Charles and Lisa were having a apple eating contest. They ate eighteen apples between the two of them. Lisa ate two more apples than Charles. How many apples did Lisa eat?

A.
6


B.
10


C.
12


D.
15

Answer:

P(Got a B) = [tex]\frac{19}{36}[/tex]

P(Female AND got a C) = [tex]\frac{1}{6}[/tex]

P(Female or Got an B =  [tex]\frac{7}{8}[/tex]

P(got a 'B' GIVEN they are male) =  [tex]\frac{9}{19}[/tex]

Step-by-step explanation:

Given:

       A B C Total

Male 6 18 3 27

Female 13 20 12 45

Total 19 38 15 72

We know that the probability = The number of favorable outcomes ÷ The total number of possible outcomes.

Total number = 72

1) If one student is chosen at random, Find the probability that the student got a B:

Got B = 38

P(Got a B) = [tex]\frac{38}{72}[/tex]

Simplifying the above probability, we get

P(Got a B) = [tex]\frac{19}{36}[/tex]

2) Find the probability that the student was female AND got a "C":

Female AND got a C = 12

P(Female AND got a C ) = [tex]\frac{12}{72}[/tex]

Simplifying the above probability, we get

P(Female AND got a C) = [tex]\frac{1}{6}[/tex]

3) Find the probability that the student was female OR got an "B":

Female OR got an B = Total number of female + students got B - Female got 20

= 45 + 38 - 20

= 73 - 20

= 63

P(Female OR got an B ) =  [tex]\frac{63}{72}[/tex]

P(Female or Got an B =  [tex]\frac{7}{8}[/tex]

4) If one student is chosen at random, find the probability that the student got a 'B' GIVEN they are male:

Total number of students who got B  = 38

Student got a B given they are male = 18

P(got a 'B' GIVEN they are male) =  [tex]\frac{18}{38}[/tex]

P(got a 'B' GIVEN they are male) =  [tex]\frac{9}{19}[/tex]

The probability study includes: Probability of a student getting 'B' is 0.528; a student being female and getting 'C' is 0.167; being female or getting 'B' is 0.875 and the probability of a male getting 'B' is 0.667.

To answer these questions, we need to use the concept of probability in mathematics. Probability is calculated by dividing the number of desired outcomes by the total number of outcomes.

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Answer:

The  equation  7 m + 15  m = 44 is the equation that can be used to determine the number of small cars rented and the number of large cars rented.

where, m : Number of large cars rented

Step-by-step explanation:

The number of people small car can hold  = 5

The number of people large car can hold  = 7

Let us assume the number of large cars rented  = m

So, the number of smaller cars rented = 3 x ( Number of large cars rented)

= 3 m

Now, the number of people in m large cars  = m x ( Capacity of 1 large car)

= m x ( 7)  = 7 m

And, the number of people in 3 m small  cars  = 3 m x ( Capacity of 1 small car) = 3 m x ( 5)  = 15  m

Total people altogether going for the plan = 44

⇒ The number of people in ( Small +Large) car  = 44

or, 7 m + 15  m = 44

Hence,  the equation  7 m + 15  m = 44 is the equation that can be used to determine the number of small cars rented and the number of large cars rented.

We define variable as

Let x be the number of small cars and y be the number of large cars.

Since ,

Each small car can hold 5 people and each large car can hold 7 people.

i.e. Number of people in x cars = 5x

Number of people in y cars = 7y

The students rented 3 times as many small cars as large cars, implies

y=3(x)

They altogether can hold 44 people.

i.e. 5x+7y=44

Thus , the system of equations that could be used to determine the number of small cars rented and the number of large cars rented :

[tex]y=3(x)[/tex]

[tex]5x+7y=44[/tex]

Answer:

a) To determine the minimum sample size we need to use the formula shown in the picture 1.

E is the margin of error, which is the distance from the limits to the middle (the mean) of the confidence interval. This means that we have to divide the range of the interval by 2 to find this distance.

E = 0.5/2 = 0.25

Now we apply the formula

n = (1.645*0.80/0.25)^2 = 27.7 = 28

The minimum sample size would be 28.

b) To answer the question we are going to make a 90% confidence interval. The formula is:

(μ - E, μ + E)

μ is the mean which is 127. The formula for E is shown in the picture.

E = 0.80*1.645/√8 = 0.47

(126.5, 127.5)

This means that the true mean is going to be contained in this interval 90% of the time. This is why it doesn't seem possible that the population mean is exactly 128.

(a) Minimum sample size needed for a 90% confidence interval is 7.

(b) With a sample mean of 127 ounces, 128 ounces seems unlikely for the population mean.

To solve this problem, we can use the formula for the confidence interval of the population mean:

[tex]\[ \text{Confidence Interval} = \text{Sample Mean} \pm Z \left( \frac{\text{Population Standard Deviation}}{\sqrt{\text{Sample Size}}} \right) \][/tex]

Where:

- Sample Mean = 127 ounces

- Population Standard Deviation = 0.80 ounce

- Z = Z-score corresponding to the desired confidence level

- Sample Size = n

(a) To determine the minimum sample size required for a 90% confidence interval:

We first need to find the Z-score corresponding to a 90% confidence level. We'll use a Z-table or a calculator. For a 90% confidence level, the Z-score is approximately 1.645.

[tex]\[ \text{Margin of Error} = Z \left( \frac{\text{Population Standard Deviation}}{\sqrt{\text{Sample Size}}} \right) \][/tex]

Given that the margin of error is 0.5 ounce, we can rearrange the formula to solve for the sample size:

[tex]\[ 0.5 = 1.645 \left( \frac{0.80}{\sqrt{n}} \right) \][/tex]

Solving for ( n ):

[tex]\[ \sqrt{n} = \frac{1.645 \times 0.80}{0.5} \][/tex]

[tex]\[ \sqrt{n} = 2.632 \][/tex]

[tex]\[ n = (2.632)^2 \][/tex]

[tex]\[ n \approx 6.92 \][/tex]

Since the sample size must be a whole number, we round up to the nearest whole number. Therefore, the minimum sample size required is 7.

(b) To determine if it's possible that the population mean could be exactly 128 ounces with a sample mean of 127 ounces, a sample size of 8, and a 90% confidence level:

[tex]\[ \text{Confidence Interval} = 127 \pm 1.645 \left( \frac{0.80}{\sqrt{8}} \right) \][/tex]

[tex]\[ \text{Confidence Interval} = 127 \pm 1.645 \left( \frac{0.80}{\sqrt{8}} \right) \][/tex]

[tex]\[ \text{Confidence Interval} = 127 \pm 1.645 \left( \frac{0.80}{2.828} \right) \][/tex]

[tex]\[ \text{Confidence Interval} = 127 \pm 1.645 \times 0.283 \][/tex]

[tex]\[ \text{Confidence Interval} = 127 \pm 0.466 \][/tex]

The confidence interval is ( (126.534, 127.466) ).

Since 128 ounces is not within the confidence interval, it seems unlikely that the population mean could be exactly 128 ounces.

Answer:

C. g(0)=8; g(1)=12

Step-by-step explanation:

Answer:

The answer to your question is g(0) = 12; g(1) = 8. Letter D

Step-by-step explanation:

  g(x) = 12(2/3)[tex]^{x}[/tex]

a) Substitute 0 in the equation and simplify

   g(0) = 12(2/3) ⁰

   g(0) = 12(1)

   g(0) = 12

b) Substitute 1 in the equation and simplify

   g(1) = 12(2/3)¹

   g(1) = 12(2/3)

   g(1) = 24/3

   g(1) = 8

Answer:

A 68 percent confidence interval for the proportion of persons who work less than 40 hours per week is (0.251, 0.351), or equivalently (25.1%, 35.1%)

Step-by-step explanation:

We have a large sample size of n = 83 respondents. Let p be the true proportion of persons who work less than 40 hours per week. A point estimate of p is [tex]\hat{p} = 0.301[/tex] because about 30.1 percent of the sample work less than 40 hours per week. We can estimate the standard deviation of [tex]\hat{p}[/tex] as [tex]\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{0.301(1-0.301)/83} = 0.0503[/tex]. A [tex]100(1-\alpha)%[/tex] confidence interval is given by [tex]\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n[/tex], then, a 68% confidence interval is [tex]0.301\pm z_{0.32/2}0.0503[/tex], i.e., [tex]0.301\pm (0.9944)(0.0503)[/tex], i.e., (0.251, 0.351). [tex]z_{0.16} = 0.9944[/tex] is the value that satisfies that there is an area of 0.16 above this and under the standard normal curve.

Final answer:

To calculate the 68 percent confidence interval for the proportion of persons working less than 40 hours per week from a sample of 83 respondents with a sample proportion of 30.1 percent, we use the formula for the confidence interval for a proportion. The resulting interval is approximately 25.06% to 35.14%.

Explanation:

To calculate a 68 percent confidence interval for the proportion of persons who work less than 40 hours per week from a sample of 83 respondents, where 30.1 percent work less than 40 hours, we use the formula for a confidence interval for a proportion:

In this formula:

Plugging the values into the formula we get:

0.301±1*sqrt((0.301(0.699)/83))

Calculating the square root part, we have:

0.301±1*sqrt((0.301*0.699)/83)
= 0.301±1*sqrt(0.210699/83)
= 0.301±1*sqrt(0.002539)
= 0.301±1*0.05039
= 0.301±0.05039

The confidence interval is thus:

0.301-0.05039 to 0.301+0.05039
= 0.25061 to 0.35139

Hence, with a 68 percent confidence level, we can say that the true proportion of the population that works less than 40 hours per week is estimated to be between 25.06% and 35.14%.

Answer:

  225 adult tickets were sold

Step-by-step explanation:

You are asked to find the number of adult tickets sold. It is convenient to let a variable represent that quantity. We can call it "a" to remind us it is the number of adult tickets (not student tickets).

The number of student tickets is 67 more, so can be represented by (a+67). The total number of tickets sold is the sum of the numbers of adult tickets and student tickets:

  (a) + (a+67) = 517

  2a + 67 = 517 . . . . . collect terms

  2a = 450 . . . . . . . . . subtract 67

  a = 225 . . . . . . . . . . divide by 2

There were 225 adult tickets sold.

_____

Check

The number of student tickets sold is ...

  a+67 = 225 +67 = 292

And the total number of tickets sold is ...

  225 + 292 = 517 . . . . . answer checks OK

[tex]\boxed{18-24i=30(cos(5.36)+isin(5.36))}[/tex]

Unlike 0, we can write any complex number in the trigonometric form:

[tex]z=r(cos\alpha+isin\alpha)[/tex]

We have the complex number:

[tex]18-24i[/tex]

So [tex]r[/tex] can be found as:

[tex]r=\sqrt{x^2+y^2} \\ \\ \\ Where: \\ \\ x=18 \\ \\ y=-24 \\ \\ r=\sqrt{18^2+(-24)^2} \\ \\ r=\sqrt{324+576} \\ \\ r=\sqrt{900} \\ \\ r=30[/tex]

Now for α:

[tex]\alpha=arctan(\frac{y}{x}) \\ \\ Since \ the \ complex \ number \ lies \ on \ the \ fourth \ quadrant: \\ \\ \alpha=arctan(\frac{-24}{18})=-53.13^{\circ}  \ or \ 360-53.13=306.87^{\circ}[/tex]

Finally:

[tex]Convert \ into \ radian: \\ \\ 360^{\circ}\times \frac{\pi}{180}=5.36rad \\ \\ \\ Hence: \\ \\ \boxed{18-24i=30(cos(5.36)+isin(5.36))}[/tex]

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Answer:

(a) [tex]y = 83 (1.048)^x[/tex]

(b) 2020

Step-by-step explanation:

(a) Let the exponential equation that shows the population in thousand after x years,

[tex]y = ab^x[/tex]

Also, suppose the population is estimated since 2010,

So, x = 0, y = 83 thousands,

[tex]83 = ab^0[/tex]

[tex]\implies a = 83[/tex]

Again by 2015 the population had grown to 105 thousand,

i.e. y = 105, if x = 5,

[tex]\implies 105 = ab^5[/tex]

[tex]\implies 105 = 83 b^5[/tex]

[tex]\implies b = (\frac{105}{83})^\frac{1}{5}=1.0481471103\approx 1.048[/tex]

Hence, the required function,

[tex]y = 83 (1.048)^x[/tex]

(b) if y = 135,

[tex]135 = 83(1.048)^x[/tex]

[tex]\implies x = 10.375\approx 10[/tex]

Hence, after approximately 10 years since 2010 i.e. in 2020 the population would be 135.

Final answer:

By assuming exponential growth, we derived the equation P = 83,000 * e^(0.048t) and determined that the town's population would exceed 135,000 around the year 2021.

Explanation:

To find an exponential equation for the town's population and determine in what year the population will exceed 135 thousand, we start by assuming the population growth follows the form P = P0 * e^(rt), where P is the final population, P0 is the initial population, r is the rate of growth, and t is the time in years. For this town, in 2010 (t=0), the population was 83,000 (P0=83,000), and by 2015 (t=5), the population grew to 105,000.

Substituting these values into the formula, we have 105,000 = 83,000 * e^(5r). Solving for r, we find that r ≈ 0.048. Thus, the exponential growth equation is roughly P = 83,000 * e^(0.048t).

To determine when the population will exceed 135,000, we set P > 135,000 and solve for t. This gives us the inequality 135,000 < 83,000 * e^(0.048t). Solving this, we find that t ≈ 10.24 years after 2010, which rounds up to the year 2021 when the population will exceed 135 thousand.

Answer:35.3

Step-by-step explanation:

100-24.95-39.75=35.3

Final answer:

Carmen spent a total of $64.70 on a book and a backpack and had $35.30 left from her original $100 after her purchases.

Explanation:

To calculate how much money Carmen had left after making her purchases, we first need to add the cost of the book and the backpack to find the total amount spent. She spent $24.95 on the book and $39.75 on the backpack. Add these two amounts together to find the total spent:

Next, subtract the total spent from the original $100 bill:

Therefore, Carmen has $35.30 left after her purchases.

Answer:

9/49

Step-by-step explanation:

For the first die, probability of obtaining 4= 2/7

For the second die, the probability of obtaining 3= 2/7

Other outcomes for each die appear with a probability of 1/7

There are six cases of obtaining the sum of the two die as 7.

E1 = 1 and 6

E2 = 2 and 5

E3 = 3 and 4

E4 = 4 and 3

E5 = 5 and 2

E6 = 6 and 1

Pr(E1) = 1/7*1/7

= 1/49

Pr(E2) = 1/7*1/7

= 1/49

Pr(E3) = 1/7*1/7

= 1/49

Pr(E4) = 2/7*2/7

= 4/49

Pr(E5) = 1/7*1/7

= 1/49

Pr(E6)= 1/7*1/7

= 1/49

P(E) = Pr(E1) + Pr(E2) + Pr(E3) + Pr(E4) + Pr(E5) + Pr(E6)

= 1/49 + 1/49 + 1/49 + 4/49 + 1/49 + 1/49

= 9/49

When rolling loaded dice with particular probabilities of obtaining 3 and 4, the probability of obtaining a sum of 7 when two dice are rolled is 8/49.

Your question pertains to the concept of probability in mathematics, particularly when dealing with loaded dice and the probabilities of certain outcomes. In your scenario, you want to determine the probability of obtaining a sum of 7 given particular probabilities of rolling a 4 on the first die and a 3 on the second die.

In general, obtaining a sum of 7 is possible with the following combinations: 1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1. With your loaded dice, however, only the combinations 4 and 3 and 3 and 4 are relevant. Given that the probability of rolling a 4 on the first die is 2/7, and the probability of rolling a 3 on the second die is also 2/7, the probability of rolling a 7 is (2/7)*(2/7)=4/49. As you are considering both the combinations 4 and 3, and 3 and 4, you need to multiply this result by 2 which gives a final probability of 8/49.

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To determine the length of an unknown side in a right triangle, we can use the sin, cos, tan ratios or the Pythagorean theorem, depending on which sides and/or angles we already know. However, the specific information about measures and sides in the two right triangles in the question is not clear, hence a specific answer cannot be provided.

The subject of this question is trigonometry, a branch of mathematics that studies relationships involving lengths and angles of triangles. As the triangles are right-angled, we can use properties specific to this type of triangle. When given the measure of an angle in a right triangle, we can find the lengths of the sides using trigonometric ratios such as sine, cosine, and tangent.

In this particular problem, we're given the measure of an angle and lengths of some sides but the information about the measures of the triangles and their sides are not clearly mentioned. To provide an answer, we'd need clear information on which triangle contains the given angle and sides.

However, to give you an idea of how this type of problem can be solved, let's consider that the measure of the angle α is 30 degrees, the measure of side a is 6 units, the measure of side b is 3 units, and the measure of side c (the hypotenuse) is 12 units (These are just hypothetical names to represent the sides). If we are asked to find the length of an unknown side, we can use the Sin, Cos or Tan ratios or even the Pythagorean theorem (a² + b² = c²), depending on the sides that we know.

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Answer:

Sample size should be atleast 617

Step-by-step explanation:

given that a 95% confidence interval for the mean of a population is to be constructed and must be accurate to within 0.3 unit.

i.e. margin of error <0.3

Std devition sample = 3.8

For 95% assuming sample size large we can use z critical value

Z critical = 1.96

we have

[tex]1.96 * std error <0.3\\1.96(\frac{3.8}{\sqrt{n} } )<0.3\\\sqrt{n} >24.83\\n>616.3634[/tex]

Sample size should be atleast 617 to get an accurate margin of error 0.3

To construct a 95% confidence interval with an accuracy of 0.3 units, the smallest sample size (n) is 25.

To construct a 95% confidence interval with an accuracy of 0.3 units, we need to determine the sample size (n). We can use the formula:

n = (Z * σ) / E

Where Z is the Z-score corresponding to the desired confidence level (in this case, 95% corresponds to a Z-score of 1.96), σ is the preliminary sample standard deviation (3.8), and E is the desired accuracy (0.3).

Substituting the values into the formula:

n = (1.96 * 3.8) / 0.3 = 24.8

Rounding up to the nearest whole number, the smallest sample size n that provides the desired accuracy is 25.

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To minimize the cost of a rectangular milk carton that holds 128 cubic cm, we need to calculate the dimensions that minimize the surface area cost. By setting up an optimization problem and using calculus, we can find the values of length, width, and height that satisfy the volume constraint and result in the lowest cost.

Minimizing Cost for a Rectangular Milk Carton

To find the dimensions of a rectangular milk carton with a given volume that minimize the cost of materials used, we can set up an optimization problem using calculus. First, we know the volume of the milk carton must be $$128 cm^3$, which gives us the constraint:

V = lwh = 128

Next, we need to express the cost function in terms of the dimensions of the box. The sides of the box cost $$1 cent per square cm, while the top and bottom cost $2 cents per square cm. Thus, the total cost, C, in cents, is:

C = 2lw + 2wh + 2lh + (4 * l * w)

To minimize the cost, we would take the partial derivatives with respect to l, w, and h, set them equal to zero, and solve the system of equations while taking into account the volume constraint. This involves the method of Lagrange multipliers or directly substituting the volume constraint into the cost function to eliminate one variable and then taking the derivative with respect to the other variables.

By finding the derivative of the cost function and setting it to zero, you can determine the values of l, w, and h that will result in the minimum cost while respecting the volume constraint. Since this is an applied problem, it is important to check that the resulting values are practical, meaning they should be positive and make sense for a milk carton.

Answer:

  the cylinder volume is greater

Step-by-step explanation:

The volume of a cube with x=1 is ...

  V(1) = 1^3 = 1

The graph shows y ≈ 1.5 for x=1. Since 1.5 > 1, the volume of the cylinder is greater.

When x = 1, the volume of the cylinder is greater than the volume of the cube.

To determine which volume is greater when x = 1, we can calculate the volume of the cube and the volume of the cylinder at x = 1 and compare them.

For the cube with side length x, the volume is given by V(x) = x^3. So, when x = 1:

V(cube) = (1)^3 = 1

For the cylinder with radius x and height 0.5x, the volume is given by the formula for the volume of a cylinder: V(cylinder) = πr^2h, where r is the radius and h is the height.

When x = 1:

r = 1 (radius)

h = 0.5(1) = 0.5 (height)

V(cylinder) = π(1^2)(0.5) = π(0.5) = 0.5π

Now, we need to compare the volumes.

V(cube) = 1

V(cylinder) = 0.5π ≈ 1.57 (rounded to two decimal places)

So, when x = 1, the volume of the cylinder is greater than the volume of the cube.

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Answer:

The total cost is 1.25 dollars.

Step-by-step explanation:      

The reaction between HCl and CaCO₃ is giving by:

2HCl(aq) + CaCO₃(s)   →   CaCl₂(aq) + CO₂(g) + H₂O(l) (1)

0.500L     M: 100.01g/mol        

0.400M

According to equation (1), 2 moles of HCl react with 1 mol of CaCO₃, so to neutralize HCl, we need the next amount of CaCO₃:

[tex] m CaCO_{3} = (\frac{1 \cdot mol HCl}{2}) \cdot M_{CaCO_{3}} = (\frac{0.500L \cdot 0.400 \frac {mol}{L}}{2}) \cdot 100.01 \frac{g}{mol} = 10.001 g [/tex]      

The CaCO₃ mass of each tablet is:

[tex] m CaCO_{3} = 1 g_{tablet} \cdot \frac{40g CaCO_{3}}{100g_{tablet}} = 0.4g [/tex]

Hence, the number of tablets that we need to neutralize the HCl is:

[tex] number_{tablets} = ( \frac{1 tablet}{0.4 g CaCO_{3}}) \cdot 10.001g CaCO_{3} = 25 [/tex]  

Finally, if every 80 tablets costs 4.00 dollars, 25 tablets will cost:

[tex] cost = (\frac {4 dollars}{80 tablets}) \cdot 25 tablets = 1.25 dollars [/tex]

So, the total cost to neutralize the HCl is 1.25 dollars.

I hope it helps you!

Answer:

Option d

Step-by-step explanation:

given that a, b, c, and d be non-zero real numbers.

[tex]ax(cx + d) = -b(cx+d) \\acx^2+x(ad+bc)+bd =0[/tex]

we can factorise this equation by grouping

[tex](acx^2+xad)+)xbc+bd) =0\\ax(cx+d) +b(cx+d) =0\\(ax+b)(cx+d) =0[/tex]

Equate each factor to 0 to get

[tex]x=\frac{-b}{a} , \frac{-d}{c}[/tex]

Ratio of one solution to another would be

[tex]\frac{-b}{a} / \frac{-d}{c} \\=\frac{ad}{bc}[/tex]

So ratio would be ad/bc

Out of the four options given, option d is equal to this

So option d is right

Answer: The company currently has 17 employees

Step-by-step explanation:

Let X represent the number of current employees in the company

From the first information, it can be expressed mathematically as

X + 3 ≥ 20

X ≥ 20 - 3

X ≥ 17

From the second information, it can be expressed mathematically as

X - 3 < 15

X < 15 + 3

X < 18

From the above solutions, it can be deduced that

17 ≤ X < 18

The only number that fulfils this criteria is 17.

Therefore, X = 17

Answer:

A youre welcome

Step-by-step explanation:

Answer:

  2.  100°              22.  7

  18.  6                  23.  9

  19.  8                  24.  65°

  20.  55°             25.  AB = (1/2)DF

  21.  6                 26.  AB ║ DF

Step-by-step explanation:

Please be aware that the triangle measurements shown for problems 18–20 and 22–24 cannot exist. In the first case, the angle is closer to 48.6° (not 35°), and in the second case, the angle is closer to 51.1°, not 25°. So, you have to take the numbers at face value and not think too deeply about them. (This state of affairs is all too common in geometry problems these days.)

_____

2. You have done yourself no favors by marking the drawing the way you have. Look again at the given conditions. You will find that x+2x must total a right angle, so x=30°. Angle P is the complement of 40°, so is 50°. Then the sum of x and angle P is 30° +50° = 80°, and the angle of interest is the supplement of that, 100°.

__

18–20. Perpendicular bisector NO means  ∆NOL ≅ ΔNOM. Corresponding parts have the same measures, and angle L is the complement of the marked angle.

__

21. ∆ODA ≅ ∆ODB by hypotenuse-angle congruence (HA), so corresponding parts are the same measure. DB = DA = 6.

__

22–24. ∆VPT ≅ ∆VPR by LL congruence, so corresponding measures are the same. Once again, the angle in question is the complement of the given angle.

__

25–26. You observe that A is the midpoint of DE, and B is the midpoint of FE, so AB is what is called a "midsegment." The features of a midsegment are that it is ...

Answer:

8. [tex]\displaystyle \frac{9[x + 5]}{x - 14}[/tex]

7. [tex]\displaystyle -\frac{2x - 1}{2[3x - 5]}[/tex]

6. [tex]\displaystyle \frac{2[x - 4]}{5[x + 3]}[/tex]

5. [tex]\displaystyle \frac{2x + 7}{x + 3}[/tex]

4. [tex]\displaystyle 3x^{-1}[/tex]

Step-by-step explanation:

All work is shown above from 8 − 4.

I am joyous to assist you anytime.

Answer:

c) 2.415

Step-by-step explanation:

Given that X represent the number on the face that lands up when a fair six-sided number cube is tossed.

The expected value of X is 3.5, and the standard deviation of X is approximately 1.708.

When another die is rolled let Y represent the number on the face that lands up when a fair six-sided number cube is tossed.

The expected value of Y is 3.5, and the standard deviation of Y is approximately 1.708.

Also we find that X and Y are independent

Let U = X+Y

Then we have U as the random variable representing the sum shown by two dice

Since X and Y are independent

[tex]Var(x+y) = Var(x) +Var(y)\\= 1.708^2 *2\\= 5.83333[/tex]

Std dev for sum

= [tex]\sqrt{5.8333} \\=2.4152[/tex]

Hence option C 2.415 is correct

Answer:

2.415

Step-by-step explanation:

Its not as complicated as that other response. You never subtract standard deviations, you always add them, and you don't add them directly, you have to square them to make them variances, add them, and find the square root of it.

In this problem, you do [tex]\sqrt{(1.708^{2}) + (1.708^{2})}[/tex]

Put it in the calculator and you get 2.415

Answer:

84 cases

Step-by-step explanation:

Given that:

Number of trucks: 28

Paper cases each truck can load = 3

Total cases of white paper = 283

So the cases delivered will be = 28 *3 = 84 cases will be delivered today

i hope it will help you!

Answer:

13,57,31

Step-by-step explanation:

Given that X, Y , Z be three random variables which satisfy the following conditions:

Var(X) = 4, Var(Y ) = 9, Var(Z) = 16. Cov(X, Y ) = −2, Cov(Z, X) = 3,

Var(y,z) =0 since given as independent

To find

[tex]a) Cov (x+2y, y-z)\\ \\= cov (x,y) +cov (2y,y) -cov (x,z) -cov(2y,z)\\= cov (x,y) +2cov (y,y) -cov (x,z) -2cov(y,z)\\=-2+2 var(y) -3-0\\= -2+18-3\\=13[/tex]

b) [tex]Var(3X − Y ).\\= 9Var(x)+var(y) -6 covar (x,y)\\= 36 +9+12\\= 57[/tex]

c) Var(X + Y + Z)[tex]=Var(x) = Var(Y) +Var(z) +2cov (x,y) +2cov (y,z) +2cov (x,z)\\= 4+9+16+(-4) +6\\= 31[/tex]

Note:

Var(x+y) = var(x) + Var(Y) +2cov (x,y)

Var(x+2y) = Var(x) +4Var(y)+4cov (x,y)

Answer:

Step-by-step explanation:

Answer:

A. A - 0.2, B - 0.3, C - 0.2, D - 0.15, E - 0.15

B. Grade B

Step-by-step explanation:

A. Total number of students = 4 + 6 + 4 + 3 + 3 = 20

So the relative frequency of each grade can be computed as the ratio of number of students with the grade and the total number of students.

A : 4/20 = 0.2

B : 6/20 = 0.3

C : 4/20 = 0.2

D : 3/20 = 0.15

E : 3/20 = 0.15

B. Among the grades, grade B has the highest relative frequency of 0.3. Hence Grade B received the most number of students.

Answer:

Step-by-step explanation:

Factor the given function:

        (x + 2)(x + 5)

f(x) = --------------------

         (x + 5)(x + 4)

The factor (x + 5) can be crossed out in numerator and denominator.  There is a 'hole' at x = -5 (which comes from setting (x + 5) = 0 and solving for x).

If we do cancel the (x + 5) factors, we are left with

           (x + 2)

f(x) =  ------------- .  We can't just substitute x = -5 because the original

           (x + 4)         function is not defined for that x value.  However, we

can locate the hole by letting x have the value of a litle more than -5, obtaining:

           ( -5+ + 2)

f(-5+) =  -------------, which comes out to approx. -3/-1, or +3 .  

            (-5+ + 4)    

So the hole location is (-5, 3).

f(x) = --------------------

         (x + 5)(x + 4)

Answer:

$ 55135.978

Step-by-step explanation:

At most, the present value of annuity must be paid. So we must find the present value of the annuity

Given in the problem, we have:

Periodic Payment = PMT = $10000

Rate of interest annually = i = 6.35 %= [tex]\frac{6.35}{100}[/tex]=0.0635

no. of periods= n=7

So to solve this, we need to use the present value formula:

Present Value = Periodic payment [tex]\frac{1-(1+rate.of.interest)^{-n} }{rate.of.interest}[/tex]

Present Value = PMT [tex]\frac{1-(1+i)^{-n} }{i}[/tex]

Present value = 10000[tex]\frac{1-(1+0.0635)^{-7} }{0.0635}[/tex]

Present Value =10000[tex]\frac{0.35011}{0.0635}[/tex]

Present Value =10000 (5.5135978)

Present value= $ 55135.978

Which is the amount that must be paid at most to get annuities such that $10,000 annually over the 7-year period are to be received.

You should pay approximately $55,598 for a 7-year ordinary annuity with a guaranteed 6.35% annual interest rate to receive annual payments of $10,000.

To find out how much you should pay for a 7-year ordinary annuity with a guaranteed rate of 6.35% compounded annually to receive payments of $10,000 annually, we can use the present value of an annuity formula:

PV = PMT × [(1 - (1 + r)⁻ⁿ) / r]

Substituting the values into the formula:

PV = $10,000 × [(1 - (1 + 0.0635)⁻⁷) / 0.0635]

PV = $10,000 × [(1 - (1 / (1.0635)⁷)) / 0.0635]

PV = $10,000 × [(1 - (1 / 1.545677)) / 0.0635]

PV = $10,000 × [0.353032 / 0.0635]

PV = $10,000 × 5.5598

PV ≈ $55,598

Therefore, you should pay approximately $55,598 for this annuity to receive payments of $10,000 annually over 7 years.

Answer:

c) x11 + x12 = 8000

Step-by-step explanation:

Xij = Gallons of the ith component used in the  jth Gasoline type

This invariably  tells us what component is in which gasoline type.

The gasoline types are:

Gasoline  1  = 11,000 gallons

Gasoline 2 = 14,000 gallons

Assuming two(2) components types:

Component 1

Component 2

The possible combinations Xij are :

                          Gasoline 1                     Gasoline 2

Component 1          X11                                X12

Component 2         X21                               X22

From the above , it is clear that the supply constraint for component 1 across the gasoline types is given by    X11  &  X12

Mathematically, since there are 8,000 gallons of Component 1, the supply constraint is given by:

X11 + X12 = 8000

The supply constraint for component 1, given the available gallons and the usage in two types of gasoline, is represented by the equation x11 + x12 = 8000.

The supply constraint for component 1 when formulating an equation that represents the total usage of component 1 in both types of gasoline should reflect the total availability of component 1. Since there are 8,000 gallons of component 1 available, the correct mathematical representation of the supply constraint for component 1 is:

x11 + x12 = 8000

This equation means that the sum of gallons of component 1 used in gasoline type 1 (x11) and gasoline type 2 (x12) must equal the total available gallons of component 1, which is 8,000 gallons.

Answer:

Value of x is 20 and y is 120.

Step-by-step explanation:

Given,

m∠A = 70°, m∠B = 60°, m∠C = 50°,m∠D = (3x + 10)°, m∠E= (1/3y + 20)°, and m∠F = (z² + 14)°

Also,

ΔABC ≅ ΔDEF,

Since, the corresponding parts of congruent triangles are always congruent or equal.

⇒ m∠A = m∠D, m∠B = m∠E and m∠C = m∠F

When m∠A = m∠D

[tex]\implies 70 = 3x + 10[/tex]

[tex]70 - 10 = 3x[/tex]

[tex]60 = 3x[/tex]

[tex]\implies x =\frac{60}{3}=20[/tex]

When, m∠B = m∠E,

[tex]\implies 60 = \frac{1}{3}y + 20[/tex]

[tex]60 - 20 =\frac{1}{3}y[/tex]

[tex]40 =\frac{1}{3}y[/tex]

[tex]\implies y =3\times 40=120[/tex]

Answer:

There are 436937109262510348800 ways to selected 12 countries.

Step-by-step explanation:

Let [tex]x_1, x_2, x_3, x_4,x_5[/tex] the countries selected from the block of 45. For [tex]x_1[/tex] there are 45 options of countries to select. For [tex]x_2[/tex] there are 44 options of countries to select because there can be no repeated countries. With the same reasoning, for [tex]x_3[/tex] there are 43 options, for [tex]x_4[/tex] there are 42 options and for [tex]x_5[/tex] there are 41 options.

Let [tex]x_6, x_7, x_8, x_9[/tex] the countries selected from the block of 57 countries. For [tex]x_6[/tex] there are 57 options of countries to select. For [tex]x_7[/tex] there are 56 options of countries to select because there can be no repeated countries. With the same reasoning, for [tex]x_8[/tex] there are 55 options and for [tex]x_9[/tex] there are 54 options.

Let [tex]x_{10}, x_{11}, x_{12}[/tex] the countries selected from the block of 69 remain countries. For [tex]x_{10}[/tex] there are 69 options of countries to select. For [tex]x_{11}[/tex] there are 68 options of countries to select because there can be no repeated countries and for [tex]x_{12}[/tex] there are 67 options.

For find the total of options o ways to select 12 countries we multiply the above options.

[tex]45*44*43*42*41*57*56*55*54*69*68*67= 436937109262510348800[/tex]

Final answer:

The number of ways to select 12 countries from the United Nations to serve on a council is determined by calculating the combinations for each block of countries and then multiplying these values together.

Explanation:

The student question deals with determining the number of ways to select 12 countries from the United Nations to serve on a council, given specific blocks of countries to choose from. To calculate this, we use the principles of combinations, as the order of selection does not matter. The number of ways to select 5 countries from the first block of 45 is given by the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of items to choose from, k is the number of items to choose, and '!' denotes factorial. Applying this to the first block:

C(45, 5) = 45! / (5! * (45-5)!) = 1,221,759 ways

Similarly, for the second block of 57 countries, from which 4 must be chosen:

C(57, 4) = 57! / (4! * (57-4)!) = 496,674 ways

For the remaining 3 countries from the last block of 69:

C(69, 3) = 69! / (3! * (69-3)!) = 54,834 ways

To find the total number of combinations, we multiply the number of ways for each block, as each selection is independent of the others:

Total ways = C(45, 5) * C(57, 4) * C(69, 3) = 1,221,759 * 496,674 * 54,834

After calculating this product, we obtain the total number of ways to select 12 countries from the United Nations for the council.