Which equation has a graph that passes through the origin and has a slope of -10?​

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

Answer:

Base = 21 while Height = 16

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 )

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:

182/3,3 8/3, 152/3

Step-by-step explanation:

a+b+c=124

a trừ c=  10

4b=c

Answer:

a=29,b=79,c=19

Step-by-step explanation:

a=c+10

b=4c

=> a+b+c=c+10+4c+c=124

=> c=19

=> a= 29, b=79

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.

Answer:

a) 75

b) 4.33

c) 0.75

d) [tex]3.2 \times 10^{-13}[/tex] probability that no one in a simple random sample of 100 young adults owns a landline

e) [tex]6.2 \times 10^{-61}[/tex] probability that everyone in a simple random sample of 100 young adults owns a landline.

f) Binomial, with [tex]n = 100, p = 0.75[/tex]

g) [tex]4.5 \times 10^{-8}[/tex] probability that exactly half the young adults in a simple random sample of 100 do not own a landline.

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they do not own a landline, or they do. The probability of an young adult not having a landline is independent of any other adult, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.

This means that [tex]p = 0.75[/tex]

(a) On average, how many young adults do not own a landline in a random sample of 100?

Sample of 100, so [tex]n = 100[/tex]

[tex]E(X) = np = 100(0.75) = 75[/tex]

(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100(0.75)(0.25)} = 4.33[/tex]

(c) What is the proportion of young adults who do not own a landline?

The estimation, of 75% = 0.75.

(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?

This is P(X = 100), that is, all do not own. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 100) = C_{100,100}.(0.75)^{100}.(0.25)^{0} = 3.2 \times 10^{-13}[/tex]

[tex]3.2 \times 10^{-13}[/tex] probability that no one in a simple random sample of 100 young adults owns a landline.

(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?

This is P(X = 0), that is, all own. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{100,0}.(0.75)^{0}.(0.25)^{100} = 6.2 \times 10^{-61}[/tex]

[tex]6.2 \times 10^{-61}[/tex] probability that everyone in a simple random sample of 100 young adults owns a landline.

(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?

Binomial, with [tex]n = 100, p = 0.75[/tex]

(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?

This is P(X = 50). So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 50) = C_{100,50}.(0.75)^{50}.(0.25)^{50} = 4.5 \times 10^{-8}[/tex]

[tex]4.5 \times 10^{-8}[/tex] probability that exactly half the young adults in a simple random sample of 100 do not own a landline.

Answer:

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

Answer:

y > -1

Step-by-step explanation:

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

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]

Answer:

[tex]5xy[/tex]

Step-by-step explanation:

[tex]\mathrm{Factor\:}:5x^2y[/tex]

[tex]5\cdot \:x\cdot \:x\cdot \:y[/tex]

[tex]\mathrm{Factor\:}:10xy^2[/tex]

[tex]2\cdot \:5\cdot \:x\cdot \:y\cdot \:y[/tex]

Common factor:-

[tex]5\cdot \:x\cdot \:y[/tex]

OAmalOHopeO

Answer: D

Step-by-step explanation:

D

Answer:

D

Step-by-step explanation:

it is between 0 and 1

Answer:

[tex]x=2[/tex]

Step-by-step explanation:

[tex]\frac{3x-2}{7}-\frac{5x-8}{4}=\frac{1}{14}[/tex]

In order to factor an integer, we need to divide it by the ascending sequence of primes 2, 3, 5.

The number of times that each prime divides the original integer becomes its exponent in the final result.

In here,  Prime number 2 to the power of 2 equals 4.

[tex]\frac{3x-2}{7}-\frac{5x-8}{2^{2} }=\frac{1}{14}[/tex]

First, We need to add fractions-

Rule:-

[tex]\frac{A}{B} +\frac{C}{D} =\frac{\frac{LCD}{B}+\frac{LCD}{D}C }{LCD}[/tex]

LCD = [tex]7 \cdot 2^{2}[/tex]

[tex]\frac{4(3x-2)+7(-(5x-8))}{7*2^{2} } =\frac{1}{14}[/tex]

[tex]x=2[/tex]

OAmalOHopeO

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

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.

Answer:

[tex]x = y^3+5y[/tex]

Step-by-step explanation:

Complete question

[tex]\frac{x - 5y}{y^3} - 1=0\\[/tex]

Required

Solve for x

We have:

[tex]\frac{x - 5y}{y^3} - 1=0[/tex]

Collect like terms

[tex]\frac{x - 5y}{y^3} = 1[/tex]

Multiply through by [tex]y^3[/tex]

[tex]x - 5y = y^3[/tex]

Make x the subject

[tex]x = y^3+5y[/tex]

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:

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

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 right answer is:

(a) 0.1456

(b) 18.125, 69.1202, 8.3139

Step-by-step explanation:

Given:

N = 24

n = 5

r = 7

The improperly drilled gearboxes "X".

then,

⇒ [tex]P(X) = \frac{\binom{7}{x} \binom {17}{5-x}}{\binom{24}{5}}[/tex]

(a)

P (all gearboxes fit properly) = [tex]P(x=0)[/tex]

                                               = [tex]\frac{\binom{7}{0} \binom{17}{5}}{\binom{24}{5}}[/tex]

                                               = [tex]0.1456[/tex]

(b)

According to the question,

[tex]X = 91+5[/tex]

Mean will be:

⇒ [tex]\mu = E(x)[/tex]

       [tex]=E(91+5)[/tex]

       [tex]=9E(1)+5[/tex]

       [tex]=9.\frac{nr}{N}+5[/tex]

       [tex]=9.\frac{5.7}{24} +5[/tex]

       [tex]=18.125[/tex]

Variance will be:

⇒ [tex]\sigma^2=Var(X)[/tex]

         [tex]=V(9Y+5)[/tex]

         [tex]=81.V(Y)[/tex]

         [tex]=81.n.\frac{r}{N}.\frac{N-r}{N}.\frac{N-n}{N-1}[/tex]

         [tex]=81.5.\frac{7}{24}.\frac{24-7}{24}.\frac{24-5}{24-1}[/tex]

         [tex]=69.1202[/tex]

Standard deviation will be:

⇒ [tex]\sigma = \sqrt{69.1202}[/tex]

       [tex]=8.3139[/tex]      

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:

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

Answer:

D on edge

Step-by-step explanation:

Answer:

Check the image

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:

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:

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

Step-by-step explanation:

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))

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

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 .