a) Read section 1.5 in the Yakir textbook. If you were a teacher and had 30 students in your class and wanted to know the class average on the first quiz, would you use a parameter or a statistic

Answer:

y > -1

Step-by-step explanation:

the line is going across the y axis and is everything above -1

Answer:

Step-by-step explanation:

speed is calculated using formula v=d/t

m= 7.5m

t= 3 min

v=?

v= 7.5m/3min

v= 2.5m/min

9514 1404 393

Answer:

  7.92 years

Step-by-step explanation:

We want to find t such that ...

  3 = 2^(t/5)

where 2^(t/5) is the annual multiplier when doubling time is 5 years.

Taking logs, we have ...

  log(3) = (t/5)log(2)

  t = 5·log(3)/log(2) ≈ 7.92 . . . years

It will take about 7.92 years for the population to triple.

Water drunk by Bobby in 10 hours = 5 units

So, water drink by Bobby in 1 hour

= 5/10 units

= 1/2 units

= 0.5 units

Answer:

1/2 water

Step-by-step explanation:

We can use a ratio to solve

5 waters        x waters

-----------   = ------------

10 hours             1 hours

Using cross products

5*1 = 10 *x

5 = 10x

Divide by 10

5/10 = x

1/2 waters =x

Answer:

f(g(1)) = - 2

Step-by-step explanation:

Find g(1) then use the value obtained to find f(x)

g(1) = 1 ← value of y when x = 1 (1, 1 ) , then

f(1) = - 2 ← value of y when x = 1 (1, - 2 )

Answer: 3

happy learning

Answer:

B.

Step-by-step explanation:

From the point (-1,0) the next point on the graph is up 3, right 1, making the slope a positive 3.

The work required for the given task of pumping all of the oil out over the top of the tank is 7,500 ft·lb

The known parameters;

The length of the rectangular tank, l = 4 feet

The width of the tank, w = 2 feet

The depth of the tank, h = 5 feet

The weight density of the oil with which the tank is filled, ρ × g = 50 lb/ft³

The unknown parameter

The work required to pump all of the oil out over the top of the tank

Method;

Calculate the force required to lift each slice (layer) of the oil to the top multiplied by the distance, y, the slice moves and summing the result as an integration as follows;

The volume of each slice, [tex]\mathbf{V_i}[/tex] = l × w × dy

The force required to move each slice, [tex]\mathbf{F_i}[/tex] = ρ × g × l × w × dy

The work done, [tex]\mathbf{W_i}[/tex], in moving the slice a distance, y, is given as follows;

[tex]\mathbf{W_i}[/tex] = ρ × g × l × w × y × dy

Therefore, the total work done, W, in pumping all the water located from y = 0, to y = 5, to the top of the tank, is given as follows;

[tex]\mathbf{W = \int\limits^5_0 {(\rho \times g \times l \times w \times y) } \, dy}[/tex]

Therefore;

W = (ρ × g × l × w × y²)/2

Plugging in the values, gives;

W = (50 lb/ft³ × 4 ft. × 3 ft. × (5 ft.)²)/2 = 7,500 ft·lb

The work required to pump all of the oil out over the top of the tank, W = 7,500 ft·lb.

Learn more about the use of integration to calculate the amount of work required for a given task here;

https://brainly.com/question/14318035

Answer:

they are parallel lines so have the same slope.   Difference would be the where they intersect the x and y axis

Step-by-step explanation:

Answer: D

Step-by-step explanation:

D

Answer:

D

Step-by-step explanation:

it is between 0 and 1

Answer:

$465.6 should be budgeted.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean $440 and standard deviation $20.

This means that [tex]\mu = 440, \sigma = 20[/tex]

How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1?

The 100 - 10 = 90th percentile should be budgeted, which is X when Z has a p-value of 0.9, so X when Z = 1.28. Then

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.28 = \frac{X - 440}{20}[/tex]

[tex]X - 440 = 1.28*20[/tex]

[tex]X = 465.6[/tex]

$465.6 should be budgeted.

Answer:

Let the number of courses that are worth 3 credits each be x and those worth 4 credits be y. With the given information, you can write the following equations:  

x + y = 48

3x + 4y = 155

You can solve the above equations by method of elimination/substitution

x + y = 48 ⇒ x = 48 - y (Now, substitution this equation into 3x + 4y = 155)

3(48 - y) + 4y = 155

144 -3y + 4y = 155

y + 144 = 155

y = 11

Now plug this solution back into x = 48 - y  

x = 48 - 11 = 37  

Check work (by plugging the solutions back into the 3x + 4y and see if it's equal to 155):

3(37) + 4(11) = 155

Answer: There are 37 of the 3-credit course and 11 of the 4-credit course

Answer:

[tex]0.04 = 4 \div 100 [/tex]

Answer:

2^(1/6) (cos(-pi/12)+i sin(-pi/12))

2^(1/6) (cos(3pi/12)+i sin(3pi/12))

2^(1/6) (cos(7pi/12)+i sin(7pi/12))

2^(1/6) (cos(11pi/12)+i sin(11pi/12))

2^(1/6) (cos(5pi/4)+i sin(5pi/4))

2^(1/6) (cos(19pi/12)+i sin(19pi/12))

Step-by-step explanation:

Let's convert to polar form.

-2i=2(cos(A)+i sin(A) )

There is no real part so cos(A) has to be zero and since we want -2 and we already have 2 then we need sin(A)=-1 so let's choose A=-pi/2.

So z=2(cos(-pi/2)+i sin(-pi/2)).

There are actually infinitely many ways we can write this polar form which we will need.

z=2(cos(-pi/2+2pi k)+i sin(-pi/2+2pi k))

where k is an integer

Now let's find the 6 6th roots or z.

2^(1/6) (cos(-pi/12+2pi k/6)+i sin(-pi/12+2pi k/6))

Reducing

2^(1/6) (cos(-pi/12+pi k/3)+i sin(-pi/12+pi k/3))

Plug in k=0,1,2,3,4,5 to find the 6 6th roots.

k=0:

2^(1/6) (cos(-pi/12+pi (0)/3)+i sin(-pi/12+pi (0)/3))

=2^(1/6) (cos(-pi/12)+i sin(-pi/12))

k=1:

2^(1/6) (cos(-pi/12+pi/3)+i sin(-pi/12+pi/3))

2^(1/6) (cos(3pi/12)+i sin(3pi/12))

k=2:

2^(1/6) (cos(-pi/12+2pi/3)+i sin(-pi/12+2pi/3))

2^(1/6) (cos(7pi/12)+i sin(7pi/12))

k=3:

2^(1/6) (cos(-pi/12+3pi/3)+i sin(-pi/12+3pi/3))

2^(1/6) (cos(11pi/12)+i sin(11pi/12))

k=4:

2^(1/6) (cos(-pi/12+4pi/3)+i sin(-pi/12+4pi/3))

2^(1/6) (cos(15pi/12)+i sin(15pi/12))

2^(1/6) (cos(5pi/4)+i sin(5pi/4))

k=5:

2^(1/6) (cos(-pi/12+5pi/3)+i sin(-pi/12+5pi/3))

2^(1/6) (cos(19pi/12)+i sin(19pi/12))

Answer:

yip that's all

Step-by-step explanation:

not more than 276 means less or equal to 276,

46 × y ≤ 276

y ≤ 6

Answer:

46y<276

Step-by-step explanation:

no more than means less than or equal to.

Answer:

right scalene

Step-by-step explanation:

Since all three sides have different lengths , this is a scalene triangle

(isosceles means two sides have the same lengths and equilateral means all three sides have the same length)

We have a right angle indicated by the box in the corner

the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.

The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.

In terms of hyperbola, F1F2=2c, c=20.

The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.

Use formula c^2=a^2+b^2c

2

=a

2

+b

2

to find b:

\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}

(20)

2

=(18)

2

+b

2

,

b

2

=400−324=76

.

The branches of hyperbola go in y-direction, so the equation of hyperbola is

\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1

b

2

y

2

−

a

2

x

2

=1 .

Substitute a and b:

\dfrac{y^2}{76}- \dfrac{x^2}{324}=1

76

y

2

−

324

x

2

=1 .

Answer:

team a and team c scored 74 points which is 48 points more than team b, scoring 26 points.

Step-by-step explanation:

Answer:

The AP is 1, 11/2, 10, 29/2, 19, ....

Step-by-step explanation:

Let the first term be a and d be the common difference of the arithmetic progression.

ATQ, a+2d+a+6d=38, 2a+8d=38 and a+8d=37. Solving this, we will get a=1 and d=9/2. The AP is 1, 11/2, 10, 29/2, 19, ....

Answer:

Just to recap, an equation has no solution when it results in an incorrect "equation".

For  example:

Equation: x+3 = x+4

Subtract x: 3 = 4???

But clearly, 3 is not equal to 4, so this equation has NO SOLUTION.

Now onto our problem:

13y+2-2y = 10y+3-y

11y+2 = 9y+3

2y=1

y=1/2

9(3y+7)-2 = 3(-9y+9)

27y+61 = -27y+27

54y = -34

y = -34/54

32.1y+3.1+2.4y-8.2=34.5y-5.1

34.5-5.1=34.5y-5.1

5.1=5.1

infinite solutions

5(2.2y+3.4) = 5(y-2)+6y

11y+17 = 11y-10

17 = -10??

That's not true, so the option "5(2.2y+3.4) = 5(y-2)+6y" has no solution.

Let me know if this helps

You're looking for a solution in the form

[tex]y(x) = \displaystyle \sum_{n=0}^\infty a_nx^n[/tex]

Differentiating, we get

[tex]y'(x) = \displaystyle \sum_{n=0}^\infty na_nx^{n-1} = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]

[tex]y''(x) = \displaystyle \sum_{n=0}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=1}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n[/tex]

Substitute these for y' and y'' in the differential equation:

[tex]\displaystyle \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n - \sum_{n=0}^\infty (n+1)a_{n+1}x^n = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1)a_{n+2}-(n+1)a_{n+1}\bigg)x^n = 0[/tex]

Then the coefficients of y are given by the recurrence

[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\frac{a_{n+1}}{n+2}&\text{for }n\ge0\end{cases}[/tex]

or

[tex]a_n = \dfrac{a_{n-1}}n[/tex]

But we cannot assume that [tex]a_0[/tex] and [tex]a_1[/tex] depend on each other; we can only guarantee that the recurrence holds for n ≥ 1, so that

[tex]a_2=\dfrac{a_1}2 \\\\ a_3=\dfrac{a_2}3=\dfrac{a_1}{3\times2} \\\\ a_4=\dfrac{a_3}4=\dfrac{a_1}{4\times3\times2} \\\\ \vdots \\\\ a_n=\dfrac{a_1}{n!}[/tex]

So in the power series solution, we split off the constant term and we're left with

[tex]y(x) = a_0 + a_1 \displaystyle \sum_{n=1}^\infty \frac{x^n}{n!}[/tex]

so that the fundamental solutions are

[tex]y_1=1[/tex]

and

[tex]y_2=x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots[/tex]

Answer:

Probability-Between    .8574 = 85.74%

Step-by-step explanation:

Z1=-2.14 Z2=1.14

*x-1 35

*x-2  58

*µ 50

*σ 7

9514 1404 393

Answer:

  C. a line that shows the set of all solutions to the equation.

Step-by-step explanation:

Any graph shows the set of all solutions to the equation being graphed.

The graph of a linear function is a straight line.

9514 1404 393

Answer:

  C. EGD

Step-by-step explanation:

A major arc is typically named using the end points and a point on the arc. Here, the end points are E and D, and points on the major arc include C, G, and F. The major arc ED could be named any of

Of course, the reverse of any of these names could also be used: DCE, DGE, DFE.

Answer:

[tex]\displaystyle AL = 2sinh(10)[/tex]

General Formulas and Concepts:

Pre-Calculus

Calculus

Differentiation

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

Derivative Rule [Chain Rule]:                                                                                 [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Basic Power Rule:

Exponential Differentiation

Integration

Parametric Integration

Vector Value Functions

Arc Length Formula [Vector]:                                                                               [tex]\displaystyle AL = \int\limits^b_a {\sqrt{[i'(t)]^2 + [j'(t)]^2 + [k'(t)]^2}} \, dt[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \vec{r}(t) = <10\sqrt{2}t , e^{10t} , e^{-10t} >[/tex]

Interval [0, 1]

Step 2: Find Arclength

Topic: AP Calculus BC (Calculus I + II)

Unit: Vector Value Functions

Book: College Calculus 10e

Step-by-step explanation:

a) The volume of the prism is

[tex]V = (n^2 - 1)×(n^2 - 1)×(5n)[/tex]

[tex]\:\:\:\:= (n^4 - 2n^2 + 1)(5n)[/tex]

[tex]\:\:\:\:=5n^5 - 10n^3 + 5n[/tex]

b) If the dimensions L of the prism are tripled, the new volume will be

[tex]V' = (3L)^3 = 27L^3 = 27V[/tex]

so it will increase by a factor of 27.

Answer:

The student is completely incorrect because there is no solution to this inequality.

Answer:

D on edge

Step-by-step explanation:

Answer:

Step-by-step explanation:

87°32' = 86°92'

(86°92')/2 = 43°46'

B = 13/(16cos(43°46')) = 1.125

Answer:

Step-by-step explanation:

Answer:

7 groups can be made each with five people :)

Answer:

h = 6[tex]\sqrt{3}[/tex]

Step-by-step explanation:

The given is the special right triangle with angle measures : 90-60-30

and the side lengths for the given angles are represented by :

2a-a[tex]\sqrt{3}[/tex]-a

the side length that sees 60 degrees is represented by a[tex]\sqrt{3}[/tex] (h in this case)

the area of a triangle is calculated by multiplying height and base and that is divided by 2

a[tex]\sqrt{3}[/tex]*a/2 = 18[tex]\sqrt{3}[/tex] multiply both sides by 2

a^2[tex]\sqrt{3}[/tex] = 36[tex]\sqrt{3}[/tex] divide both sides by [tex]\sqrt{3}[/tex]

a^2 = 36 find the roots for both sides

a = 6

since h sees angle measure 60 and is represented by a[tex]\sqrt{3}[/tex]

h = 6[tex]\sqrt{3}[/tex]