Two jets leave an air base at the same time and travel in opposite directions. One jet travels 100 miles an hour faster than the other. If the two jets are 3924 miles apart after3 hours, what is the rate of each jet?

Answer: The distance between A and B is 300 miles.

Step-by-step explanation:

Hi, to solve this problem we have to analyze the information given.

We know that when he was 15 miles away from point B, he was traveling at 60mph. if we apply the formula : time= distance /speed;

So, he traveled that distance in 15 minutes.

That means that he returned to point A in 3.75 hours (4 hours -15minutes) at a speed of 80 mph.

Applying the formula again to calculate the distance:

Answer:

Range of the average number of tours is between 150 and 200 including 150 and 200.

Step-by-step explanation:

Given:

The profit function is modeled as:

[tex]p(x)=-2x^2+700x-10000[/tex]

The profit is at least $50,000.

So, as per question:

[tex]p(x)\geq50000\\-2x^2 + 700x-10000\geq 50000\\-2x^2+700x-10000-50000\geq 0\\-2x^2+700x-60000\geq 0\\\\\textrm{Dividing by 2 on both sides, we get}\\\\-x^2+350x-30000\geq 0[/tex]

Now, rewriting the above inequality in terms of its factors, we get:

[tex]-1(x-150)(x-200)\geq 0\\(x-150)(x-200)\leq 0[/tex]

Now,

[tex]x<150,(x-150)(x-200)>0\\x>200,(x-150)(x-200)>0\\For\ 150\leq x\leq200,(x-150)(x-200)\leq 0\\\therefore x=[150,200][/tex]

Therefore, the range of the average number of tours he must arrange per day to earn a monthly profit of at least $50,000 is between 150 and 200 including 150 and 200.

Answer:

About 1.65 meals per hour

Step-by-step explanation:

9 days of work in May * 8 hours per day = 72 hours of work in May per employee

72 hours * 13 employees = 936 hours worked for all employees in May

1540 meals in May/ 936 hours worked for all employees in May = about 1.65 meals per hour

Answer: she runs 132 feets in each lap and 660 feets in 5 laps

Step-by-step explanation:

Stella runs laps around the edge of the yard. This means she runs round the entire shape of the yard.

Miss bridgeyard is 24 ft by 42 ft. This means that the length and width of Miss bridgeyard are not the same. Therefore, Miss bridgeyard has the shape of a rectangle. The distance that stella covers in one lap is the perimeter of the rectangular Miss bridgeyard.

Perimeter of a rectangle = 2( L + W )

If length,L = 42 feets and

Width ,W = 24 feets, the perimeter would be

2(42+24)/= 2×66 = 132 feets

She runs a distance of 132 feets in one lap.

Distance in 5 laps would be

132 × 5 = 660 feets

Answer:

Option D.

Step-by-step explanation:

Given information: KLMN is parallelogram, K(7,7), L(5,3), M(1,1) and N(3,5).

Diagonals of a parallelogram bisect each other.

If diagonals of a parallelogram are perpendicular to each other then the parallelogram is a rhombus.

If a line passes through two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex], then the rate of change is

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Slope of KM is

[tex]m_1=\frac{1-7}{1-7}=1[/tex]

Slope of LN is

[tex]m_2=\frac{5-3}{3-5}=-1[/tex]

The product of slopes of two perpendicular lines is -1.

Find the product of slopes.

[tex]m_1\cdot m_2=1\cdot (-1)=-1[/tex]

The product of slopes of KM and NL is -1. It means diagonals are perpendicular and KLMN is a rhombus.

Therefore, the correct option is D.

The probability of drawing two pink marbles from a bag with replacement can be calculated using the multiplication rule of independent events as P(pink and pink) = P(pink) x P(pink), where P(pink) is the probability of drawing a pink marble.

The subject of this problem is probability in Mathematics, particularly with replacement. The scenario involves drawing two pink marbles from a bag with replacement. This means that after the first marble is drawn, it is put back into the bag before the second one is drawn.

The probability of drawing a pink marble on the first draw is calculated by dividing the number of pink marbles by the total number of marbles. Similarly, the probability of drawing a pink marble on the second draw, with replacement, stays the same because the total number of marbles in the bag is the same as in the first draw.

To calculate the final probability, you use the multiplication rule of independent events (events where the outcome of the first event does not affect the outcome of the second event). According to this rule, the probability of both events happening is the product of the probabilities of each event. Hence, if P(pink) represents the probability of drawing a pink marble, the probability of drawing two pink marbles (with replacement) is P(pink and pink) = P(pink) x P(pink).

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Answer: x = 2, y = -4 , z = 1

Step-by-step explanation:

-x + y + 3z = -3 - - - - - - - - - - 1

x - 2y - 2z = 8 - - - - - - -- - - - 2

3x - y - 4z = 6 - - - - - - - - - - - - - 3

Let us use the method of elimination

We would add equation 1 to equation 2. It becomes

-y+z= 5 - - - - - - - - - - - - - - -4

Multiply equation 2 by 3 and equation 3 by 1

3x - 6y -6z = 24- - - - - - - - - - 5

3x - y - 4z = 6 - - - - - - - - - - - - -6

Subtracting equation 6 from equation 5

-5y -2z = 18 - - - - - - - - - - 7

Substituting z = 5 + y into equation 7, it becomes

-5y -2(5+y) = 18

-5y -10-2y = 18

-5y -2y = 18+10

-7y = 28

y = 28/-7 = -4

z = 5 + y

z = 5 -4 = 1

We would substitute y = -4 and z = 1 into equation 2

It becomes

x - 2×-4 - 2×1 = 8

x+8-2 = 8

x +6 = 8

x = 8-6 = 2

x = 2, y = -4 , z = 1

Let us check by substituting the value into equation 1

-x + y + 3z = -3

-2-4+ 3= -3

-6 + 3 = -3

-3 = -3

Answer:

  11 miles

Step-by-step explanation:

After the first hour, they walk 3 more hours at 2 miles per hour. So, the total distance is ...

  5 mi + (3 h)(2 mi/h) = 5 mi + 6 mi = 11 mi

The blue team will walk 11 miles in 4 hours.

Answer:

x  =  140  ft

w = 70 ft

A(max)  =  9800 ft²

Step-by-step explanation:

We have:

280 feet of fence to enclose three sides of a rectangular area

perimeter of the rectangle ( 3 sides ) is

p  =  L  =  x  +2w       w   = (L - x ) / 2       w   =  ( 280  -  x ) / 2

where:

x is the longer side

w is the width

A(x,w)  = x*w         ⇒   A(x)  =  x* ( 280 - x ) / 2  ⇒ A(x)  = (280x -x²)/2

Taking derivatives on bth sides of the equation

A´(x)  = ( 280 -2x)*2 /4          A´(x)  = 0      ( 280 -2x)  =  0

280 -2x  = 0     x = 280/2

x  =  140  ft

And   w  = ( 280 - x ) / 2  ⇒  w  =(  280  -140  )/ 2

w = 70 ft

A(max)  =  9800 ft²

Answer:

Step-by-step explanation:

Answer:

x = 8; y = 2.5.

Step-by-step explanation:

As we know , when two complex numbers are equal their real as well as imaginary part are equal.

So comparing on both sides ,

2x = 16   and    5 = 2y

x = 8       and    y = 2.5.

So ,   x = 8; y = 2.5.

Tommy must clean at least 76 pools, in addition to mowing 41 lawns, to earn the $1500 he needs to buy a car.

Tommy has a summer job mowing lawns at $20 per lawn and cleaning pools at $9 per pool to save up for a car. To determine how many pools he needs to clean to reach his goal of $1500, we need to calculate his earnings from mowing lawns first and then see how much more he needs to earn from pool cleaning.

First, we calculate Tommy's lawn mowing earnings:
41 lawns × $20 per lawn = $820

After mowing lawns, Tommy will need an additional $1500 - $820 to buy the car. This difference is $680.

Next, to find out how many pools Tommy needs to clean, we divide the remaining amount by the amount he earns per pool:
$680 ÷ $9 per pool ≈ 75.56

Since Tommy can't clean a fraction of a pool, he will need to clean at least 76 pools to make enough money to buy the car. Therefore, the answer is 76 pools.

Option C

Domain: {-5, 1, 3, 7); Yes, it is a function

The given relation is :-

{(-5, 2), (7, 7), (3,6), (1, 7)}

It is of form (x, y)

The domain is the set of all the values of  "x" . The range is the set of all the values of  "y"

We need to find domain :-

The domain is the set of all possible x-values which will make the function "work", and will output real y-values.

Domain is the set of "x" values , in the given relation these are:-

Domain is :-  { -5, 7, 3, 1}

And Range is :- {2, 7, 6, 7}

Since there is one value of y for every value of  "x"

A relation from a set X to a set Y is called a function if each element of X is related to exactly one element in Y.

Hence, the relation is a function

The option C) is correct  

Answer:

Option C) The symbol y represents the predicted value of price.

Step-by-step explanation:

We are given the following in the question:

We find a regression equation with x representing the ratings and y representing price.

The equation has a slope of 130 and a y-intercept of 350.

Comparing with the slope intercept form:

[tex]y = mx + c\\\text{where m is the slope and c is the y intercept}[/tex]

Thus, we can write the equation as:

[tex]y = 130x + 350[/tex]

Here, y is the predicted variable that is the price, c is the price of hotel when a rating of 0 is given.

Thus, symbol y represents:

C) The symbol y represents the predicted value of price.

The equation of the regression line is y = 130x + 350. The symbol 'y' in this equation represents the predicted price of a hotel based on its rating.

The equation of a line in slope-intercept form is given by y = mx + b, where m is the slope of the line and b is the y-intercept. In this case, we have been provided with a slope of 130 and a y-intercept of 350. Therefore, the equation of the regression line is y = 130x + 350. This equation is the model, created using regression analysis, predicting the price of hotels based on their ratings.

The symbol y in this situation refers to the predicted value of price for a hotel depending on its rating. Hence, the correct answer for part (b) is 'C) The symbol y represents the predicted value of price'.

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Answer:

Part (A): it would take 2 seconds to reach maximum height of 264 foot.

Part (B): Ball will hit the ground in about 6.1 seconds

Part (C): S(0) represents the initial height of the base ball or the baseball was thrown  at a height of 200 ft.

Step-by-step explanation:

Consider the provided function.

[tex]s(t)=-16t^2+64t+200[/tex]

Part (A) After how many seconds does the ball reach it's maximum height? What is the maximum height?

The coefficient of t² is a negative number, so the graph of the above function is a downward parabola.

From the given function a=-16, b=64 and c=200

The downward parabola attain the maximum height at the x coordinate of the vertex. [tex]x=\frac{-b}{2a}[/tex]

Substitute the respectives.

[tex]x=\frac{-64}{2(-16)}=2[/tex]

Substitute x=2 in the provided equation.

[tex]s(t)=-16(2)^2+64(2)+200=264[/tex]

Hence, it would take 2 seconds to reach maximum height of 264 foot.

Part (B) How many seconds does it take until the ball finally hits the ground?

Substitute s(t)=0 in the provided equation.

[tex]-16t^2+64t+200=0[/tex]

Use the formula [tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex] to find the solutions of the quadratic equation.

[tex]t=\frac{-64+\sqrt{64^2-4\left(-16\right)200}}{2\left(-16\right)}\\t=\pm\frac{4+\sqrt{66}}{2}\\t\approx-2.1\ or\ 6.1[/tex]

Reject the negative value as time can't be a negative number.

Hence, ball will hit the ground in about 6.1 seconds

Part (C) Find s(0) and describe what this means.

Substitute x=0 in the provide equation.

[tex]s(0)=-16(0)^2+64(0)+200[/tex]

[tex]s(0)=200[/tex]

S(0) represents the initial height of the base ball or the baseball was thrown  at a height of 200 ft.

Part (D) Use your results from parts (a) through (c) to graph the quadratic function.

Use the starting points (0,200), maximum point (2,264) and the end point (6.1,0) in order to draw the graph of the function.

Connect the points as shown in figure.

The required figure is shown below.

Answer:

  v = 2√2i -2√2j

Step-by-step explanation:

  v = ||v||·cos(θ)i +||v||·sin(θ)j

  v = 4cos(315°)i +4sin(315°)j . . . . . . fill in the numbers

  v = 2√2i -2√2j . . . . . . . . . . . . . . . put in desired form

Answer:

a) No

b) No

c) No

d) No

Step-by-step explanation:

Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:

1. u+v∈V

2. u+v=v+u

3. (u+v)+w=u+(v+w).

4. Exist 0∈V such that u+0=u

5. For each u∈V exist −u∈V such that u+(−u)=0.

6. if c is an escalar and u∈V, then cu∈V

7. c(u+v)=cu+cv

8. (c+d)u=cu+du

9. c(du)=(cd)u

10. 1u=u

let's check each of the properties for the respective operations:

Let [tex]u=(u_1,u_2,u_3), v=(v_1,v_2,v_3)[/tex]

Observe that  

1. u+v∈V

2. u+v=v+u, because the adittion of reals is conmutative

3. (u+v)+w=u+(v+w). because the adittion of reals is associative

4. [tex](u_1,u_2,u_3)+(0,0,0)=(u_1+0,u_2+0,u_3+0)=(u_1,u_2,u_3)[/tex]

5. [tex](u_1,u_2,u_3)+(-u_1,-u_2,-u_3)=(0,0,0)[/tex]

then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.

a)

6. [tex]c(u_1,u_2,u_3)=(cu_1,u_2,cu_3)\in V[/tex]

7.

[tex]c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),u_2+v_2,c(u_3+v_3))\\=(cu_1+cv_1,u_2+v_2,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv[/tex]

8.

[tex](c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,u_2,(c+d)u_3)=\\=(cu_1+du_1,u_2,cu_3+du_3)\neq (cu_1+du_1,2u_2,cu_3+du_3)=cu+du[/tex]

Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(ax,y,az)[/tex]

b)  6. [tex]c(u_1,u_2,u_3)=(cu_1,0,cu_3)\in V[/tex]

7.

[tex]c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),0,c(u_3+v_3))\\=(cu_1+cv_1,0,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv[/tex]

8.

[tex](c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,0,(c+d)u_3)=\\=(cu_1+du_1,0,cu_3+du_3)=(cu_1,0,cu_3)+(du_1,0,du_3) =cu+du[/tex]

9.

[tex]c(du)=c(d(u_,u_2,u_3))=c(du_1,0,du_3)=(cdu_1,0,cdu_3)=(cd)u[/tex]

10

[tex]1u=1(u_1,u_2,u3)=(1u_1,0,1u_3)=(u_1,0,u_3)\neq(u_1,u_2,u_3)[/tex]

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(ax,0,az)[/tex]

c) Observe that [tex]1u=1(u_1,u_2,u3)=(0,0,0)\neq(u_1,u_2,u_3)[/tex]

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(0,0,0)[/tex].

d)  Observe that [tex]1u=1(u_1,u_2,u3)=(2*1u_1,2*1u_2,2*1u_3)=(2u_1,2u_2,2u_3)\neq(u_1,u_2,u_3)=u[/tex]

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product [tex]a(x,y,z)=(2ax,2ay,2az)[/tex].

None of the given definitions make ( V ) a vector space because they fail to satisfy the necessary vector space axioms.

To determine whether ( V ) is a vector space under the given definitions of scalar multiplication, we need to check if each definition satisfies the vector space axioms.

Definition (a): [tex]\( a(x,y,z) = (ax,y,az) \)[/tex]

Additive Identity: Yes, [tex]\( 1(x,y,z) = (x,y,z) \)[/tex].

Scalar Distributive (over vectors): [tex]\( a((x_1,y_1,z_1)+(x_2,y_2,z_2)) = a(x_1+x_2, y_1+y_2, z_1+z_2) = (a(x_1+x_2), y_1+y_2, a(z_1+z_2)) \).[/tex]

Scalar Distributive (over scalars): [tex]\( (a+b)(x,y,z) = ((a+b)x,y,(a+b)z) = (ax+bx,y,az+bz) \).[/tex]

Associative: [tex]\( a(b(x,y,z)) = a(bx,y,bz) = (abx,y,abz) = (ab)(x,y,z) \).[/tex]

Conclusion: Does not satisfy scalar distributive over vectors.

Definition (b): [tex]\( a(x,y,z) = (ax,0,az) \)[/tex]

Additive Identity: Yes, \( 1(x,y,z) = (x,0,z) \).

Scalar Distributive (over vectors): [tex]\( a((x_1,y_1,z_1)+(x_2,y_2,z_2)) = a(x_1+x_2,y_1+y_2,z_1+z_2) = (a(x_1+x_2),0,a(z_1+z_2)) = (ax_1+ax_2,0,az_1+az_2) \)[/tex]

Scalar Distributive (over scalars): [tex]\( (a+b)(x,y,z) = ((a+b)x,0,(a+b)z) = (ax+bx,0,az+bz) \).[/tex]

Associative: [tex]\( a(b(x,y,z)) = a(bx,0,bz) = (abx,0,abz) = (ab)(x,y,z) \).[/tex]

Conclusion: Does not satisfy scalar distributive over vectors.

Definition (c): [tex]\( a(x,y,z) = (0,0,0) \)[/tex]

Additive Identity: Yes, [tex]\( 1(x,y,z) = (0,0,0) \).[/tex]

Scalar Distributive (over vectors): [tex]\( a((x_1,y_1,z_1)+(x_2,y_2,z_2)) = (0,0,0) \).[/tex]

Scalar Distributive (over scalars): [tex]\( (a+b)(x,y,z) = (0,0,0) \).[/tex]

Associative: [tex]\( a(b(x,y,z)) = (0,0,0) \).[/tex]

Conclusion: Does not satisfy any of the scalar distributive properties.

Definition (d): [tex]\( a(x,y,z) = (2ax,2ay,2az) \)[/tex]

Additive Identity: No, [tex]\( 1(x,y,z) = (2x,2y,2z) \).[/tex]

Scalar Distributive (over vectors): [tex]\( a((x_1,y_1,z_1)+(x_2,y_2,z_2)) = a(x_1+x_2, y_1+y_2, z_1+z_2) = (2a(x_1+x_2), 2a(y_1+y_2), 2a(z_1+z_2)) = (2ax_1+2ax_2, 2ay_1+2ay_2, 2az_1+2az_2) \).[/tex]

Scalar Distributive (over scalars): [tex]\( (a+b)(x,y,z) = (2(a+b)x, 2(a+b)y, 2(a+b)z) = (2ax+2bx, 2ay+2by, 2az+2bz) \).[/tex]

Associative: [tex]\( a(b(x,y,z)) = a(2bx,2by,2bz) = (4abx,4aby,4abz) \neq (2ab)(x,y,z) \).[/tex]

Conclusion: Does not satisfy scalar multiplication associativity.

For this case we have a function of the form [tex]y = f (x)[/tex], where:

[tex]f (x) = 10-x[/tex]

We must find the value of the function when:

[tex]x = -2,0,2[/tex]

[tex]f (-2) = 10 - (- 2) = 10 + 2 = 12[/tex]

[tex]f (0) = 10-0 = 10[/tex]

[tex]f (2) = 10-2 = 8[/tex]

Thus, we have that the function has a value of [tex]y = {12,10,8}[/tex] when [tex]x = {- 2,0,2}[/tex]

Answer:

[tex]y = {12,10,8}[/tex]

Answer:

a) d(x)=[tex]\sqrt{17x^{2} -12x+5}[/tex]

b)d'(x)=[tex]\frac{17x-6}{\sqrt{17x^{2} -12x+5} }[/tex]

c)The critical point is x=[tex]\frac{6}{17}[/tex]

d)Closest point is ([tex]\frac{6}{17}[/tex],[tex]\frac{7}{17}[/tex]

Step-by-step explanation:

We are given the line

[tex]y=4x-1[/tex]

Let a point Q([tex]x,y[/tex]) lie on the line.

Point P is given as P(2,0)

By distance formula, we have the distance D between any two points

A([tex]x_{1},y_{1}[/tex]) and B([tex]x_{2},y_{2}[/tex]) as

D=[tex]\sqrt{(x_{1}-x_{2})^2 + (y_{1}-y_2)^2}[/tex]

Thus,

d(x)=[tex]\sqrt{(x-2)^2+(y-0)^2}[/tex]

But we have, [tex]y=4x-1[/tex]

So,

d(x)=[tex]\sqrt{(x-2)^2+(4x-1)^2}[/tex]

Expanding,

d(x)=[tex]\sqrt{17x^2-12x+5}[/tex]  - - - (a)

Now,

d'(x)= [tex]\frac{\frac{d}{dx} (17x^2-12x+5)}{2(\sqrt{17x^2-12x+5}) }[/tex]

i.e.

d'(x)=[tex]\frac{17x-6}{\sqrt{17x^{2} -12x+5} }[/tex] - - - (b)

Now, the critical point is where d'(x)=0

⇒ [tex]\frac{17x-6}{\sqrt{17x^{2} -12x+5} }[/tex] =0

⇒[tex]x=\frac{6}{17}[/tex]    - - - (c)

Now,

The closest point on the given line to point P is the one for which d(x) is minimum i.e. d'(x)=0

⇒[tex]x=\frac{6}{17}[/tex]

as [tex]y=4x-1[/tex]

⇒y=[tex]\frac{7}{17}[/tex]

So, closest point is ([tex]\frac{6}{17},\frac{7}{17}[/tex])   - - -(d)

Answer:

(x-4)² - 11

Step-by-step explanation:

You find half of 8 which is 4 and half of x² which is x. this forms (x - 4).

However this would expand as

x²-8x+16 which isn't the expression. So to make it 5, you have to take away 11 leaving you with

(x-4)²-11

Answer:

(x - 4)^2 - 11.

Step-by-step explanation:

x^2 - 8x + 5

Note that x^2 - 8x = (x - 4)^2 - 16 so we have:

(x - 4)^2 - 16 + 5

= (x - 4)^2 - 11.

To get (x - 4)^2 - 16 I used the identity:

x^2 + ax = ( x + a/2)^2 - a^2/4    with a = -8.

Answer:

The answer is b.)  -5.2 degrees

Step-by-step explanation:

to find the mean of this problem you have to add all numbers and then divide it by how many numbers there is.

so you have to add  -42+ -17+14+-4+23 and that'll equal -26

so you take -26 and divide it by 5 because thats how many numbers their are to divide

-26 divided by 5 is (-5.2)

Answer:

The probability is 0.7946

Step-by-step explanation:

Let's call F the event that 16 of the 30 tortillas are failures, A the event that you choose a type 1 part and B the event that you choose a type 2 part.

So, the probability that you picked a Type 1 part given that 16 of the 30 tortillas are failures is calculated as:

P(A/F)=P(A∩F)/P(F)

Where P(F) = P(A∩F) + P(B∩F)

Then, the probability that a type 1 part created 16 failures can be calculated using the binomial distribution as:

[tex]P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}[/tex]

Where x is the number of failures, n is the total number of tortillas and p is the failure rate, so:

[tex]P(16)=\frac{30!}{16!(30-16)!}*0.4^{16}*(1-0.4)^{30-16}=0.0489[/tex]

Therefore, The probability P(A∩F) that you choose a type 1 part and this part created 16 square tortillas is:

(0.3)(0.0489) = 0.0147

Because 0.3 is the probability to choose a type 1 part and 0.0489 is the probability that a type 1 part created 16 square tortillas.

At the same way, the probability that a type 2 part created 16 failures is:

[tex]P(16)=\frac{30!}{16!(30-16)!}*0.75^{16}*(1-0.75)^{30-16}=0.0054[/tex]

Therefore, P(B∩F) is:  (0.7)(0.0054) = 0.0038

Finally, P(F) and P(A/F) are equal to:

P(F) = 0.0147 + 0.0038 = 0.0185

P(A/F) = 0.0147/0.0185 = 0.7946

Answer:

What is the question?????

The total weight of the smaller and the bigger cat Alberto has is 26 1/12 pounds.

A fraction is written in the form of p/q, where q ≠ 0.

Fractions are of two types they are proper fractions in which the numerator is smaller than the denominator and improper fractions where the numerator is greater than the denominator.

Given, Alberto has 2 cats.

The smaller cat weighs 10 3/4 pounds and the larger cat weighs 15 1/3 pounds.

Therefore, The weights of the cats together is the sum of their individual

weights which is,

= (10 3/4 + 15 1/3) pounds.

= (43/4 + 46/3) pounds.

= [(3×43 + 4×46)/12] pounds.

= (129 + 184)/12 pounds.

= 313/12 pounds.

= 26 1/12 pounds.

So, Together the cats weigh 26 1/12 pounds.

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Answer: The correct answer is D). Between 150 and 200; exclusive

Step-by-step explanation:

Given profit function p(x) of a tour operator is modeled by

p(x)=[tex](-2)x^{2} +700x-10000[/tex]

Where, x is the average number of tours he arranges per day.

To find number of tours to arrange per day to get monthly profit of at least 50,000$:

Now, he should make at-least 50000$ profit.

we can write as p(x)>50000$

[tex](-2)x^{2} +700x-10000\geq50000[/tex]

[tex](-2)x^{2} +700x-60000\geq0[/tex]

Roots are x is 150 and 200

(x-150)(x-200)>0

Case 1 : x>150 and x>200

x>150 also satisfy the x>200.

Case2: x<100 and x<200

x<200 also satisfy the x<100

Thus, the common range is 150<x<200

The correct answer is D). Between 150 and 200; exclusive

Answer:  between 150 and 200; inclusive

Step-by-step explanation:

The answer is 'inclusive' NOT 'exclusive.'

Answer:

x =  3.53 ft

y - 3.53 ft

z = 3.53 ft

Step-by-step explanation:

given details

volume = 44 ft^3

let cardboard dimension is x and y and height be z

we know that area of given cardboard without lid is given as

A = xy + 2xy + 2yz

xyz   = 44 ft^3

To minimize area we have

A = xy + 2x (44/xy) + 2y(44/xy)

A = xy + (44/y) + (44/x)

we have

[tex]Ax = y - \frac{44}{x^2}[/tex]

[tex]0 = yx^2 = 44[/tex]................1

[tex]Ay = x - \frac{44}{y^2}[/tex]

[tex]0 = x - \frac{44}{y^2}[/tex]

[tex]xy^2 = 44[/tex] ..............2

from 1 and 2

[tex]yx^2 = xy^2[/tex]

xy(y-x) = 0

xy = 0 or y = x

from geometry of probelem

x ≠ 0 and y ≠ 0

so y = x

x^3 = 44

x =  3.53 ft = y

z = 44/xy = 3.53

To find the dimensions of the box that requires the least amount of cardboard, we need to minimize the surface area of the box. Since it doesn't have a lid, the box will have an open top. Let's call the length of the box 'x' and the width and height 'y'. The dimensions of the box that requires the least amount of cardboard are x = 44 ft and y = 0 ft.

To find the dimensions of the box that requires the least amount of cardboard, we need to minimize the surface area of the box. Since it doesn't have a lid, the box will have an open top. Let's call the length of the box 'x' and the width and height 'y'.

The volume of the box is given as 44 ft3, so we have the equation x * y * y = 44.

To minimize the surface area, we can differentiate the surface area function with respect to x or y, set it equal to zero, and solve for the corresponding variable.

Let's differentiate the surface area function with respect to x to find the critical point:

0 = 2y2 + 2xy * dy/dx

Since the box has an open top, the length, x, cannot be zero. Therefore, we can solve the equation 2y2 + 2xy * dy/dx = 0 for dy/dx. This gives us:

dy/dx = -y/x

Now, we can substitute this into the equation for the surface area:

S = x * y2 + 2xy * dy/dx

Simplifying, we get:

S = x * y2 - 2y2

To find the critical point, we set the derivative equal to zero:

0 = y2 - 2y2

0 = -y2

Since y is squared, it cannot be negative. Therefore, the only possible critical point is when y is zero, which means the dimensions of the box are x = 44 ft and y = 0 ft.

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(i) The length of the longer side is 3 cm

(ii) The length of the longer side is 6 cm

(iii) The length of the longer side is 15 cm

Step-by-step explanation:

A rectangle has sides in the ratio 1 : 3, we need to find the length of the longer side if:

Let us use the ratio method to solve the problem

(i)

∵ The ratio of the two sides of the rectangle is 1 : 3

∵ The length of the shorter side is 1 cm

→  Shorter    :    Longer

→  1               :    3

→  1               :    x

By using cross multiplication

∴ 1 × x = 1 × 3

∴ x = 3

∵ x represents the length of the longer side

∴ The length of the longer side = 3 cm

The length of the longer side is 3 cm

(ii)

∵ The ratio of the two sides of the rectangle is 1 : 3

∵ The length of the shorter side is 2 cm

→  Shorter    :    Longer

→  1               :    3

→  2               :    x

By using cross multiplication

∴ 1 × x = 2 × 3

∴ x = 6

∵ x represents the length of the longer side

∴ The length of the longer side = 6 cm

The length of the longer side is 6 cm

(iii)

∵ The ratio of the two sides of the rectangle is 1 : 3

∵ The length of the shorter side is 5 cm

→  Shorter    :    Longer

→  1               :    3

→  5               :    x

By using cross multiplication

∴ 1 × x = 5 × 3

∴ x = 15

∵ x represents the length of the longer side

∴ The length of the longer side = 15 cm

The length of the longer side is 15 cm

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Answer:

(2.23, 7,57)

Step-by-step explanation:

equation of this circle is

x^2 + (y - 2)^2 = 36

y = 2.5x + 2

Substitute for y in the equation of the circle:-

x^2 + (2.5x + 2 - 2)^2 = 36

x^2 + 6.25x^2 = 36

x^2 = 36 / 7.25  

x = +/-   6  /  2.693  =  +/- 2.228

when x = 2.228 y = 2.5(2.228) + 2 =  7.57    to nearest hundredth

when x = -2.228 y = 2.5(-2.228) + 2 =   -3.57

So they intersect at 2 points but the intersect in the first quadrant is at (2.23, 7,57)        to nearest hundredth.

Answer:

$7,661

Step-by-step explanation:

Closing balance = Opening balance + purchases - Issued items

Given

Office supplies account on January 1 = $6,791 - Opening balance

Purchases = $3,205

Supplies on hand on January 31 = $2,155 - Closing balance

Substituting into the formula above

2155 = 6791 + 3025 - Issued items

Issued items = 6791 + 3025 - 2155

                     = $7,661

The amount to be used for the appropriate adjusting entry is $7,661

Final answer:

The adjusting entry for the used office supplies for the month of January is $7,841, which is calculated by subtracting the supplies on hand at the month's end from the sum of the starting balance and purchases made during the month.

Explanation:

To calculate the adjusting entry for office supplies, you need to calculate the cost of supplies that were used during the month. Start with the balance of supplies on hand at the beginning of the month, add the purchases made during the month, and then subtract the balance of supplies on hand at the end of the month.

The calculation is as follows:

Starting balance on January 1: $6,791

Add purchases during January: $3,205

Subtract ending balance on January 31: $2,155

The adjusting entry for supplies used = (Starting balance + Purchases) - Ending balance
= ($6,791 + $3,205) - $2,155
= $9,996 - $2,155
= $7,841

Therefore, the adjusting entry to record the office supplies used would be for $7,841.