with respect to t. Hence:

[tex]\displaystyle \frac{d}{dt}\left[I\right] = \frac{d}{dt}\left[I_0e^{-kt}\right][/tex]

Differentiate. Since I₀ is a constant:

[tex]\displaystyle \frac{dI}{dt} = I_0\left(\frac{d}{dt}\left[ e^{-kt}\right]\right)[/tex]

Using the chain rule:

[tex]\displaystyle \frac{dI}{dt} = I_0\left(-ke^{-kt}\right) = -kI_0e^{-kt}[/tex]

We have:

[tex]\displaystyle \frac{dI}{dt}+kI=0[/tex]

Substitute:

[tex]\displaystyle \left(-kI_0e^{-kt}\right) + k\left(I_0e^{-kt}\right) = 0[/tex]

Distribute and simplify:

[tex]\displaystyle -kI_0e^{-kt} + kI_0e^{-kt} = 0 \stackrel{\checkmark}{=}0[/tex]

Hence proven.

Answer:

(3x+2) by (2x+1)

Step-by-step explanation:

A cardboard is a rectangle, and has two dimensions. Given a quadratic equation, you should find a way to split it in two.

The easiest way to do so is through factoring. (There are many ways to do this, take a look at the plethora of sources offered on the internet.)

When the expression 6x^2 + 7x + 2 is factored, it is (3x+2)(2x+1). Hence, these are your dimensions.

Answer:

See answers below

Step-by-step explanation:

Given the following functions:

r(x) = x - 6

s(x) = 2x²

r(s(x)) = r(2x²)

Replacing x with 2x² in r(x) will give;

r(2x²) = 2x² - 6

r(s(x)) = 2x² - 6

(r-s)(x) = r(x) - s(x)

(r-s)(x) = x - 6 - 2x²

Rearrange

(r-s)(x) = - 2x²+x-6

(r+s)(x) = r(x) + s(x)

(r-s)(x) = x - 6 + 2x²

Rearrange

(r-s)(x) = 2x²+x-6

Answer:

t = 2

Step-by-step explanation:

5(6t-3)=5t+35

Step 1 distribute the 5 by multiplying 5 to what is inside of the parenthesis

5(6t) - 5(3) = 5t + 35

Outcome: 30t - 15 = 5t + 35

Step 2 add 15 to both sides

30t - 15 + 15 = 5t + 35 + 15

Outcome: 30t = 5t + 50

Step 3 subtract 5t from both sides

30t - 5t = 5t - 5t + 50

Outcome: 25t = 50

Step 4 divide both sides by 25

25t/25 = 50 we're left with t = 2

Let f(x) = ax³ + 9x² + 4x – 10

g(x) = 0⇒x - 3 = 0

⇒ x = 0 + 3

⇒ x = 3

On dividing f(x) by x - 3, it leaves a remainder 5.

Now keeping, f(3) = 5

⇒a(3)³ + 9(3)² + 4(3) - 10 = 5

⇒ a × 27 + 9 × 9 + 4 × 3 - 10 = 5

⇒ 27a + 81 + 12 - 10 = 5

⇒ 81 + 12 - 10 - 5 = 27a

⇒ 81 + 12 - 15 = 27a

⇒ 93 - 15 = 27a

⇒ 78 = 27a

⇒ a = 27/78

⇒ a = 0.3461

Answer:

s = 7√6.

Step-by-step explanation:

From the 30-60-90 triangle we know that cos 30 = √3/2.

cos 30 = s / 14√2

√3/2 =  s / 14√2

2s = 14√2√3

s = 14√2√3 / 2

s = 7√6.

Answer:

See explanation.

General Formulas and Concepts:

Algebra I

Algebra II

Pre-Calculus

Calculus

Differentiation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Basic Power Rule:

Derivative Rule [Quotient Rule]:                                                                           [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]

Parametric Differentiation:                                                                                     [tex]\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle x = 2t - \frac{1}{t}[/tex]

[tex]\displaystyle y = t + \frac{4}{t}[/tex]

Step 2: Find Derivative

Here we see that if we increase our values for t, our derivative would get closer and closer to 0.5 but never actually reaching it. Another way to approach it is to take the limit of the derivative as t approaches to infinity. Hence  [tex]\displaystyle \frac{dy}{dx} < \frac{1}{2}[/tex].

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametrics

Book: College Calculus 10e

Answer:

i think its the 3 line. they are congruent.

Answer:

its the 3rd option

Step-by-step explanation:

first of all AA means angle-angle which means we are using their angles to compare them

second, those lines on the sides are only there to tell you they are in the exact same angle, and the two boxes on the bottom show that the angle of both triangles is 90° therefore they are the same

Explanation:

P(A) represents the probability of event A

P(not A) is the probability that event A doesn't happen

We only have two choices: Either A happens or it doesn't

So that means P(A) + P(not A) = 1

The "1" represents a 100% chance, aka certainty.

This question is incomplete, the complete question is;

Many people grab a granola bar for breakfast or for a snack to make it through the afternoon slump at work. A Kashi GoLean Crisp Chocolate Caramel bar weighs 45 grams. The mean amount of protein in each bar is 8 grams. Suppose the amount of protein in the bars have a normal distribution with a standard deviation of 0.32 grams and a random Kashi bar is selected.  

Suppose the amount of protein is at least 8.6 grams. What is the probability that it is more than 8.7 grams

Answer:

the required probability is 0.472

Step-by-step explanation:

Given the data in the question;

Let x represent the random variable that shows the protein in the bar.

{ normal distribution }

mean μ = 8

standard deviation σ =  0.32

Now, Suppose the amount of protein is at least 8.6 grams. What is the probability that it is more than 8.7 grams

first we get the z-score for x = 8.6

z = x - μ / σ

z = ( 8.6 - 8 ) / 0.32

z = 0.6 / 0.32

z = 1.875

so

P( x ≥ 8.6 ) = P( z ≥ 1.875 ) = 1 - 0.9696 = 0.0304

Also for, x = 8.7

z = x - μ / σ

z = ( 8.7 - 8 ) / 0.32

z = 0.7 / 0.32

z = 2.1875

so

P( x > 8.7 ∩ x ≥ 8.6 ) = P( x > 8.7 ) = P( z > 2.1875 ) = 1 - 0.98565 = 0.01435

Now, the required probability will be;

P( x > 8.7 | x ≥ 8.6 ) = [P( x > 8.7 ∩ x ≥ 8.6 )] / [ P( x ≥ 8.6 ) ]

= 0.01435 / 0.0304

= 0.472

Therefore, the required probability is 0.472

9514 1404 393

Explanation:

There are several ways you can go at this. Here are a couple. All proofs will start with the given relations being repeated as part of the proof. Here are the next steps.

∠AED ≅ ∠BAE . . . . alternate interior angles are congruent

∠AED +∠AEB = 90° . . . . angle sum theorem

∠BAE +∠AEB = 90° . . . . substitution property of equality

∠CEF ≅ ∠BCE . . . . alternate interior angles are congruent

∠CEF +∠CEB = 90° . . . . angle sum theorem

∠BCE +∠CEB = 90° . . . . substitution property of equality

∠BAE +∠AEB = ∠BCE +∠CEB . . . . substitution property of equality

∠BAE +∠AEB = ∠BCE +∠AEB . . . . substitution property of equality

∠BAE = ∠BCE . . . . addition property of equality

∠ABE = ∠CBE = 90° . . . . BE ⊥ AC

BE ≅ BE . . . . reflexive property of congruence

ΔBEA ≅ ΔBEC . . . . ASA congruence theorem

∠BAE ≅ ∠BCE . . . . CPCTC

Answer:

a) 0.3056 = 30.56% probability that 3 females and 2 males are selected.

b) 0.0306 = 3.06% probability that all five students selected are males.

c) 0.0131 = 1.31% probability that all five students selected are females.

d) 0.9869 = 98.69% probability that at least one male is selected.

Step-by-step explanation:

The students are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

In this question:

15 + 13 = 28 students, which means that [tex]N = 28[/tex]

5 are selected, which means that [tex]n = 5[/tex]

13 females, which means that [tex]k = 13[/tex]

a. 3 females and 2 males are selected?

3 females, so this is P(X = 3).

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 3) = h(3,28,5,13) = \frac{C_{13,3}*C_{15,2}}{C_{28,5}} = 0.3056[/tex]

0.3056 = 30.56% probability that 3 females and 2 males are selected.

b.all five students selected are males?

0 females, so this is P(X = 0).

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 0) = h(0,28,5,13) = \frac{C_{13,0}*C_{15,5}}{C_{28,5}} = 0.0306[/tex]

0.0306 = 3.06% probability that all five students selected are males.

c. all five students selected are females?

This is P(X = 5). So

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 5) = h(5,28,5,13) = \frac{C_{13,5}*C_{15,0}}{C_{28,5}} = 0.0131[/tex]

0.0131 = 1.31% probability that all five students selected are females.

d.at least one male is selected?

Less than five females, so:

[tex]P(X < 5) = 1 - P(X = 5) = 1 - 0.0131 = 0.9869[/tex]

0.9869 = 98.69% probability that at least one male is selected.

9514 1404 393

Answer:

Step-by-step explanation:

To find the initial amount, put 0 where t is in the formula and do the arithmetic.

  A(0) = 523(1/2)^0 = 523(1) = 523

The initial amount is 523 grams.

__

To find the amount remaining after 100 years, put 100 where t is in the formula and do the arithmetic.

  A(100) = 523(1/2)^(100/30) ≈ 523(0.0992123) ≈ 52

About 52 grams will remain after 100 years.

An

Step-by-step explon:

9514 1404 393

Answer:

  B. a = plus-or-minus 4

Step-by-step explanation:

  3a² -21 = 27 . . . . . . . given

  3a² = 48 . . . . . . . . . . add 21 to both sides (desired form)

  a² = 16 . . . . . . . . . . .  divide both sides by 3

  a = ±4 . . . . . . . take the square root

Answer:

5 cases

10 additional units

Step-by-step explanation:

Given that :

Total number of units needed = 60 units

Total number of units per case = 14

Hence, the total number of cases required will be :

Number of units needed / number of units per case

Number of cases required = 60 / 14 = 4.285 (this means that 5 cases are required as 4 cases won't be up to 60 units)

With 5 cases, we have exceeded the required units needed :

Additional units will be : (14 * 5) - 60

Additional units = 70 - 60 = 10 units

Answer:

Symmetric with respect to the x-axis

Symmetric with respect to the y-axis

Symmetric with respect to the origin

From the test the person wants, and the sample data, we build the test hypothesis, find the test statistic,  and use this to reach a conclusion both using the critical value and the p-value.

Doing this, the conclusions are:

---------------------------------------------------------

Hypothesis:

A machine in a factory must be repaired if it produces more than 10% defectives in production.

At the null hypothesis, we test if it does not have to be repaired, that is,  the proportion is of at most 10%. So

[tex]H_0: p \leq 0.1[/tex]

At the alternative hypothesis, we test if it does have to be repaired, that is, the proportion is greater than 10%. So

[tex]H_1: p > 0.1[/tex]

------------------------------------------------------

Decision rule:

0.01 significance level, using a left-tailed test(testing if the mean is more than a value), which means that:

----------------------------------------------------------

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.

0.1 is tested at the null hypothesis:

This means that [tex]\mu = 0.1, \sigma = \sqrt{0.1*0.9}[/tex]

A random sample of 100 items from a day's production contains 15 defectives.

This means that [tex]n = 100, X = \frac{15}{100} = 0.15[/tex]

Value of the test-statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{0.15 - 0.1}{\frac{\sqrt{0.1*0.9}}{\sqrt{100}}}[/tex]

[tex]z = 1.67[/tex]

----------------------------------------------

Decision: Critical region

The test statistic is [tex]z = 1.67 < z_c[/tex], meaning that there is not enough evidence to conclude that the proportion of defectives is above 10%, that is, it does not support his decision to repair at the 0.01 significance level.

Decision: p-value

The p-value of the test is the probability of finding a sample proportion above 0.15, which is 1 subtracted by the p-value of z = 1.67.

Looking at the z-table, z = 1.67 has a p-value of 0.9525.

1 - 0.9525 = 0.0475.

The p-value of the test is 0.0475 > 0.01, meaning that there is not enough evidence to conclude that the proportion of defectives is above 10%, that is, it does not support his decision to repair at the 0.01 significance level.

A similar problem is found at https://brainly.com/question/24326664

Answer:

h = 6.2 units

Step-by-step explanation:

Given triangle ABC is a right triangle with the measures of the two sides,

BC = [tex]\frac{10}{2}[/tex] = 5 units

AC = 8 units

By applying Pythagoras theorem in the given triangle,

AC² = AB² + BC²

8² = AB² + 5²

AB² = 64 - 25

AB = √39

AB = 6.24 units

AB ≈ 6.2 units

Answer:

u = 2

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

cos theta = adj side / hypotenuse

cos 45 = sqrt(2) / u

u cos 45 = sqrt(2)

u = sqrt(2) / cos 45

u = sqrt(2) / 1/ sqrt(2)

u = sqrt(2) * sqrt(2)

u =2

u=2

Answer:

Solution given:

Cos 45°=base/hypotenuse

[tex]\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{u}[/tex]

doing crisscrossed multiplication

[tex]\sqrt{2}*\sqrt{2}=1*u[/tex]

u=2

Step-by-step explanation:

1. The inputs are the dice values. Since the cube is six sided, the possible. values are (1,2,3,4,5,6).

2. The outputs values are points if we roll the number. If we land in a special space, we must respect that rule so our outputs are

(2,2,8,5,6,8).

3. I cant show a mapping diagram on brainly. Draw a mapping diagram and make sure to connect the x values and y values of the following.Also make sure to start on square 1.

4. The domain of the function is the same as the input. The dice values, 1,2,3,4,5,6

5.The range are the. values that occur if we roll the number about square 1.

2,2,8,5,6,8

6. No, the player could have rolled 3 and landed on 8. The player also could have rolled 6.

7. Yes, every one x value corresponds with one y value.

The question is incomplete. The complete question is :

In a​ poll, 1100 adults in a region were asked about their online vs.​ in-store clothes shopping. One finding was that 43​% of respondents never​ clothes-shop online. Find and interpret a ​95% confidence interval for the proportion of all adults in the region who never​ clothes-shop online.

Solution :

95% confidence interval for p is :

[tex]$\hat p - Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}} < p < \hat p + Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$[/tex]

[tex]$0.43 - 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}} < p < 0.43 + 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}}$[/tex]

0.401 < p < 0.459

Therefore, 95% confidence interval is from 0.401 to 0.459

Answer:

I think the answer is 5/6

Step-by-step explanation:

There are three even numbers and two uneven numbers less than four. Therefore, on a standard die, the.probability of Rollin a number that is even or less than for is 5/6.

Answer:

Step-by-step explanation :

Let % increase in administrative secretary be = x

Let % increase in new faculty receive be = y

Administrative secretary salary = 28,000

New faculty receive Salary = 40,000

(8)*(x/100)* (28000) + (7)*(y/100)*(40000) = 5,000

2240x +2800 y = 5,000

224x +280 y = 500

56x +70y = 125

Therefore, x and y should be chosen such that it satisfy the above equation.

Answer:

didn't get the question did u forget to put the sequence ???????

pls write the question fully so that I can help you