Lisa owns a "Random Candy" vending machine, which is a machine that picks a candy out of an assortment in a random fashion. Lisa controls the probability of picking each candy. The machine has too much of the candy "Coffee Toffee," so Lisa wants to program it so that the probability of getting "Coffee Toffee" twice in a row is greater than \dfrac{4}{3} 3 4 ​ start fraction, 4, divided by, 3, end fraction times the probability of getting a different candy in one try. Write an inequality that models the situation. Use ppp to represent the probability of getting "Coffee Toffee" in one try.

Answer:

blank 1 = 24

blank2= 30

Step-by-step explanation:

Given:

2(4x+3)(3x+5)

=2(12x^2+20x+9x+15)

=2(12^2+29x+15)

=24x^2+58x+30

Hence blank 1= 24 and blank2= 30 !

Answer:

a=b

greater than

Step-by-step explanation:

Answer:

volumes of the cones are equal when (a=b)

If a > b, then the cross-sectional area of cone a is (GREATER THAN) is  the cross-sectional area of cone B at every level parallel to their respective bases.

Step-by-step explanation:

The correct answer is F. Quantitative data are numerical in​ nature, while qualitative data are categorical in nature.

Explanation:

In research and all the different fields that apply to it, the word "data" refers to information, values or knowledge that can be used to understand a specific situation or phenomenon. Additionally, data can be of two different types quantitative and qualitative, these differ in their nature, the phenomenons they described and the way they should be analyzed. Indeed quantitative data refers mainly to numerical data or information about quantities such as statistics that are especially useful in mathematics, science and similar that focus on numbers. On the other hand, qualitative data refers to data based on categories or qualities and because of this qualitative data is used in humanistic research, although both types of data can be combined to study a phenomenon. Considering this, the key difference between both types of data is "Quantitative data are numerical in​ nature, while qualitative data are categorical in nature".

Answer:

it is F

Step-by-step explanation:

F. Quantitative data are numerical in​ nature, while qualitative data are categorical in nature.

Answer:

The range in shelter A is 22, and the range in shelter B is 18.

Step-by-step explanation:

The range is the largest value minus the smallest value.

For shelter A, the range is 30 − 8 = 22.

For shelter B, the range is 28 − 10 = 18.

Answer:

The range in shelter A is 22, and the range in shelter B is 18.

Step-by-step explanation:

The box plots show the weights, in pounds, of the dogs in two different animal shelters. The range in shelter A is 22, and the range in shelter B is 18 correctly compares the ranges of the data.

Answer:

Converges to -25.

Step-by-step explanation:

[tex]\sum_{k=1}^{\infty} -5 \cdot (\frac{4}{5})^{k-1}[/tex] converges since [tex]r=\frac{4}{5}<1[/tex].

The sum is given by [tex]\frac{a_1}{1-r}[/tex] where [tex]a_1[/tex] is -5.

[tex]\frac{-5}{1-\frac{4}{5}}=\frac{-5}{\frac{1}{5}}=-5(5)=-25[/tex].

Answer:

420 ways

Step-by-step explanation:

According to the given statement:

We want to divide the 10 dogs into three groups, one with 3 dogs, one with 5 dogs, and one with 2 dogs. How many ways can we form the groups such that fluffy is in the 3-dog group and nipper is in the 5-dog group.

In this way we have 8 dogs left.

2 spaces left in 3 dogs group

4 spaces left in 5 dogs group

and 2 spaces in 2 dogs group

Therefore:

= 8!/2!4!2!

= 8*7*6*5*4*3*2*1/2*4*3*2*2

= 8*7*6*5/2*2

= 1680/4

=420

It means there are 420 ways to from the groups....

[tex]

A(4,-2) \\

B(0, 10) \\

AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\

AB=\sqrt{(0-4)^2+(10-(-2))^2} \\

AB=\sqrt{16+144} \\

AB=\sqrt{160}\approx\boxed{12.65} \\

[/tex]

The answer is D.

Hope this helps.

r3t40

Answer:

x≤ 60 - (3÷1/15)

Step-by-step explanation:

Number of blankets to be made = 3

Darcie crochets 1/15 blanket per day

To crochet one carpet she needs = 15 days

To crochet three carpets she needs = 15*3 = 45 days

Number of days she had = 60 days

She can skip days = 60-45 = 15 days

Let x be the number of days to complete her work.

Thus the equation becomes x≤ 60 - (3÷1/15)

You can solve for x to determine the number of days, Darcie can skip crocheting, you will get the answer 15....

The answer explains how to set up and solve an inequality to determine the number of days Darcie can skip crocheting while still meeting her goal of donating blankets.

To determine the number of days Darcie can skip crocheting and still meet her goal, we need to set up an inequality based on the information given.

First, calculate the total number of blankets Darcie needs to crochet: 3 blankets.

Set up the inequality: 1/15 * (60 - x) ≥ 3, where x represents the number of days she can skip crocheting.

Solve the inequality: 60 - x ≥ 45, x ≤ 15.

Therefore, Darcie can skip crocheting for up to 15 days and still meet her goal of donating 3 blankets.

Answer:

The right answer us "is always" as its all angles are 60 degree.

Answer:

7 rpm = 0.73 rad/s

Step-by-step explanation:

R = Radius of merry-go-round = 1.8 m

[tex]I_M[/tex]= Moment of inertia of merry-go-round = 255 kg m²

[tex]I_C[/tex]= Moment of inertia of child

ω = 9 rev/min

m = Mass of child = 24 kg

From the conservation of angular momentum

[tex]I\omega=I'\omega '\\\Rightarrow I\omega=(I_M+I_C)\omega'\\\Rightarrow \omega'= \frac{I\omega}{(I_M+I_C)}\\\Rightarrow \omega'=\frac{I\omega}{(I_M+mR^2)}\\\Rightarrow \omega'=\frac{255\times 9}{(255+24\times 1.8^2)}\\\Rightarrow \omega'=6.9\ rev/min[/tex]

∴ New angular speed of the merry-go-round is 7 rpm = [tex]7\times \frac{2\pi}{60}=\mathbf{0.73\ rad/s}[/tex]

Answer:

47.88 in^2 of paper.

Step-by-step explanation:

The amount of paper needed for the base = 4.2^2 = 17.64 in^2.

There are 4 triangular faces.

The area of each triangular face = 1/2* base * height

= 1/2 * 4.2 * 3.6

=  7.56 in^2

That is a total of 4 * 7.56 = 30.24 in^2.

So the total amount of paper need to cover the entire pyramid

= 17,64 + 30.24

= 47.88 in^2.

Answer:

5.2 Km/Min

Step-by-step explanation:

Speed/Rate= Distance/Time

Speed/Rate = 114.4/22

Speed/Rate = 5.2 KM/min

Answer:

5.3 km/min

Step-by-step explanation:

Distance = 114.4 kilometers

Time = 22 minutes

Rate = 114.4 km /22 min

Rate = 5.2 km/min

First lets change the fractions so they have common denominators:

3/5 = 6/10

The bank was 1/10, after adding the pennies it was 6/10

6/10 - 1/10 = 5/10 = 1/2

This means 400 pennies filled 1/2 the piggy bank.

There are 2 halves (1/2) to a whole ( full piggy bank).

The piggy bank can hold 400 x 2 = 800 pennies.

Answer:

  D.)  AB≅A′B′, ∠A≅∠A′, and ∠C≅∠C′

Step-by-step explanation:

Option A identifies two sides and the angle not between them. The two triangles will be congruent in that case only if the angle is opposite the longest side, which is not true in general.

Option B: same deal as Option A.

Option C identifies three congruent angles, which will prove the triangles similar, but not necessarily congruent.

Option D identifies two angles (sufficient for similarity) and one side, sufficient (with similarity) for congruence. The applicable congruence theorem is AAS.

The statements that could be used to prove that ΔABC and ΔA′B′C′ are congruent are: A.) ∠A≅∠A′, AC≅A′C′, and BC≅B′C′; C.) ∠A≅∠A′, ∠B≅∠B′, and ∠C≅∠C′; and D.) AB≅A′B′, ∠A≅∠A′, and ∠C≅∠C′.

The statements that could be used to prove that ΔABC and ΔA′B′C′ are congruent are:

A.) ∠A≅∠A′, AC≅A′C′, and BC≅B′C′

C.) ∠A≅∠A′, ∠B≅∠B′, and ∠C≅∠C′

D.) AB≅A′B′, ∠A≅∠A′, and ∠C≅∠C′

In order for two triangles to be congruent, their corresponding angles and sides need to be congruent. In this case, option A states that ∠A≅∠A′, AC≅A′C′, and BC≅B′C′, which satisfies the conditions for congruence. Option C has congruent angles but does not mention congruent sides, so it does not prove congruence. Option D mentions congruent sides but does not mention congruent angles, so it also does not prove congruence.

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Answer:

-4

Step-by-step explanation:

[√2(cos(3π/4) + i sin(3π/4))]⁴

(√2)⁴ (cos(3π/4) + i sin(3π/4))⁴

4 (cos(3π/4) + i sin(3π/4))⁴

Using De Moivre's Theorem:

4 (cos(4 × 3π/4) + i sin(4 × 3π/4))

4 (cos(3π) + i sin(3π))

3π on the unit circle is the same as π:

4 (cos(π) + i sin(π))

4 (-1 + i (0))

-4

Answer:

BD = 10

Step-by-step explanation:

Since AC is a tangent to the circle at B then ∠ABD = 90°

Using Pythagoras' identity in the right triangle ABD with hypotenuse AD

The square on the hypotenuse is equal to the sum of the squares on the other two sides, that is

BD² + AB² = AD²

BD² + 24² = 26²

BD² + 576 = 676 ( subtract 576 from both sides )

BD² = 100 ( take the square root of both sides )

BD = [tex]\sqrt{100}[/tex] = 10

Answer:

  C, D, B, A

Step-by-step explanation:

The greater the angle the tangent line makes with the positive x-axis, the greater the slope. Angles increase in the counterclockwise direction, so the question here is equivalent to asking for the tangent lines to be put in clockwise (decreasing slope) order.

That order is C, D, B, A.

An exponential function may be given by:

f(x) = A(1+r)^x

A is the initial amount and r is a decimal representing the growth rate.

We can see that 1+r = 1.2, and we solve for r:

1 + r = 1.2

r = 0.2

The growth rate is 0.2, or 20%

Choice B

The rate of growth of Mason's daily sales is 20%.

The rate of growth of Mason's daily sales can be found by determining the percentage increase in the sales from one day to the next. To find this, we can compare the sales on two consecutive days and calculate the ratio of the second day's sales to the first day's sales. Let's consider the sales on the first day (x) and the sales on the second day (x+1).

Given the sales model f(x) = 50(1.2)^x, we can substitute x and x+1 into the equation to get the sales on the first and second day, respectively. The ratio of the second day's sales to the first day's sales is:

f(x+1) / f(x) = (50(1.2)^(x+1)) / (50(1.2)^x) = 1.2.

So, the rate of growth is 1.2, which represents a 20% increase in sales from one day to the next.

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Answer:

option 4 ⇒ 236 lb.

Step-by-step explanation:

Best explanation of the question is as shown in the attached figure.

we will use the parallelogram method to calculate resultant force.

to get the length of the resultant force ⇒ use the cosines law

The cosine law is a² = b² + c² - 2 * b * c * cos (∠A)

Applying at the question where b = F₁  , c = F₂  and  ∠A = ∠x

Given that F₁ = 100 pounds  , F₂ = 150 pounds  and  ∠x = 180° - 40° = 140°

∴ (Resultant force)² = 100² + 150² - 2 * 100 * 150 * cos (∠140) = 55481

∴ Resultant force = √55481 = 235.54 ≅ 236 pounds

The answer is option 4 ⇒ 236 lb.

Answer:

  $1.45

Step-by-step explanation:

To answer the question, we would like to have an equation that has the cost of a game as its only variable. That is, we would like to eliminate the cost of a ride from the system of equations we must write.

Let g and r represents the costs of a game and a ride, respectively. Then the expenses of the two carnival-goers can be described by ...

  5g +6r = 11.75

  7g +8r = 16.15

We note that the ratio of coefficients in the variable (r) that we want to eliminate is 3:4. So we can subtract 4 times the first equation from 3 times the second to eliminate that variable.

  3(7g +8r) -4(5g +6r) = 3(16.15) -4(11.75)

  g = 1.45 . . . . . . simplify

The cost to play one game was $1.45.

Answer:

$1.45

Step-by-step explanation:

Answer:

A = 14 units ^2

Step-by-step explanation:

The area of a triangle is

A = 1/2 b h  where b is the base and h is the height

b = the length of EF  which is 7 units

h = D to the line EF which is 4 units

A = 1/2 (7*4)

A = 14 units ^2

There are 52,920 ways to select 5 appetizers, 4 main courses, and 5 desserts for the banquet.

To solve this problem, we can use the concept of combinations, as we're selecting items without considering the order.

For appetizers:

We need to choose 5 appetizers out of 6 available. This can be calculated using the combination formula: nCr = n! / [r! * (n-r)!], where n is the total number of items, and r is the number of items to be chosen.

So, for appetizers, it's [tex]6C_5 = 6! / [5! * (6-5)!] = 6 ways[/tex].

For main courses:

Similarly, we need to choose 4 main courses out of 7 available. So, it's

[tex]7C_4 = 7! / [4! * (7-4)!] = 35 ways[/tex].

For desserts:

We need to choose 5 desserts out of 10 available. So, it's

[tex]10C_5 = 10! / [5! * (10-5)!] = 252 ways[/tex].

To find the total number of ways:

We multiply the number of ways for each category since these choices are independent.

Total ways = (6 ways for appetizers) * (35 ways for main courses) * (252 ways for desserts) = 52920 ways.

Thus, there are 52,920 ways to select 5 appetizers, 4 main courses, and 5 desserts for the banquet.

Complete Question:

A catering service offers 6 appetizers, 7 main courses, and 10  desserts. A customer is to select 5 appetizers, 4 main courses, and 5 desserts for a banquet. In how many ways can this be done?

Answer:

b ≈ 17

Step-by-step explanation:

Using trigonometric ratio, sine we can find side b.

sine x = opposite/hypotenuse

sin 45 = 12/b

b ≈ 17

Answer:

b = 12[tex]\sqrt{2}[/tex]

Step-by-step explanation:

Using the sine ratio in the right triangle and the exact value

sin45° = [tex]\frac{1}{\sqrt{2} }[/tex]

From the triangle

sin45° = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{12}{b}[/tex] = [tex]\frac{1}{\sqrt{2} }[/tex]

Cross- multiply

b = 12[tex]\sqrt{2}[/tex]

Answer:

  2√3

Step-by-step explanation:

You recognize this as a 30°-60°-90° triangle, so you know the hypotenuse (R) is twice the length of the shortest side (√3).

The magnitude of R is 2√3.

_____

In case you haven't memorized the ratios for a 30°-60°-90° triangle, you can use trigonometry and the fact that ...

  Sin = Opposite/Hypotenuse

  sin(30°) = √3/R

  R = √3/sin(30°) = √3/(1/2) = 2√3

Of course, doing this on your calculator will give a numerical answer, which you may not want.

Answer:

In general you can choose the vertices at any arbitrary points but for easier computations and calculations we can choose 1 vertex at origin with co-ordinates [tex](0,0)[/tex] and it's adjacent vertex either on x-axis with co-ordinates [tex](x,0)[/tex] or on y-axis with ordinates [tex](0,y)[/tex]

Thus the coordinates of vertices become

Answer:

Step-by-step explanation:

The way you have written the first question may be what is confusing you. It should be written as

bn = 3*b_(n-1) + 2

So b2 =

b2 =3*b_(2 -1) + 2

b2 = 3*b1 + 2

b2 = 3*5 + 2

b2 = 15 + 2

b2 = 17

===========

b3 = 3b_(n _1) + 2

b_2 = 17 (from the step above)

b3 = 3*17 + 2

b3 = 51 + 2

b3 = 53

==========

b_4 = 3*b_3 + 2

b_4 = 3*53 + 2

b_4 = 159 + 2

b_4 = 161

Do you see how this works? You take the previous term, multiply by 3 and add 2 to get the current term. This one builds up rather quickly.

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Next Question

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tn = a + (n - 1)*d

t6 = a + (6 - 1)*d

t4 = a + 5d

4 = a + 5d    

t10 = a + 9d

Subtract t4 [4 = a + 3d ] from t10 written bellow

- 4 = a +  9d

4  = a + 5d

- 8 = 4d              Divide by 4

-8/4 = 4d/4

-2 = d

t6 = a + 5d

4 = a + 5*(-2)

4 = a - 10               Add 10 to both sides.

4 + 10 = a - 10 + 10

14 = a

tn = 14 + (n - 1)*d

Answer: d

Please don't use red. It is really hard to read.

Answer:

A.) −2y + 10

Step-by-step explanation:

−x − 2y = −10

Add 2y to both sides.

-x = 2y - 10

Multiply both sides by -1.

x = -2y + 10

Let's solve the system using the substitution method. We'll start by isolating \( x \) in the second equation.
The second equation is given by:
\[ -x - 2y = -10 \]
To isolate \( x \), we'll follow these steps:
1. Add \( 2y \) to both sides of the equation, which gives us:
\[ -x = 2y - 10 \]
2. Now, we multiply both sides by \( -1 \) to solve for \( x \), which gives us:
\[ x = -1(-2y + 10) \]
\[ x = 2y - 10 \]
The expression that we would substitute into the first equation for \( x \) is \( 2y - 10 \).
Therefore, the correct answer is:
C.) \( 2y - 10 \)

[tex]\bf \textit{Sum and Difference Identities} \\\\ tan(\alpha + \beta) = \cfrac{tan(\alpha)+ tan(\beta)}{1- tan(\alpha)tan(\beta)} \qquad tan(\alpha - \beta) = \cfrac{tan(\alpha)- tan(\beta)}{1+ tan(\alpha)tan(\beta)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ tan(x-3\pi )=\cfrac{tan(x)-tan(3\pi )}{1+tan(x)tan(3\pi )} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\bf tan(3\pi )\implies \cfrac{sin(3\pi )}{cos(3\pi )}\implies \cfrac{0}{-1}\implies 0\qquad therefore \\\\[-0.35em] ~\dotfill\\\\ tan(3\pi )=\cfrac{tan(x)-tan(3\pi )}{1+tan(x)tan(3\pi )}\implies tan(x-3\pi )=\cfrac{tan(x)-0}{1+0} \\\\\\ tan(x-3\pi )=\cfrac{tan(x)}{1}\implies tan(x-3\pi )=tan(x)[/tex]

Answer:

-17

Step-by-step explanation:

Plug in -2 as your x value for the g(x) equation and simplify.

[tex]g(-2)=\sqrt{-2+6} \\g(-2)=\sqrt{4} \\g(-2)=2[/tex]

Next, plug in your g(x) value (2) to the f(x) equation for x and simplify.

[tex]f(2)=-7(2)-3\\f(2)=-14-3\\f(2)=-17[/tex]

Answer:

[tex]\large\boxed{-2x^3y^2+9x^2y^3-3xy^4-6x^4y+y^5}[/tex]

Step-by-step explanation:

[tex](-2x^3y^2+4x^2y^3-3xy^4)-(6x^4y-5x^2y^3-y^5)\\\\=-2x^3y^2+4x^2y^3-3xy^4-6x^4y+5x^2y^3+y^5\qquad\text{combine like terms}\\\\=-2x^3y^2+\underline{4x^2y^3}-3xy^4-6x^4y+\underline{5x^2y^3}+y^5\\\\=-2x^3y^2+9x^2y^3-3xy^4-6x^4y+y^5[/tex]