At a high school, the probability that a student takes a science class and a history class is 0.48. The probability that a student takes a science class is 0.82, and the probability that the student takes a history class is 0.64. What is the probability (rounded to the nearest hundredth) that a student takes a history class given that the student is taking a science class?

The value of inverse of function is,

⇒ f ⁻¹ (x) = x² / 36

We have to given that;

Function is,

⇒ f (x) = 6√x

Now, We can find the inverse of the function is,

⇒ f (x) = 6√x

⇒ y = 6√x

Solve for x;

⇒ y/6 = √x

⇒ (y/6)² = x

⇒ x = y² / 36

⇒ f ⁻¹ (x) = x² / 36

Thus, The value of inverse of function is,

⇒ f ⁻¹ (x) = x² / 36

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Answer:6

Step-by-step explanation:

Answer:

[tex]573.9 \text{ in}^3[/tex]

Step-by-step explanation:

First, we can find the volume of the rectangular prism using the formula:

[tex]V_\square = l \cdot w \cdot d[/tex]

where [tex]l[/tex] is the prism's length, [tex]w[/tex] is its width, and [tex]d[/tex] is its depth.

Plugging the given dimensions into the formula:

[tex]V_\square = 8 \cdot 12 \cdot 6[/tex]

[tex]V _\square= 96 \cdot 6[/tex]

[tex]\boxed{V_\square = 576 \text{ in}^3}[/tex]

Next, we can find the volume of the half-sphere using the formula:

[tex]V_\circ = \dfrac{2}{3} \pi r^3[/tex]

where [tex]r[/tex] (or [tex]d/2[/tex]) is the half-sphere's radius.

Plugging the given diameter value into the formula:

[tex]V_\circ = \dfrac{2}{3} \pi (2/2)^3[/tex]

[tex]V_\circ = \dfrac{2}{3} \pi (1)^3[/tex]

[tex]\boxed{V_\circ=\dfrac{2}{3}\pi \text{ in}^3}[/tex]

Finally, we can find the volume of the composite figure by subtracting the volume of the half-sphere from the volume of the rectangular prism.

[tex]V = V_\square - V_\circ[/tex]

[tex]V = 576 \text{ in}^3 - \dfrac{2}{3}\pi \text{ in}^3[/tex]

We can evaluate this using a calculator.

[tex]\boxed{V\approx 573.9 \text{ in}^3}[/tex]

18 will be the result of the given problem.

Applying the order of operations (PEMDAS), we first calculate the value inside the parentheses, which is 3 + 6 = 9.

Then we multiply 4 by 9, which gives 36.

Finally, we divide 36 by 2, which gives the final answer of 18. Therefore:

4 x (3 + 6) ÷ 2 = 18

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The probability of randomly choosing two boys is 0.174

If the selection is done at random, then the probability of choosing a boy is equal to the quotient between the number of boys and the total number of people.

There are 12 boys and 16 + 12 = 28 students in total, then the probability is:

P = 12/28

Now, for the second selection the probability is computed in the same way, but now there are 11 boys and 27 students (one was already choosen), then:

P = 11/27

the joint probability (for the two selections) is given by the product between the individual probabilities, we will get:

P = (12/28)*(11/27)

P = 0.174

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The area of triangle UVW to the nearest square inch is approximately 2 square inches.

The area of triangle UVW can use the Heron's formula:

Area = √(s(s-a)(s-b)(s-c))

s is the semiperimeter of the triangle and a, b and c are the lengths of its sides.

First, we can calculate the semiperimeter s:

s = (u + v + w)/2

s = (5.5 + 4.4 + 9.6)/2

s = 9.75

The length of each of the three sides of the triangle:

a = v = 4.4 inches

b = u = 5.5 inches

c = w = 9.6 inches

Then, we can plug these values into the Heron's formula and simplify:

Area = √(s(s-a)(s-b)(s-c))

Area = √(9.75(9.75-4.4)(9.75-5.5)(9.75-9.6))

Area = √(9.75(5.35)(4.25)(0.15))

Area ≈ √(3.296)

Area ≈ 1.81

Rounding this to the nearest square inch, we get:

Area ≈ 2 square inches

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C. The total profit, if all shares for both stocks are sold at the close of business on day 7, is $587.62

D. The best day to sell TechCo shares would be day 7, as it had the highest closing price. For HealthCo, day 7 would also be the best day to sell.

E. There is no difference in profit or loss compared to part C.

Part 2

A. Diversity can be evaluated based on the portfolio in question

B. By analyzing the types of assets present in the portfolio, it is possible to determine where potential risks may originate.

C. The individual carrying a particular portfolio can be characterized based on their risk tolerance, financial goals, and stage of life.

Part 1:

A. TechCo (TCO) and HealthCo (HCO) have their shares priced at $100 and $50 respectively.

Name of Stock Symbol Quantity of Shares Purchased Purchase Price per Share Total Purchase Value

TechCo TCO 75 $100 $7,500

HealthCo HCO 150 $50 $7,500

B. Here is an illustration of the potential ending values:

Name of Stock Closing Price Day 1 Closing Price Day 2 Closing Price Day 3 Closing Price Day 4 Closing Price Day 5 Closing Price Day 6 Closing Price Day 7

TechCo $101 $99 $100 $98 $102 $101 $103

HealthCo $51 $50 $52 $49 $51 $52 $54

C. If we sell all shares on day 7, we can calculate the profit/loss and the brokerage fee:

TechCo: 75 shares * $103/share = $7,725

HealthCo: 150 shares * $54/share = $8,100

Total = $7,725 + $8,100 = $15,825

Brokerage fee = $15,825 * 1.5% = $237.38

Profit = $15,825 - $15,000 (initial investment) - $237.38 (brokerage) = $587.62

D. The best day to sell TechCo shares would be day 7, as it had the highest closing price. For HealthCo, day 7 would also be the best day to sell.

E. Since we've already assumed selling on the best day, there is no difference in profit or loss compared to part C.

Part 2:

A. Diversity can be evaluated based on the portfolio in question. A portfolio that encompasses investments in a variety of sectors, asset classes, and regions would exhibit a high level of diversity.

B. By analyzing the types of assets present in the portfolio, it is possible to determine where potential risks may originate.

An investment portfolio that primarily consists of stocks or other assets with high risks would be considered as an aggressive portfolio. A portfolio that primarily consists of bonds or low-risk assets is considered to be a conservative one.

A balance between the two would be a moderate portfolio.

C. The individual carrying a particular portfolio can be characterized based on their risk tolerance, financial goals, and stage of life.

For example, a young professional with a high-risk tolerance might carry an aggressive portfolio to maximize growth, while a retired individual might have a conservative portfolio to preserve their capital.

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[tex]x=\textit{lbs of aluminum at 30\%}\\\\ ~~~~~~ 30\%~of~x\implies \cfrac{30}{100}(x)\implies 0.3 (x) \\\\\\ y=\textit{lbs of aluminum at 54\%}\\\\ ~~~~~~ 54\%~of~y\implies \cfrac{54}{100}(y)\implies 0.54 (y) \\\\\\ \textit{48 lbs of aluminum at 52\%}\\\\ ~~~~~~ 52\%~of~48\implies \cfrac{52}{100}(48)\implies 0.52 (48)\implies 24.96 \\\\[-0.35em] ~\dotfill[/tex]

[tex]\begin{array}{lcccl} &\stackrel{lbs}{quantity}&\stackrel{\textit{\% of lbs that is}}{\textit{AL only}}&\stackrel{\textit{lbs of}}{\textit{AL only}}\\ \cline{2-4}&\\ \textit{1st alloy}&x&0.3&0.3x\\ \textit{2nd alloy}&y&0.54&0.54y\\ \cline{2-4}&\\ mixture&48&0.52&24.96 \end{array}~\hfill \begin{cases} x + y = 48\\\\ 0.3x+0.54y=24.96 \end{cases} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{\textit{using the 1st equation}}{x+y=48\implies y=48-x} \\\\\\ \stackrel{\textit{substituting on the 2nd equation from above}}{0.3x~~ + ~~0.54(48-x)~~ = ~~24.96}\implies 0.3x+25.92-0.54x=24.96 \\\\\\ 0.30x-0.54x=-0.96\implies -0.24x=-0.96\implies x=\cfrac{-0.96}{-0.24} \\\\\\ \boxed{x=4}\hspace{9em}y=48-x\implies \boxed{y=44}[/tex]

Answer:

Answer is -2/21r + (-19)

Step-by-step explanation:

(-1/7r - 9 -2/3r) - (-5/7r + 10)

= -1/7r - 9 - 2/3r + 5/7r - 10  (distributing the negative sign)

= (-1/7r + 5/7r) + (-2/3r) - 9 - 10  (grouping the like terms)

= 4/7r - 2/3r - 19  (combining the like terms)

= (12r - 14r)/21r - 38r/21r  (getting a common denominator)

= -2r/21r - 38r/21r

= (-2r - 38r)/21r

= -40r/21r

= -40/21

Therefore, the simplified expression is B. -2/21r + (-19).

Hüseyin concluded: "I was able to map quadrilateral EFGH onto quadrilateral EJGK using a sequence of rigid transformations and dilations, so the figures are similar."  The error that he make is option A: Huseyn  used a vertical stretch, so the quadrilaterals aren't similar.

Hüseyin's transformations is one that keeps the EFGH's shape and proportions, so EJGK is similar to EFGH. Similar figures have congruent angles and proportional sides.

The error is seen when he uses vertical stretch, that is violating similarity condition of quadrilaterals EFGH and EJGK. A vertical stretch would alter the proportions of the sides and angles in quadrilateral EFGH, making it non-similar to that of  EJGK.

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After simplification,

(2x + 1)/2(x^2)

Given,

(x/4)+(1/8)/(x^2/4)

Simplifying,

Firstly take LCM of 4 and 8,

LCM of 4 and 8 will be 8.

Then,

(2x +1) /(8) / x^2 /4

Further,

Take reciprocal of the denominator,

(2x +1) /(8) (4/x^2)

(2x + 1)/2(x^2)

Hence after simplification the expression becomes (2x + 1)/2(x^2).

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Jodi will need 10 half-cup measurements of flour to make 2 batches of cookies.

To double the recipe, Jodi needs to multiply the amounts of each ingredient by 2.

The original recipe calls for 2½ cups of flour. Doubling this amount gives:

2½ cups * 2 = 5 cups of flour

We may divide the total amount of flour required (5 cups) by the quantity in each 12-cup measurement to get the number of 1-1/2 cups Jodi needs.

5 cups / ½ cup = 10

Therefore, Jodi will need 10 half-cup measurements of flour to make 2 batches of cookies.

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Answer: £234

Step-by-step explanation:

First, we need to subtract the deposit (£350) from the total cost of the trip (£1754):

£1754 - £350 = £1404

Then we need to divide the remaining amount of money (£1404) by the number of months Janice will be on holiday (6 months):

£1404 ÷ 6 months = £234 each month

Hope this helps!

Answer:

Set your calculator to Degree mode.

10² = 6.78² + 4² - 2(6.78)(4)(cos x)

100 = 61.9684 - 54.24(cos x)

38.0316 = -54.24(cos x)

cos x = -.701173

[tex]x = {cos}^{ - 1} ( - .701173)[/tex]

x = 134.52°

right isosceles triangle.

Right Scalene Triangle

Angles

Acute angle:  An angle whose measure is less than 90 degrees.

Right angle:  An angle whose measure IS 90 degrees.  As a consequence, the two rays that form the angle are perpendicular.

Obtuse angle:  An angle whose measure is greater than 90 degrees.

Triangles (defined by side lengths)

Equilateral Triangle:  A triangle with all three sides congruent (all three sides have the same length)

Isosceles Triangle:  A triangle that contains two congruent sides (two sides with the same length).

Scalene Triangle: A triangle with three different length sides.

Triangles (defined by angle measures)

Acute Triangle:  A triangle whose angles are all acute angles

Right Triangle:  A triangle that contains one right angle.

Obtuse Triangle:  A triangle that contains one obtuse angle.

By angle

Our triangle has an angle that appears to be a right angle (the two sides look perpendicular).  This would be a Right Triangle (when classified by angles).

By side length

All three sides of this triangle are different in length.  This would be a Scalene Triangle (when classified by side length).

Combining the attributes:

This is both a Right Triangle and a Scalene Triangle.  Together, it is a Right Scalene Triangle.

Right Isosceles

To be a Right Isosceles Triangle, the triangle needs to be both a Right Triangle, and an Isosceles Triangle.

A Right triangle contains one right angle, and an Isosceles Triangle contains a pair of sides that are the same length.

While this triangle does contain a right angle, all three sides appear to be different length (scalene).

This is not a Right Isosceles Triangle.

Acute Isosceles

To be an Acute Isosceles Triangle, the triangle needs to be both an Acute Triangle, and an Isosceles Triangle.

An Acute Triangle contains all angles that are acute angles (less than 90 degrees), and an Isosceles Triangle contains a pair of sides that are the same length.

However, this triangle contains a right angle (equal to 90 degrees, not less than), so it cannot be an Acute Triangle, and this triangle does not contain a pair of sides that are the same length, so it cannot be an Isosceles Triangle.

This is not an Acute Isosceles Triangle.

Acute Scalene

To be an Acute Scalene Triangle, the triangle needs to be both an Acute Triangle, and a Scalene Triangle.

An Acute Triangle contains all angles that are acute angles (less than 90 degrees), and a Scalene Triangle has sides that are all different length.

While this triangle does have sides that are all different length, it contains a right angle, whose measure IS 90 degrees (not less than 90 degrees).  So it cannot be an Acute Triangle

This is not an Acute Scalene Triangle.

To find the volume of the figure, we use the formula: (2cm x 8cm x 7cm) + (1cm x 6cm x 8cm). The correct answer is (B)

To find the volume of a joined shape, you have to solve for the volume of each component of the shape and sum them together.

In this case, we have 2 shapes joined and both are cuboids.

Volume of a cuboid = length x breath x height

Volume₁ = 8 x 7 x 2 = 2 x 8 x 7

Volume₂ = 8 x 1 x 6 = 1 x 6 x 8

Total volume = (2cm x 8cm x 7cm) + (1cm x 6cm x 8cm)

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The implications of Constructivist approach in teaching mathematics is that it helps the students to develop a better way to comprehend the content of what they are being taught.

A constructive approach of teaching mathematics believes that students construct knowledge rather than just passively take in information.

A constructive approach of teaching also believes that students integrate new knowledge with existing knowledge to create a deeper understanding of the mathematics being taught to them.

Therefore, constructive approach of teaching can help the students to comprehend the content of what they are being taught.

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The end behavior of the quadratic function is f(x) = (x + 5)² - 2

We have the table of values which represents a quadratic function f(x).

A quadratic equation is represented as;

f(x) = a(x - h)² + k

Where, Vertex = (h, k)

If we plot the graph according to the given table, the vertex will be,

(h, k) = (-5, -2)

Substitute, (h, k) = (-5, -2) in f(x) = a(x - h)² + k

we have, f(x) = a(x + 5)² - 2

Also, the graph, we have the point (0, 23)

This means that

a(0 + 5)² - 2 = 23

25a = 25

a = 1

Substitute a = 1 in f(x) = a(x + 5)² - 2

f(x) = (x + 5)² - 2

Hence, the equation is polynomial f(x) = (x + 5)² - 2

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The distribution of the weight of one dozen eggs is approximately normally distributed with a mean of 68 grams and a standard deviation of 4.2 grams.

The distribution of the weight of one dozen eggs is also approximately Normally distributed. The mean of the distribution is the same as the mean of the individual eggs, which is 68 grams.

The standard deviation of the distribution is the square root of the sum of the variances of the individual eggs and the sample size, which is:

= 2.2 * ✓12

= 4.2 grams.

Therefore, the distribution of the weight of one dozen eggs is approximately normally distributed with a mean of 68 grams and a standard deviation of 4.2 grams.

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Part A: The equation of the circle in standard form is x² + (y - 7)² = 2².

Part B: Center: (h, k) = (0, 7), Radius: 2

Part C: Please check the attachment.

In this problem we find the definition of the equation of a circle in general form, whose standard form must be found:

(x - h)² + (y - k)² = r²

Where:

This can be done by completing the square. First, write the equation of the circle:

x² + y² = 14 · y - 45

Second, complete the square by algebra properties:

x² + y² - 14 · y + 45 = 0

x² + (y² - 14 · y + 49) = 4

x² + (y - 7)² = 2² (PART A)

Third, write the coordinates of the center and the radius of the circle:

Center: (h, k) = (0, 7), Radius: 2 (PART B)

Finally, we proceed to graph the resulting expression.

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Cameron and Rebecca are organizing a lunch for their friends. Within the dishes to be served, they will prepare a tomato, carrot and broccoli salad. The number of tomatoes and carrots are 10-partners and the total amount of vegetables is 12. If we increase the number of broccoli by three and decrease the number of tomatoes by two, the amount of tomatoes and broccoli will also be 10-partners. What is the number of tomatoes, carrots and broccoli bought by Cameron and Rebecca?

Cameron and Rebecca purchased 7 tomatoes, 3 carrots and 2 broccoli.

Number of tomatoes = x

Number of carrots = 10 - x

Number of broccoli = y

System of Equations:

x + 10 - x + y = 12

y = 12 - 10

y = 2

x - 2 + y + 3 = 10 (If we increase the number of broccoli by three and decrease the number of tomatoes by two, the amount of tomatoes and broccoli will also be 10-partners)

x + 1 + y = 10

x + y = 9

y = 9 - x

Substitution of y:

x + 10 - x + y = 12

x + 10 - x + 9 - x = 12

-x + 19 = 12

-x = 12 - 19

x = 7 ⇒ 10 - x = 3

Cameron and Rebecca purchased 7 tomatoes, 3 carrots and 2 broccoli.

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398.223 redondeado a la decena más cercana es 400.

Redondear un número a la decena más cercana implica determinar a qué decena se acerca más. En este caso, estamos redondeando el número 398.223.

Para hacer esto, observamos el dígito en la posición de las decenas, que es el 9 en 398.223. El dígito a la derecha de la posición de las decenas es el lugar de las centésimas, que en este caso es 8.

Para redondear a la decena más cercana, seguimos la regla general de redondeo: si el dígito en el siguiente lugar de valor más bajo (en este caso, el lugar de las centésimas) es 5 o mayor, redondeamos hacia arriba el dígito en el lugar de valor objetivo (en este caso, las decenas). Si el dígito es menor que 5, redondeamos hacia abajo.

En nuestro caso, el dígito en el lugar de las centésimas (8) es mayor que 5, por lo que redondeamos hacia arriba el dígito en el lugar de las decenas (9) al número siguiente más alto. Por lo tanto, 398.223 redondeado a la decena más cercana se convierte en 400.

En resumen, redondear 398.223 a la decena más cercana nos da 400 porque el dígito en el siguiente lugar de valor más bajo (8) es mayor que 5.

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Answer: the awnser is b

Step-by-step explanation:

Comparing the median of each box-and-whisker plot, the type of transportation that is LEAST likely to take more than 30 minutes is: d. train.

In order to determine the transportation that is LEAST likely to take more than 30 minutes, we have to compare the median of each data set represented on the box-and-whisker plot for each transportation.

The box-and-whisker plot that has the lowest median would definitely represent the the transportation that is LEAST likely to take more than 30 minutes, since median represents the typical minutes or center of the data.

Therefore, from the box-and-whisker plots given, the one for train has the lowest median. Therefore train would LEAST likely take more than 30 minutes.

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For the 4-hour infusion, 1500ml of the 20% acetylcysteine solution must be made up and combined with 500ml of glucose.

To calculate the volume of acetylcysteine that needs to be added to the glucose, we can use the following formula:

The volume of acetylcysteine (in ml) = (Total amount of acetylcysteine needed in mg / Concentration of acetylcysteine in mg/ml) x 100

We have been asked to add 3000mg of acetylcysteine to 500 ml of glucose. If the concentration of the ampoule containing acetylcysteine is 20%, this means that there are 20g of acetylcysteine in 100 ml of solution, or 200mg/ml.

Using the formula above, we can calculate the volume of acetylcysteine that needs to be added as follows:

Volume of acetylcysteine = (3000mg / 200mg/ml) x 100 = 1500ml

Therefore, 1500ml of the 20% acetylcysteine solution needs to be drawn up and added to the 500ml of glucose to make the 4-hour infusion.

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The simplified expression of [tex]\sqrt[4]{567(x^9)(y^{11})}[/tex] is [tex]3x^2y^2\sqrt[4]{7xy^3}[/tex].

Given expression:

[tex]\sqrt[4]{567(x^9)(y^{11})}[/tex].

We can simplify the expression using the property that says for any non-negative numbers a and b, and any integer n ≠ 0,

[tex](a\times b)^{n} = a^n \times {b^n}[/tex]

Using this property, we can simplify the expression as follows:

[tex](567x^9y^{11})^{1/4} = (567)^{1/4} \times (x^9)^{1/4} \times (y^{11})^{1/4}[/tex]

Now we need to simplify,

[tex](81\times 7)^{1/4}x^{9/4}y^{11/4}\\=(81)^{1/4}(7)^{1/4}(x^{\frac{8}{4}+\frac{1}{4})(y^{\frac{8}{4}+\frac{3}{4})[/tex]

Also,

[tex]{3^4}^{1/4}(7)^{1/4}(x^{2+\frac{1}{4}})(y^{2+\frac{3}{4}})\\=3^1 7^{1/4}x^2x^{1/4}y^2y^{1/4}\\=3x^2y^2(7^{1/4}x^{1/4}y^{3/4})\\=3x^2y^2(7xy^3)^{1/4}\\=3x^2y^2\sqrt[4]{7xy^3}[/tex]

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The minimum number of tickets the agent must sell to have a 95% chance of selling at least one prize-winning ticket is 368 tickets

To unravel this issue, we are able to utilize the concept of likelihood. Let's begin with discovering the likelihood of not offering any prize-winning tickets with one ticket:

- The likelihood of not winning with one ticket is 499/500 (since there's as it were one winning ticket out of 500).

- The likelihood of not winning with n tickets is (499/500)[tex]^n[/tex](since the probabilities are autonomous and we duplicate them together).

Presently, we need to discover the least number of tickets the specialist must offer to have a 95% chance of offering at the slightest one prize-winning ticket. In other words, we need to discover the least esteem of n such that:

- The likelihood of not winning with n tickets is less than or rises to 5% (or 0.05).

- This can be identical to (499/500)[tex]^n[/tex]≤ 0.05.

We will illuminate for n utilizing logarithms:

- Take the normal logarithm of both sides:

ln[(499/500)[tex]^n[/tex]] ≤ ln(0.05).

- Utilize the logarithmic property of types to bring down the example:

n ln(499/500) ≤ ln(0.05).

- Fathom for n by partitioning both sides by ln(499/500) and taking the ceiling work (since n must be an entire number):

n = ceil[tex][ln(0.05) / ln(499/500)].[/tex]

Employing a calculator, we get 

n = ceil[tex][ln(0.05) / ln(499/500)][/tex]≈ 368

In this manner, the specialist must offer at the slightest 368 tickets to have a 95% chance of offering at the slightest one prize-winning ticket. 

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To solve for f, we can rearrange the equation:

1975f = t1758 + Q2

First, we can isolate f by dividing both sides of the equation by 1975:

f = (t1758 + Q2) / 1975

So the final answer is:

f = (t1758 + Q2) / 1975