A box has candies in it: are taffy, are peppermint, and are caramel. (Each candy falls into only one of these categories.) Brian wants to select two candies to eat for dessert. The first candy will be selected at random, and then the second candy will be selected at random from the remaining candies. What is the probability that the first candy selected is taffy and the second candy is caramel?
Do not round your intermediate computations. Round your final answer to three decimal places.

The length of the intervals of the given frequency table is

500 ft.

The given list shows the lengths in feet of the 25 longest bridges in the United States.

we have to determine the median and mode of the bridge data.

The median is the middle number of the data set arranged in ascending order.

1010, 1053, 1200, 1207, 1380, 1470, 1495, 1596, 1600, 1600, 1600, 1632, 1750, 1800, 1850, 2000, 2150, 2150, 2300, 2310, 2800, 3500, 3800, 4200, 4260.

The median is 1750.

Mode is the number that appears most often in the data set.

1010, 1053, 1200, 1207, 1380, 1470, 1495, 1596, 1600, 1600, 1600, 1632, 1750, 1800, 1850, 2000, 2150, 2150, 2300, 2310, 2800, 3500, 3800, 4200, 4260

The mode is 1600.

Therefore the correct option is Median: 1750, Mode: 1600

So the length of the intervals of the given frequency table is

500 ft.

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It is true that the line segment AB equals 26

The given parameters are:

AB = x +16

BC = 4x + 11

AC = 77

The two-column proof is as follows:

AC = AB + BC                     Line segment formula

77 = x + 16 + 4x + 11            Substitution property of equation

77 = 5x + 27                        Addition property of equation

5x = 50                               Subtraction property of equation  

x = 10                                    Division property of equation

AB = 10 +16                          Substitution property of equation

AB = 26

Hence, the line segment AB has been proved to equal 26

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Answer:

175 ft^2

Step-by-step explanation:

split the pentagon into 5 triangles with base length 10ft and height 7ft. each triangle then has an area of 10 * 7 * 1/2 = 35 ft^2

then the pentagon has an area 35*5 = 175 ft^2

Answer:

Yes, x = 0 is a solution to the given equation

Step-by-step explanation:

Hi Student!

The goal of this question is to determine x = 0 is a solution of the expression that was provided.  The first step that we must take is input 0 into all of the x's that we have in the expression.  Then we just simplify both sides and determine if the end expression is true and if it is then x = 0 is a solution.

Plug in the values

Simplify both sides

Looking at the final expression, we can see that 1 is indeed equal to 1 and since the expression is true, we can say that x = 0 is a solution of the expression that was provided in the problem statement.

Answer:

B

Step-by-step explanation:

The dimensions of the the plot of land is a length of 19 meters and width of 12 meters.

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the length of the field and y represent the width of the field, hence:

Deion has 62 m of fencing:

2(x + y) = 62

x + y = 31

y = 31 - x     (1)

The area of the land is 228 square meters. hence:

xy = 228

x(31 - x) = 228

x² - 31x + 228 = 0  

x = 12 or x = 19

when x = 12; y = 31 - 12 = 19

when x = 19; y = 31 - 19 = 12

The dimensions of the the plot of land is a length of 19 meters and width of 12 meters.

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The domain of the function is x > 0.

The correct option is (B)

Domain of a function is defined by the set of x-values of the graph of a function.

as, domain of a function is defined by the set of x-values of the graph of a function.

From the graph we can see that,

x-values are starts from x > 0 [Since x ≠ 0]

Also, moves towards infinity,

Hence, domain of the function be, (0, ∞) or x > 0

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Answer:

B

Step-by-step explanation:

An x-intercept of the continuous function in the table is (-1, 0)

The x-intercept of a line is the point where the line crossed the x-axis or the  point where the value of y is zero.

From the table, the x-intercept are all the point where the value of f(x) is zero. Hence the Which is an x-intercept of the continuous function in the table is (-1, 0)

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[tex]\text{Nth term of an arithmetic series} = a +(n-1)d \\\\\text{Nth term of an geometric series}= ar^{n-1}\\\\\text{where,}\\\\\text{a = first term.}\\\\\text{d = common difference.}\\\\\text{r = common ratio.}[/tex]

The probability that a student is a female or major in civil engineering is 62%

At a certain college, 49% of the students are female, and 21% of the students major in civil engineering. Furthermore, 8% of the students both are female and major in civil engineering. What is the probability that a randomly selected female student majors in civil engineering?

Let A represent Female and B represents civil engineering.

The above representation means that the given parameters are:

The required probability is calculated as:

P(A or B) = P(A) + P(B) - P(A and B)

This gives

P(A or B) = 49% + 21% - 8%

Evaluate

P(A or B) = 62%

Hence, the probability that a student is a female or major in civil engineering is 62%

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Answer: x=15, y=3

Step-by-step explanation:

Subtracting the two equations, we get -3y=-9, meaning y=3.

Substituting this into the first equation, we get that x-3=12, and thus, x=15.

For the first image, I would recommend using Desmos; it is an online graphing calculator. You can find it on online and it is free; I use it all the time for my school work

I would input each function and sees which one matches the image on the test.

As for the second image, I believe the answer is C; x is greater than or equal to -1.

For the third and last image, I believe the answer is A; x is less than -1.

I hope this helps .

Answer: Yes it is possible

Example: Yeast is a single-celled fungus.

There are probably other types of fungus that are single-celled. However, some other fungi are multi-celled. You will likely need more information about the organism under the microscrope before you can classify it properly.

Answer: 0.5 or - 0.834

Step-by-step explanation: Here is the explanation!

Given the diameter of the surface of the clock, the area of the surface of the alarm clock is 3846.5cm².

Note that: Area of a circle is expressed as;

A = πr²

Where r is radius and π is constant pi ( π = 3.14 )

Given that;

A = πr²

A = 3.14 × ( 35cm )²

A = 3.14 × 1225cm²

A = 3846.5cm²

Therefore, given the diameter of the surface of the clock, the area of the surface of the alarm clock is 3846.5cm².

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Answer:

y = 9

Step-by-step explanation:

given that y varies inversely as x then the equation relating them is

y = [tex]\frac{k}{x}[/tex] ← k is the constant of variation

to find k use the condition y = 12 when x = 6

12 = [tex]\frac{k}{6}[/tex] ← multiply both sides by 6 to clear the fraction

72 = k

y = [tex]\frac{72}{x}[/tex] ← equation of variation

when x = 8 , then

y = [tex]\frac{72}{8}[/tex] = 9

Answer:

67m²

Step-by-step explanation:

12m + 6m + 4m + 36 m 8m = 67m²

Answer:

[tex]x=\frac{3}{4},\:x=-1[/tex]

Keys:

For this problem, you need the quadratic formula(listed below).

When you see ± in a quadratic equation, you must know there is going to be at least 2 solutions.

Step-by-step explanation:

solving for x₁ and x₂

[tex]4x^2+x=3\\4x^2+x-3=3-3\\4x^2+x-3=0\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot 4\left(-3\right)}}{2\cdot 4}\\[/tex]

[tex]1^2=1\\=\sqrt{1-4\cdot \:4\left(-3\right)}\\=\sqrt{1+4\cdot \:4\cdot \:3}\\=\sqrt{1+48}\\=\sqrt{49}\\=\sqrt{7^2}\\\sqrt{7^2}=7\\=7[/tex]

[tex]x_{1,\:2}=\frac{-1\pm \:7}{2\cdot \:4}\\x_1=\frac{-1+7}{2\cdot \:4},\:x_2=\frac{-1-7}{2\cdot \:4}\\[/tex]

solve for x₁

[tex]\frac{-1+7}{2\cdot \:4}[/tex]

[tex]=\frac{6}{2\cdot \:4}[/tex]

[tex]=\frac{6}{8}[/tex]

[tex]= \frac{6\div2}{8\div2}[/tex]

[tex]=\frac{3}{4}[/tex]

solve for x₂

[tex]\frac{-1-7}{2\cdot \:4}[/tex]

[tex]=\frac{-8}{2\cdot \:4}[/tex]

[tex]=\frac{-8}{8}[/tex]

[tex]=-\frac{8}{8}[/tex]

[tex]=-1[/tex]

Hope this helps!

If the definitions of type I and type II regions is the same as in the link provided, then as a type I region the integration domain is the set

[tex]R_{\rm I} = \left\{(x,y) \mid 0 \le x \le 3 \text{ and } \sqrt{x+1} \le y \le 2\right\}[/tex]

and as a type II region,

[tex]R_{\rm II} = \left\{(x,y) \mid 0 \le x \le y^2-1 \text{ and } 1 \le y \le 2\right\}[/tex]

where we solve y = √(x + 1) for x to get x as a function of y.

A. The area of the type I region is

[tex]\displaystyle \iint_{R_{\rm I}} dA = \int_0^3 \int_{\sqrt{x+1}}^2 dy \, dx = \int_0^3 (2 - \sqrt{x+1}) \, dx = \boxed{\frac43}[/tex]

B. The area of the type II region is of course also

[tex]\displaystyle \iint_{R_{\rm II}} dA = \int_1^2 \int_0^{y^2-1} dx \, dy = \int_1^2 (y^2-1) \, dy = \boxed{\frac43}[/tex]

I've attached a plot of the type II region to give an idea of how it was determined. The black arrows indicate the domain of x as it varies from the line x = 0 (y-axis) to the curve y = √(x + 1).

Answer:

tiookvgvc. jbjvth kivtcth jjvf h. bkbgv

Answer:

The IQR of the given distribution is

Step-by-step explanation:

The given distribution has the five-number

28, 34, 43, 59, 62

Divide these numbers in two equal parts.

(28, 34), 43,( 59, 62)

Now divide each parenthesis in two equal parts.

(28), (34), 43,( 59), (62)

It means first quartile is the average of 28 and 34. Third quartile is the average of 59 and 62.

The interquartile range (IQR) of this distribution is

Therefore the IQR of the given distribution is 29.5.

Answer:

Hi,

Step-by-step explanation:

sin(a-b)=sin(a) cos(b)+ cos(a) sin(b)

a=45° and b=30°

[tex]sin(45^o-30^o)=sin(45^o)*cos(30^o)+cos(45^o)*sin(30^o)\\\\=\dfrac{\sqrt{2}}{2}* \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}*\dfrac{1}{2}\\\\\\\boxed{sin(15^o)=\dfrac{\sqrt{2}}{4}*(1+\sqrt{3} )}\\[/tex]

(1) The missing term in the sequence, a₁₂  = 0.8.

(2) The missing term in the sequence, a₈ = 102.5.

(3) The missing term in the sequence, a₈ = 111.

(4) The missing term in the sequence, a₁₂ = -19.

(5) The missing term in the sequence, a₁₂ = 94.

(6)  The missing term in the sequence, a₆ = 40.

(7)  The missing term in the sequence, a₃₆ = -52.

(8)  The missing term in the sequence, a₂₁ = -58.

The missing term in the sequence is determined as follows;

Tₙ = a + (n - 1)d

T₄ = a + 3d

18.4 = a + 3d  ---(1)

T₅ = a + 4d

16.2 = a + 4d  ---(2)

subtract (1) from (2)

-2.2 = d

18.4 = a + 3(-2.2)

a = 25

a₁₂  = a + 11d

a₁₂  = 25 + 11(-2.2)

a₁₂  = 0.8

a₂ = a + d

57.5 = a + d -- (1)

a₅ = a + 4d

80 = a + 4d  --- (2)

solve (1) and (2)

d = 7.5

a = 50

a₈ =  a + 7d

a₈ = 50 + 7(7.5)

a₈ = 102.5

a₁₀ = a + 9d

141 = a + 9d --- (1)

a₁₃ = a + 12d

186 = a + 12d --- (2)

Subtract (1) from (2)

d = 15

a = 6

a₈ = a + 7d

a₈ = 6 + 7(15)

a₈ = 111

a₂₂ = a + 21d

-49 = a + 21d ---- (1)

a₂₅ = a + 24d

-58 = a + 24d --- (2)

subtract (1) from (2)

d = -3

a = 14

a₁₂ = a + 11d

a₁₂ = 14 + 11(-3)

a₁₂ = -19

a₄ = a + 3d

-2 = a + 3d --- (1)

a₈ = a + 7d

46 = a + 7d ---- (2)

Subtract (1) from (2)

d = 12

a = -38

a₁₂ = a + 11d

a₁₂ = -38 + 11(12)

a₁₂ = 94

a₉ = a + 8d

64 = a + 8d --- (1)

a₁₂ = a + 11d

88 = a + 11d --- (2)

Subtract (1) from (2)

d = 8

a = 0

a₆ = a + 5d

a₆ = 0 + 5(8)

a₆ = 40

a₂₀ = a + 19d

-4 = a + 19d ---- (1)

a₂₃ = a + 22d

-13 = a + 22d --- (2)

Subtract (1) from (2)

d = -3

a = 53

a₃₆ = a + 35d

a₃₆ = 53 + 35(-3)

a₃₆ = -52

a₂₈ = a + 27d

5 = a + 27d ---- (1)

a₃₃ = a + 32d

50 = a + 32d --- (2)

Subtract (1) from (2)

d = 9

a = -238

a₂₁ = a + 20d

a₂₁ = -238 + 20(9)

a₂₁ = -58

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Answer:

[tex]31.8\%[/tex]

Step-by-step explanation:

The area of the circle is [tex]A=\pi r^2=\pi(4)^2=16\pi[/tex]

The area of the triangle is [tex]A=\frac{bh}{2}=\frac{8*4}{2}=\frac{32}{2}=16[/tex]

Hence, the probability of a randomly selected point within the circle falls in the red shaded area is [tex]\frac{16}{16\pi}=\frac{1}{\pi}\approx0.318\approx31.8\%[/tex]

Answer:

22 inches

Step-by-step explanation:

3 feet = 36 inches

36-14=22

The other half is 22 inches long

The total area with land based on the information given is 3x² + 60x.

Let the width = x

Let the length = (1.5 × x) = 1.5x

Area = 1.5x × x = 1.5x²

The area along with land:

Width = x + 20

Length = 3 × x = 3x

The total area with land:

= Length × Width

= 3x(x + 20)

= 3x² + 60x.

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