50 Points! Multiple choice geometry question. Photo attached. Thank you!

Answer:

1/2 of an hour ( I think )

Step-by-step explanation:

The second longest time that was recorded was 2/3/4 (whole/numerator/denominator) of a hour while the second shortest time was 2/1/4 of a hour. When you subtract 2/1/4 from 2/3/4 you end up with 2/4. Since 2/4 can be simplified to 1/2 you would say there is a 1/2 hour different between the second longest and second shortest time spent reading.

The missing values in the quantitative reasoning given are : -2, 13 and 9

Given the rule :

We can deduce that :

For the left circle :

circle = -6 - (-4) = -6 + 4 = -2

For the right circle :

circle = 11 - (-2) = 11 + 2 = 13

For the left square :

square = 13 + (-4)

square = 13 -4 = 9

Therefore, the missing values are : -2, 13 and 9

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To find the center and radius of the circle represented by the inequality [tex]\displaystyle \sf x^{2} +y^{2} -12y-12\leq 0[/tex], we can complete the square for the y terms.

The inequality can be rewritten as:

[tex]\displaystyle \sf x^{2} +( y^{2} -12y) -12\leq 0[/tex]

To complete the square for the y terms, we need to add and subtract [tex]\displaystyle \sf ( 12/2) ^{2} =36[/tex] inside the parentheses:

[tex]\displaystyle \sf x^{2} +( y^{2} -12y+36) -36-12\leq 0[/tex]

Simplifying, we have:

[tex]\displaystyle \sf x^{2} +( y-6)^{2} -48\leq 0[/tex]

Now we can rewrite the inequality in the standard form of a circle equation:

[tex]\displaystyle \sf ( x-h)^{2} +( y-k)^{2} \leq r^{2}[/tex]

Comparing this with the obtained equation, we can identify the center and radius of the circle:

Center: [tex]\displaystyle \sf ( h,k)=( 0,6)[/tex]

Radius: [tex]\displaystyle \sf r=\sqrt{48}[/tex]

Therefore, the center of the circle is at [tex]\displaystyle \sf ( 0,6)[/tex], and its radius is [tex]\displaystyle \sf \sqrt{48}[/tex].

[tex]\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}[/tex]

♥️ [tex]\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}[/tex]

[tex]\boxed{A}\\\\ U=5(2n+22)+2\left( n+\cfrac{3}{2} \right) \\\\\\ U=10n+110+2n+3 \implies U=12n+113 \\\\[-0.35em] ~\dotfill\\\\ \boxed{B}\hspace{5em}\textit{in 2009, that's 9 years after 2000, n = 9}\\\\ U(9)=12(9)+113\implies U(9)=221 ~~ millions[/tex]

Answer:

7

Step-by-step explanation:

The line represented by the equation y = 7 is a horizontal line that passes through the y-axis at 7, so the y intercept of this line is 7

Answer:

only god knows

Step-by-step explanation:

because they didn't give us an answer on how many text messages anyone sent

The required probability is 3 √7 / 32.

Given, a square with sides of length 4 units and a red-shaded triangle with sides 3 units, 3 units and 4 units. We need to find the probability that a randomly selected point within the square falls in the red-shaded triangle.To find the probability, we need to divide the area of the red-shaded triangle by the area of the square. So, Area of square = 4 × 4 = 16 square units. Area of triangle = 1/2 × base × height.

Using Pythagorean theorem, the height of the triangle is found as: h = √(4² − 3²) = √7

The area of the triangle is: A = 1/2 × base × height= 1/2 × 3 × √7= 3/2 √7 square units. So, the probability that a randomly selected point within the square falls in the red-shaded triangle is:  P = Area of triangle/Area of square= (3/2 √7) / 16= 3 √7 / 32.

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At a temperature of 20°C, the mouse would burn approximately 1,567.44 calories each day.

According to the given model C = 0.37219T + 1,560, where C represents the number of calories an idle mouse burns each day and T represents the temperature of its environment in °C.

To find the most comfortable temperature for an idle mouse, we need to determine the temperature at which the mouse burns the least amount of calories per day.

To find this temperature, we can minimize the equation C = 0.37219T + 1,560. To do so, we take the derivative of C with respect to T and set it equal to zero:

dC/dT = 0.37219 = 0

Solving this equation, we find that the derivative is a constant value, indicating that the function C = 0.37219T + 1,560 is a linear equation with a slope of 0.37219. This means that the mouse burns the least calories at any temperature, as the slope is positive.

Therefore, there is no specific "most comfortable" temperature for an idle mouse in terms of minimizing calorie burn. However, if we consider the range of temperatures mice typically encounter, we can find a temperature where the calorie burn is relatively low.

For example, if we take a temperature of 20°C, we can calculate the calorie burn:

C = 0.37219 * 20 + 1,560

C = 7.4438 + 1,560

C ≈ 1,567.4438 calories per day

Therefore, at a temperature of 20°C, the mouse would burn approximately 1,567.44 calories each day.

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The length of the other diagonal is 11.25 cm.

Area is the measure of a region's size on a surface. The area of a plane region or plane area refers to the area of a shape or planar lamina, while surface area refers to the area of an open surface or the boundary of a three-dimensional object.

In this question, we are given the following:

The area of a kite is 180. One of the diagonals is 16.

What is the length of the other diagonal?

The details of the solution are as follows:

We know that,

The area of a kite is the product of the diagonals divided by 2:

[tex]\text{A} = \dfrac{(\text{d}^1 \times \text{d}^2)}{2}[/tex]

You can substitute what we have:

[tex]180= \dfrac{(16 \times \text{d}^2)}{2}[/tex]

And solve.

[tex]180 = 16 \times \text{d}^2[/tex]

[tex]\text{d}^2=\dfrac{180}{16}[/tex]

[tex]\text{d}^2=\bold{11.25 \ cm}[/tex]

Therefore, the length of the other diagonal = 11.25 cm.

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Out of the four children attending the party, with only two spots left, there are six different ways to select two children to fill those spots.

If there are four children and only two spots left at the party, we need to determine the number of combinations possible for selecting two children out of the four. To calculate this, we can use the concept of combinations from combinatorics.

Combinations refer to the selection of items from a larger set without considering the order. In this case, the order in which the children are selected does not matter; we only need to know which two children are chosen. The formula for combinations is given by:

C(n, r) = n! / (r! * (n - r)!)

where n is the total number of items (in this case, children) and r is the number of items we want to select (in this case, the two available spots at the party).

Using the formula, we can substitute n = 4 and r = 2:

C(4, 2) = 4! / (2! * (4 - 2)!)

Simplifying further:C(4, 2) = 4! / (2! * 2!)

Now, let's calculate the factorial terms:

4! = 4 * 3 * 2 * 1 = 24

2! = 2 * 1 = 2

Substituting the factorial terms:

C(4, 2) = 24 / (2 * 2)

Simplifying the denominator:

C(4, 2) = 24 / 4 = 6

Therefore, there are 6 different combinations possible for selecting two children out of the four to fill the two available spots at the party.

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Note the correct question is

4 childen go to a party but there is only 2 spots left. How many combinations  are there?

Answer:

41)  Yes, the relation is a function.

42)  The domain of the function is [-2, 4].

43)  The range of the function is [-1, 3].

Step-by-step explanation:

A relation is a set of ordered pairs where each input (x-value) is associated with one or more outputs (y-values).

A function is a special type of relation where each input (x-value) is associated with exactly one output (y-value).

We can determine if a graphed relation is a function by applying the Vertical Line Test. It states that if a vertical line intersects the graph at more than one point, then the relation does not pass the test and is not a valid function.

As the given graph passes the Vertical Line Test, the relation is a function.

[tex]\hrulefill[/tex]

The domain of a function is the set of all possible input values (x-values).

The range of a function is the set of all possible output values (y-values).

From inspection of the given graph, the continuous curve begins in quadrant II at point (-2, 1) and ends in quadrant IV at point (4, -1).

The endpoints of the graph are represented by closed circles, which means that the corresponding x and y values are included in the domain and range.

Therefore, the domain of the function is the x-values of the endpoints: [-2, 4].

The minimum point of the curve is endpoint (4, -1) and the maximum point is (0, 3). Therefore, the range of the function is the y-values of the minimum and maximum points: [-1, 3].

Terri have to react 1.42 seconds before the volleyball hits the ground.

Quadratics can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations. The general form of the quadratic equation is:

[tex]\text{ax}^2 + \text{bx} + \text{c} = 0[/tex]

Given data:

The height can be modeled by a quadratic equation.

[tex]h(t)=-16t^2+v_0t+h_0[/tex]

Where h is the height and t is the time.

[tex]h(t)=-16t^2+19.5t+4.5[/tex]

[tex]-16t^2+19.5t+4.5=0[/tex]

[tex]a = -16, b = 19.5, c = 4.5[/tex]

It looks like a quadratic equation. we can solve it by quadratic formula.

[tex]\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]\rightarrow t=\dfrac{-19.5\pm\sqrt{(-19.5)^2-4\times(-16)(4.5)} }{2(-16)}[/tex]

[tex]\rightarrow t=\dfrac{-19.5\pm\sqrt{380.25+288} }{-32}[/tex]

[tex]\rightarrow t=\dfrac{-19.5\pm25.851 }{-32}[/tex]

[tex]\rightarrow t=\dfrac{-19.5-25.851 }{-32}, \ t=\dfrac{-19.5+25.851 }{-32}[/tex]

[tex]\rightarrow t=1.42, \ t=-0.20[/tex]

Time cannot be in negative. So neglect t = –0.235.

Hence, Terri have to react 1.42 seconds before the volleyball hits the ground.

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Using the physics concept of projectile motion and inputting the given values into the appropriate equation, we can determine the time it takes for the volleyball to hit the ground after being served

This question is a classic use of physics, more specifically, the concept of projectile motion. Here, the volleyball can be conceived as a projectile. When Patricia serves the ball upward, the ball will first ascend and then descend due to gravity.

Let's use the following equation which is a version of kinematic equations to solve this problem, adjusting for the fact that we're dealing with an initial height of 4.5 ft and an ending height of 0 ft (when the ball hits the ground). The equation y = yo + vot - 0.5gt² , where:

Setting y=0, yo=4.5 feet, vo=19.5 feet/second, and g=32.2 feet/second², and plug these values into the equation, we'll get a quadratic equation in the form of 0 = 4.5 + 19.5t - 16.1t². Solve that equation for t to find the time it takes for the ball to hit the ground.

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Answer:

The values of b that satisfy the equation are:

b = (2√3 - 3) / 2

b = (-2√3 - 3) / 2

In other words, b can take the values (2√3 - 3) / 2 or (-2√3 - 3) / 2.

Step-by-step explanation:

To find the values of b that satisfy the equation 3(2b + 3)² = 36, we can solve for b by following these steps:

1. Divide both sides of the equation by 3:

  (2b + 3)² = 12

2. Take the square root of both sides:

  √[(2b + 3)²] = √12

  Simplifying further:

  2b + 3 = ±√12

3. Subtract 3 from both sides:

  2b = ±√12 - 3

4. Divide both sides by 2:

  b = (±√12 - 3) / 2

  Simplifying further:

  b = (±√4 * √3 - 3) / 2

  b = (±2√3 - 3) / 2

Therefore, the values of b that satisfy the equation are:

b = (2√3 - 3) / 2

b = (-2√3 - 3) / 2

In other words, b can take the values (2√3 - 3) / 2 or (-2√3 - 3) / 2.

The y-intercept of the line of best fit can be interpreted as the predicted test grade when no time is spent on homework, which in this case is approximately 72. However, it is important to consider the limitations and potential sources of error in any statistical analysis.

In statistics, linear regression is a commonly used statistical method for analyzing the relationship between two variables, such as time spent on homework and test grades. A line of best fit, also known as a regression line, is a line that summarizes the linear relationship between the variables. In this case, the line of best fit has an equation of y = 7.9 x + 72.
The y-intercept of the line is the point where the line intersects with the y-axis. It represents the value of y when x is equal to zero. In other words, it is the predicted test grade when no time is spent on homework. According to the given equation, the y-intercept is 72. This means that if a student spends no time on homework, they can still expect to receive a test grade of 72.
However, it is important to note that this interpretation assumes that the line of best fit is an accurate representation of the relationship between time spent on homework and test grades. Additionally, there may be other variables that influence test grades, such as innate ability, test-taking skills, or external factors like test anxiety or distractions during the exam.
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Given that the side of a square field is 52 m so, the area of the square field is 2704 m².

The side of a square field is given as 52 m.

Now, Let’s find the area of the square field using the given information.

As we know, area of a square can be calculated by using the formula:

A = a², where ‘a’ is the side of the square.

Now, by substituting the given value of ‘a’ in the given formula above we will get the area of the square field as,

A = (52)²

A = 2704 m²

Therefore, the area of the square field of given side i.e. 52m is 2704 m².

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The box plot D is the correct box plot that displays the given data.

To find the correct box plot that displays the given data, we need to first determine the five-number summary of the data: minimum value, first quartile (Q1), median (Q2), third quartile (Q3), and maximum value.

The minimum value is 8, and the maximum value is 18. To find the quartiles, we need to first determine the median of the data set. Since there are 18 data points, the median is the average of the middle two values:

(11 + 11)/2 = 11

The median is 11.

To find Q1, we look at the median of the lower half of the data set:

(9 + 9)/2 = 9

Q1 is 9.

To find Q3, we look at the median of the upper half of the data set:

(16 + 17)/2 = 16.5

Q3 is 16.5.

Now that we have the five-number summary, we can compare the box plots to see which one is correct.

Box plot A has the correct minimum and maximum values, but the median (Q2) is too high and the whiskers are not the correct length.

Box plot B has the correct median and whisker length, but the minimum value is too low.

Box plot C has the correct minimum, median, and whisker length, but the maximum value is too high.

Box plot D has the correct minimum, median, and maximum values, as well as the correct whisker length. (d)

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The inequality tha calculates the possible values of x is x < 2

From the question, we have the following parameters that can be used in our computation:

The figure

Where, we have

3x + 2 < 10

And, we have

2x + 6 < 10

Evaluate the expressions

So, we have

3x < 8 and 2x < 4

Evaluate

x < 8/3 and x < 2

Hence, the inequality tha calculates x is x < 2

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Answer:  [tex]\frac{8}{5}[/tex]  

Step-by-step explanation:

You can put it into a ratio when the units are the same.

Let's convert 15 ft to yards

3 ft  = 1 yd

So divide 15 by 3

15ft  = 5 yd

So the ratio you want is:

8 yds to 5 yds

To write that in fraction ratio form:

[tex]\frac{8}{5}[/tex]               >Keep as improper fraction for ratios

8

In this pattern, we have 4 12 then 20.

We can see that the difference 4 and 12 is 8.

Since the difference between 12 and 20 is also 8, the common difference of the sequence is 8.

Work Shown:

21/35 = (7*3)/(7*5) = 3/5

The simplified expression is [tex]-4x^2 + 3x + 1[/tex], which is a quadratic polynomial. Option D.

To simplify the expression [tex](8x^2 + 3x) - (12x^2 - 1)[/tex], we can distribute the negative sign to each term within the parentheses:

[tex]8x^2 + 3x - 12x^2 + 1[/tex]

Next, we can combine like terms by adding or subtracting coefficients of similar powers of x:

[tex](8x^2 - 12x^2) + 3x + 1[/tex]

Simplifying further, we have:

[tex]-4x^2 + 3x + 1[/tex]

The resulting polynomial [tex]-4x^2 + 3x + 1[/tex] is a quadratic polynomial since it has a highest power of x^2 (the exponent of x is 2), which is[tex]-4x^2.[/tex]Quadratic polynomials are polynomials of degree 2 and can be represented by a parabola when graphed.

In summary, the simplified expression [tex](8x^2 + 3x) - (12x^2 - 1)[/tex] simplifies to [tex]-4x^2 + 3x + 1[/tex] , which is a quadratic polynomial. So Option D is correct

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Note the complete question is

Step-by-step explanation:

To find the coordinates of the focus of a parabola in the general form y = ax^2 + bx + c, you can use the formula (h, k) where h = -b/(2a) and k = (4ac - b^2)/(4a).

In the given equation y = -18x^2 - 2x - 4, we can identify that a = -18, b = -2, and c = -4. Plugging these values into the formulas, we get:

h = -(-2)/(2*(-18)) = 1/18

k = (4*(-18)*(-4) - (-2)^2)/(4*(-18)) = -71/9

Therefore, the focus of the parabola is (1/18, -71/9).

None of the given answer options match the coordinates of the focus calculated, so none of the options (A, B, C, D) are correct.

Rounding to three decimal places, the upper boundary of the 95% confidence interval for the population proportion of values greater than 24 is approximately G.0.961.Therefore, closest option is G: 0.925

To find the upper boundary of a 95% confidence interval for the population proportion of values greater than 24, we need to calculate the sample proportion and then use the appropriate distribution table.

First, we count the number of values in the sample that are greater than 24. From the given sample, we have 12 values greater than 24.

Next, we calculate the sample proportion by dividing the count of values greater than 24 by the total number of values in the sample:

Sample proportion = (Number of values greater than 24) / (Total number of values in the sample)

= 12 / 16

= 0.75

Now, we need to use the appropriate distribution table to find the upper boundary of the confidence interval. Since the sample size is relatively small (16), we can use the t-distribution table.

For a 95% confidence level with a two-tailed test, we need to find the critical value corresponding to an alpha of 0.025 (0.05 divided by 2) and degrees of freedom (df) of n-1 = 16-1 = 15.

Using the t-distribution table, the critical value for a 95% confidence level with 15 degrees of freedom is approximately 2.131.

Finally, we can calculate the upper boundary of the confidence interval:

Upper boundary = Sample proportion + (Critical value * Standard error)

= 0.75 + (2.131 * sqrt((0.75 * (1-0.75)) / 16))

≈ 0.75 + (2.131 * 0.099)

≈ 0.75 + 0.211

≈ 0.961

Rounding to three decimal places, the upper boundary of the 95% confidence interval for the population proportion of values greater than 24 is approximately 0.961. Therefore, closest option is G: 0.925

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answer: 13/2 or 6 1/2

step-by-step explanation:

hihi so basically your problem is making a solvable equation so w variables and stuff

heres my explanation !

the difference of a number and 2 is b-2

twice the difference of a number and 2 would be 2(b-2)

number 9 = 9 (duh lol)
so

2(b-2) = 9

2b - 4 = 9

2b = 13

improper: b = 13/2
mixed: b = 6 1/2

The graph of the function f(x)=√(x + 1) is in the first option

A radical graph, also known as a square root graph, represents the graph of a square root function. A square root function is a mathematical function that takes the square root of the input variable.

The general form of a square root function is f(x) = √(ax + b) + c,

where a, b, and c are constants that determine the characteristics of the graph.

In the given function:

a = 1

b = 1

c = 0

The graph is plotted and attached

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In the given question, it is stated that a premium candy manufacturer makes chocolate candies that, when finished, vary in color. The hue value of a randomly selected candy follows an approximately normal distribution with mean 30 and a standard deviation of 5. A quality inspector discards 12% of the candies due to unacceptable hues (which are equally likely to be too small or too large). So, the largest hue value that the inspector would find acceptable is 36.85

We are required to find out the largest hue value that the inspector would find acceptable. We are given that the mean value is 30, standard deviation is 5, and 12% of candies are discarded due to unacceptable hues. Now, we need to find out the largest hue value that the inspector would find acceptable.

To find the largest acceptable hue value we can use the Z score formula. Z = (X - μ) / σ

Now, substituting the values in the formula we have: Z = (X - 30) / 5

This value corresponds to the percentile of the distribution. We are required to find the largest hue value that the inspector would find acceptable and given that 12% of the candies are discarded due to unacceptable hues. So, the acceptable percentile would be 100% - 12% = 88% or 0.88

Now, using the z-score table or calculator, we can find the Z value corresponding to the 88th percentile. Z = 1.17

Now, we can use this Z score value to find the corresponding X value by using the Z-score formula and solving for X.1.17 = (X - 30) / 5

Solving for X,X = 30 + 5(1.17)X = 36.85. Therefore, the largest hue value that the inspector would find acceptable is 36.85 (rounded to two decimal places).

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The rocket is at a height of 1876.1 meters after 9 seconds,the velocity of the rocket after 9 seconds is 143.8 m/s and  the acceleration of the rocket after 9 seconds is -9.8 m/s².

To find the height of the rocket when it was initially launched, we can plug in t = 0 into the equation h(t) = -4.9t² + 232t + 185.

h(0) = -4.9(0)² + 232(0) + 185

     = 0 + 0 + 185

     = 185

Therefore, the rocket was initially launched at a height of 185 meters.

To find the height of the rocket after 9 seconds, we can plug in t = 9 into the equation h(t) = -4.9t² + 232t + 185.

h(9) = -4.9(9)² + 232(9) + 185

     = -4.9(81) + 2088 + 185

     = -396.9 + 2088 + 185

     = 1876.1

Therefore, the rocket is at a height of 1876.1 meters after 9 seconds.

To find the velocity of the rocket after 9 seconds, we can take the derivative of the height function h(t) with respect to time (t) and evaluate it at t = 9.

The velocity function v(t) is the derivative of h(t) with respect to t:

v(t) = dh/dt = d/dt(-4.9t² + 232t + 185)

       = -9.8t + 232

v(9) = -9.8(9) + 232

       = -88.2 + 232

       = 143.8

Therefore, the velocity of the rocket after 9 seconds is 143.8 m/s.

To find the acceleration of the rocket after 9 seconds, we can take the derivative of the velocity function v(t) with respect to time (t) and evaluate it at t = 9.

The acceleration function a(t) is the derivative of v(t) with respect to t:

a(t) = dv/dt = d/dt(-9.8t + 232)

       = -9.8

a(9) = -9.8

Therefore, the acceleration of the rocket after 9 seconds is -9.8 m/s².

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