The whole page Tysm plsssssss step by step btw

The probability that the first roll is a total of at least 5 and the second roll is a total of at least 4  is 0.764 or 76.4%.

The probability is the chance that an expected event occurs out of many possible events.

For two independent events happening, we use the rule of multiplication of independent events based on dice probabilities.

The event that the first roll is a total of at least 5 = A

In the event that the second roll is a total of at least 4 = B

The number of faces on each die, n = 6

The probability of event A = ³⁰/₆ = ⁵/₆

The probability of event B = ³³/₃₆ = ¹¹/₁₂

Multiplying the two probabilities, P(A) and P(B) = ⁵/₆ ​× ¹¹/₁₂

= ⁵⁵/₇₂ = 0.764

Thus, there is a ⁵⁵/₇₂ or 0.764 chance that the first roll is at least 5 and the second roll is at least 4.

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Answer:10.66 years. Punch again to double check

Step-by-step explanation:

A=Pe^(rt)

11700=5850e^(.065t)

isolate the variable expression

2=e^(.065t) divide by 5850

ln2=lne^(.065t) take the natural log of both sides

ln2=.065t. Bring down the power, lne equals

t is about 10.66 years. Divide ln2 by .065

The sum expression [tex][2(4 + \frac9n)^4](\frac9n) + ... + [2(4 + \frac{9n}n)^4](\frac9n)[/tex] using the sigma notation is [tex]\sum\limits^{n}_{i=1} [2(4 + \frac{9i}n)^4](\frac9n)[/tex]

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

[tex][2(4 + \frac9n)^4](\frac9n) + ... + [2(4 + \frac{9n}n)^4](\frac9n)[/tex]

From the above expression, we can see that the different expression in each term is

9/n

So, we introduce a variable

Assuming the variable is i, the i-th term of the expression would be[tex]t(i) = [2(4 + \frac{9i}n)^4](\frac9n)[/tex]

When represented using the sigma notation, we have

[tex]\sum\limits^{n}_{i=1} [2(4 + \frac{9i}n)^4](\frac9n)[/tex]

Hence, the sum expression using the sigma notation is [tex]\sum\limits^{n}_{i=1} [2(4 + \frac{9i}n)^4](\frac9n)[/tex]

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An estimate for the mean time spent watching television is approximately 24.375 hours.

Midpoint of 0 < h ≤ 10 = (0 + 10) / 2 = 5

Midpoint of 10 < h ≤ 20 = (10 + 20) / 2 = 15

Midpoint of 20 < h ≤ 30 = (20 + 30) / 2 = 25

Midpoint of 30 < h ≤ 40 = (30 + 40) / 2 = 35

Sum of observations = (5 × 5) + (15 × 15) + (25 × 50) + (35 × 10) = 125 + 225 + 1250 + 350 = 1950

Total number of observations = 5 + 15 + 50 + 10 = 80

Mean = Sum of observations / Total number of observations = 1950 / 80 = 24.375

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The function shows that the chairs and sofas the manufacturer should produce to earn that profit is C. (6, 12) P= 1,260

The profit function is P = 50x + 80y.

In order to find the optimal production quantities that will maximize the profit function, we need to determine which vertex yields the highest profit.

From the given vertices, we can see that the profit at (0,0) is 0. The profit at (0,15) is 1,200, and the profit at (6,12) is 1,260.

Therefore, the manufacturer should produce 6 chairs and 12 sofas to earn a profit of 1,260.

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The area of the shaded region is 51.75 in².

From the figure

Radius of semicircle = √25² - 20²

                                  =√625-  400 = 15 inch

Now, Area of Trapezium = 1/02 ( a +b ) h

= 1/2 (17 + 37) x 15

= 1/2 (54) x 15

= 27 x 15

= 405 in²

Then, Area of Shaded region

= Area of Trapezium -  Area of semicircle

= (405 - πr²)

= 405 - 353.25

= 51.75 in²

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The correct numbers for Derrek and Thomas' football teams are :

According to the information given, Thomas's football team lost 3 yards in their first drive, which means they moved backward 3 yards from the line of scrimmage. This can be represented as -3 on the number line.

On the other hand, Derrek's football team gained 5 yards in their first drive, which means they moved forward 5 yards from the line of scrimmage. This can be represented as +5 on the number line.

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Answer:

yes

Step-by-step explanation:

the factor theorem says if (x + a) is a factor of a polynomial f(x) then f(- a) = 0

given (x+ 5) then if f(- 5) = 0 , (x + 5) is a factor

4[tex](-5)^{4}[/tex] + 16(- 5)³ - 6(- 5)² + 73(- 5) + 15

= 4(625)+ 16(- 125) - 6(25) - 365 + 15

= 2500 - 2000  - 150 - 350

= 500 - 500

= 0

then (x + 5) is a factor of the given polynomial

The correct option regarding which bus has the least spread among the travel times is given as follows:

Bus 14, with an IQR of 6.

To obtain measures of spread, we can use the dot plot to visualize the frequency of each observation in the data-set, and then calculate the interquartile range (IQR).

For a group of 15 students, we can find the quartiles as follows:

The first quartile (Q1) is the median of the first half of the data-set, which consists of the first seven students.

Looking at the dot plot, the fourth dot represents the median of the first half. For Bus 14, Q1 = 12. For Bus 18, Q1 = 9.

The third quartile (Q3) is the median of the second half of the data-set, which consists of the last seven students. Looking at the dot plot, the eleventh dot represents the median of the second half. For Bus 14, Q3 = 18.

For Bus 18, Q3 = 16.

We can then calculate the IQR as the difference between Q3 and Q1:

For Bus 14, IQR = Q3 - Q1 = 18 - 12 = 6.

For Bus 18, IQR = Q3 - Q1 = 16 - 9 = 7.

Based on the IQR values, we can conclude that Bus 14 is more consistent than Bus 18, as it has a lower IQR.

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The minimum grade you need to get a B in the course is given as follows:

x = 80.625.

The mean of a data-set is given by the sum of all observations in the data-set divided by the number of observations, which is also called the cardinality of the data-set.

The grades and their percentages are given as follows:

You want a grade of 80, hence the minimum test average is obtained as follows:

0.15 x 73 + 0.2 x 79 + 0.25 x 84 + 0.4x = 80

47.75 + 0.4x = 80

x = (80 - 47.75)/0.4

x = 80.625.

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By making the financially irresponsible decision of maxing out your credit card limit, you might not have money for an emergency situation.

As, There is no foolproof technique to determine how much you can spend in excess of your credit limit, but you can go over it.

When determining the risk of accepting an over-the-limit transaction, card issuers may take into account a number of criteria, including your prior payment history.

Now, maxing out credit card leads to negative results.

So, either not have money for an emergency situation or not be able to buy the things you need.

But we can still spend after maxing out the limit.

Thus, not have money for an emergency situation is correct option.

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Answer:

Positive Association

Complementary Angles

Negative Association

Clusters in Data

Supplementary Angle

Non-linear Association

Step-by-step explanation:

Match the boxes on the left to the answers:

Positive Association

Complementary Angles

Negative Association

Clusters in Data

Supplementary Angle

Non-linear Association

The calculated value of the length RP is 107 units

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

The circles and the tangent lines

Start by calculating the segment PQ using the following Pythagoras theorem

PQ² = (50 + 11)² + 11²

Evaluate

PQ² = 3842

So, we have

PQ = 62

The value of the length RP is then calculated as

RP = PQ + TU

So, we hav

RP = 62 + 45

Evaluate

RP = 107

Hence, the value of the length RP is 107 units

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The graph of the piecewise function is given by the image presented at the end of the answer.

A piece-wise function is a function that has different definitions, depending on the input of the function.

All the five definitions of the function in this problem are constant, hence the graph will be composed by horizontal lines.

The definitions are given as follows:

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Answer:

[tex]sec(-\theta)=-3.1[/tex]

[tex]sin(-\theta)=-0.62[/tex]

Step-by-step explanation:

Part a.

[tex]sec(\theta)[/tex] is related to [tex]cos(\theta)[/tex] through a reciprocal relationship [tex]sec(\theta)=\frac{1}{cos(\theta)}[/tex].

Since [tex]cos(\theta)[/tex] is an even function (reflecting left-right across the y-axis doesn't change the graph), then [tex]sec(\theta)[/tex] is also an even function.

For any input in the domain, even functions produce the same output if the opposite of the input is used.  In other words, if "f" is an even function, for all x in the domain of f, [tex]f(x)=f(-x)[/tex].

Thus, for the secant function, if a mystery value "x" is used as an input, and -3.1 is obtained as an output, then if the opposite of x, or -x, is input into the secant function, the output will also be -3.1.

[tex]sec(-\theta)=-3.1[/tex]

Part b.

The [tex]sin(\theta)[/tex] function does not reflect left-right across the y-axis to produce the same graph, so the sine function is not even.

However, the sine function can be rotated 180 degree about the origin, sometimes thought of as reflecting through the origin, to produce the same graph.  Visually, this is an "odd" function.

For any input in the domain, if the opposite of the input is used, odd functions produce the opposite of the original output.  In other words, if "g" is an odd function, for all x in the domain of g, [tex]-g(x)=g(-x)[/tex].

Thus, for the sine function, if a mystery value "x" is used as an input, and 0.62 is obtained as an output, then if the opposite of x, or -x, is input into the sine function, the output will be the opposite of 0.62, meaning -0.62.

[tex]sin(-\theta)=-0.62[/tex]

Sum

Mean = (4.1 + 3.2 + 2.8 + 2.6 + 3.7 + 3.1 + 9.4 + 2.5 + 3.5 + 3.8) / 10

Mean = 36.7 / 10

Mean = 3.67

To find the median, we need to put the values in order:

2.5, 2.6, 2.8, 3.1, 3.2, 3.5, 3.7, 3.8, 4.1, 9.4

The middle number is the median, which is 3.35 in this case (3.2+3.5)/2.

To find the mode, we look for the value that appears most often. In this case, there is no mode as no value appears more than once.

To find the first quartile (Q1), we need to find the value that separates the bottom 25% of the data from the top 75%. We can do this by finding the median of the lower half of the data:

2.5, 2.6, 2.8, 3.1, 3.2

The median of this lower half is 2.8, so Q1 = 2.8.

To find the third quartile (Q3), we need to find the value that separates the bottom 75% of the data from the top 25%. We can do this by finding the median of the upper half of the data:

3.7, 3.8, 4.1, 9.4

The median of this upper half is 3.95, so Q3 = 3.95.

To find the outlier, we can use the rule that any value more than 1.5 times the interquartile range (IQR) away from the nearest quartile is considered an outlier. The IQR is the difference between Q3 and Q1:

IQR = Q3 - Q1

IQR = 3.95 - 2.8

IQR = 1.15

1.5 times the IQR is 1.5 * 1.15 = 1.725.

The only value that is more than 1.725 away from either Q1 or Q3 is 9.4. Therefore, 9.4 is the outlier in this data set.

Answer:

Step-by-step explanation:

Mean is the average of the data vales : 38.7 (sum of all values) divided by 10 (the number of values).  Mean = 38.7/10 = 3.87

Median is the "middle" number" = put the date in order and find the middle value:

2.5, 2.6, 2.8, 3.1, 3.2, 3.5, 3.7, 3.8, 4.1, 9.4, since there is no middle data value, find the average of the 2 in the middle.

3.2 + 3.5/2 = 6.7 ÷ 2 = 3.35 - median

All values appear once so there is no mode.

First quartile is the middle of the lower set of data = 2.8

Third quartile is the middle of the upper set = 3.8

The outlier is 9.4.

How to calculate the outlier:

First you need the IQR which is the diffence of Q3 - Q1,

so the IQR is 3.8- 2.8 = 1

Outliers are the quartile + or - (1.5)(IQR)

Q1 -(1.5)(1) = 2.8 - 1.5 = 1.3

Q3 - (1.5)(1)= 3.8 + 1.5 =4.5

So anything below 1.3 or above 4.5 is an outlier.

There is one 9.4

The given polynomial is a prime polynomial.

The given polynomial is

x² - x + 35

Since we know that discriminant of quadratic polynomial

p(x)= ax² + bx + c  is b² - 4ac

Therefore, discriminant of the given polynomial be,

⇒ (-1)² - 4x1x35

⇒ - 139

Since discriminant is negative number so it has no real roots.

And an irreducible or primal polynomial is a polynomial with integer coefficients that cannot be factored into polynomials of lower degree and also with integer coefficients.

Hence it is not factored completely.

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The integral of  [tex]\int\limits^e_1 {x^{1/2}lnx } \, dx[/tex]  is (2/3)[tex]e^{3/2}[/tex] - (4/9).

To find the integral of   [tex]\int\limits^e_1 {x^{1/2}lnx } \, dx[/tex]  , we can use integration by parts. Let u = ln x and dv = [tex]x^{1/2}[/tex] dx. Then du/dx = 1/x and v = (2/3) [tex]x^{3/2}[/tex] .

Using the formula for integration by parts, we have:

∫[tex]x^{1/2}[/tex]lnx dx = uv - ∫v du/dx dx

= ln x * (2/3) [tex]x^{3/2}[/tex]  - ∫(2/3) [tex]x^{3/2}[/tex]  * (1/x) dx

= ln x * (2/3) [tex]x^{3/2}[/tex]  - (2/3) ∫ [tex]x^{1/2}[/tex]  dx

= ln x * (2/3) [tex]x^{3/2}[/tex]  - (4/9) [tex]x^{3/2}[/tex]  + C

where C is the constant of integration.

To evaluate the definite integral from 1 to e, we substitute e for x in the expression above and subtract the result when x = 1:

 [tex]\int\limits^e_1 {x^{1/2}lnx } \, dx[/tex]   = [(2/3) [tex]e^{3/2}[/tex] ln e - (4/9) [tex]e^{3/2}[/tex] ] - [(2/3)[tex]1^{3/2}[/tex]ln 1 - (4/9)[tex]1^{3/2}[/tex]]

= (2/3) [tex]e^{3/2}[/tex] - (4/9)

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Answer:

False.

Step-by-step explanation:

A integer in math is defined as a whole number. Therefore, the presence of fractions (which, in this case, is 1/2) and decimals, would mean that the term is in fact, not a integer.

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Answer:

C

Step-by-step explanation:

if the mean and median are the same, the graph will be symmetrical

Answer:

2(4x-3)

Step-by-step explanation:

Find common numbers between 8 and 6, which is 2. 2x4= 8 2x3=6

x is only on 8 so you can't put x on the outside of the bracket. so you put it with the 4 so it comes out correct

The tangent secant exterior angle measure theorem indicates that the measure of the arc DG, m[tex]\widehat{DG}[/tex] is 85°, the correct option is the option B.

B. m[tex]\widehat{DG}[/tex] = 85°

The tangent secant exterior angles theorem states that if two secant or tangent or a secant and tangent intersect on the exterior to a circle, the measure of the angle formed is half the difference of the arcs they intercept.

The possible options in the question indicates that the possible drawing in the consists of a tangent and a secant, with the angle m∠DBG, being external to the circle.

The tangent secant exterior angle measure theorem indicates that we get;

m∠DBG = (1/2) × (m[tex]\widehat{GFE}[/tex] - m[tex]\widehat{DG}[/tex])

The question indicates; m[tex]\widehat{GFE}[/tex] = 149°

m∠DBG = 32°

Therefore;

32 = (1/2) × (149° - m[tex]\widehat{DG}[/tex])

149° - m[tex]\widehat{DG}[/tex] = 2 × 32° = 64°

m[tex]\widehat{DG}[/tex] = 149° - 64° = 85°

The measure of the arc DG, m[tex]\widehat{DG}[/tex] is 85°

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Here is the data collected as a result of the experiment.
1) Numerical order is: 0.20, 0.20, 0.46, 0.46, 2.10, 2.10, 5.50, 11.80, 12.20, 12.20, 13.22, 14.65, 16.07, 17.49, 18.91, 20.33, 20.50, 20.50, 21.75, 23.18

2)  The data represents rainfall in millimeters over a 20-day period.

3) Mean = 11.69 mm.

4) The minimum rainfall is 0.20 mm ;

the maximum is 23.18 mm.

1) The data is listed as follows.

2) The data represents the amount of rainfall in millimeters over a 20 day period.

3) The mean rainfall over the 20 day period is 11.69 mm.

= (0.20 + 0.20 + 0.46 + 0.46 + 2.10 + 2.10 + 5.50 + 11.80 + 12.20 + 12.20 + 13.22 + 14.65 + 16.07 + 17.49 + 18.91 + 20.33 + 20.50 + 20.50 + 21.75 + 23.18) /20

Mean (x)  = 11.69 mm.

3) The minimum rainfall is 0.20 mm and the maximum is 23.18 mm.

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Answer:

Step-by-step explanation:

We are given a figure, consisting of a right triangle with a circle cut out of it.

To Calculate : area of the shaded region below,

First Let's find the Area of the right triangle:

Area(triangle) = 1/2 × Base × Height

= 1/2 × 56 × 56

= 1/2 × 3136

= 3136/2

= 1568 m².

Now, We are diameter of the circle = 20 m

Radius of the circle = 20/2 = 10 m

Acc. to question we have to use 3.14 as an

approximation for pi.

Area (circle) = 3.14 × 10 × 10

Area (circle) = 3.14 × 100

Area (circle) = 314 m²

Area of shaded region = Area of triangle - Area of circle =

= 1568 m² - 314 m²

= 1.254 square meters

Therefore, The approximate area of the shaded region is 1.254 square meters.

Option (a) is the required answer

Answer: First, let's convert the room's dimensions from feet to yards, since the carpet cost is given per square yard:

17 feet = 5.67 yards (rounded to two decimal places)

16 feet = 5.33 yards (rounded to two decimal places)

Next, we can calculate the area of the room in square yards by multiplying these values:

Area = 5.67 yards × 5.33 yards = 30.23 square yards (rounded to two decimal places)

Finally, we can multiply the area by the cost per square yard to get the total cost:

Total cost = 30.23 square yards × $101/square yard = $3,056.23 (rounded to two decimal places)

Therefore, it will cost $3,056.23 to carpet the room.

Answer: D

Step-by-step explanation: 60x9=540 and 540x80=43,200

Kyle has $306 in his account so, right answer is Option D.

Let Kyle's money = x

Hayden's money = y

Jeremy's money = z

Given that Kyle has three times as much money in his savings account as  Hayden.

x = 3y   .....Eqn 1

Hayden has twice as much money in his savings account  as Jeremy.

z = 2y  ..... Eqn 2

Now Combined Kyle, Hayden, and Jeremy have a total of  $612.

x+y+z=612   .....Eqn 3

Putting value of x and z from Eqn 1 and 2 in Eqn 3

6y = 612

y = 106

Putting in Eqn 1;

x = 306

Thus, Kyle has $306 in his account.

The right answer is Option D.

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Answer:

Let the width of the rectangle be w. Then the length of the rectangle is 5w.

The perimeter of a rectangle is the sum of all four sides. So, the perimeter of the rectangle is 2w+2(5w)=108 cm.

Simplifying the right side of the equation, we get 12w=108 cm.

Dividing both sides of the equation by 12, we get w=9 cm.

Since the width is 9 cm, the length of the rectangle is 5w=5(9)=45 cm.

Therefore, the length of the rectangle is 45 cm and the width is 9 cm.

1. It would be better to use the median to describe the data set.

2. It would be better to use the median to describe the data set.

3. It would be better to use the mean to describe the data set as the average salary of the workers is to be calculated.

4. It would be better to use the mean to describe the data set as the average test grade of the class is to be calculated.

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