Find the area of the shaded region below. The units are in feet.
Use 3.14 for π. Round your answer to two decimal places.

Answer:

X = - sqrt 12 / 4

X = sqrt 11 + 1 / 4

The only true statement is b. 40% of the boys surveyed have at least one pet.

To determine which statement is true, let's analyze the data provided in the table:

- Total number of boys surveyed: 45

- Total number of girls surveyed: 65

Now, let's break down the information based on the provided table:

1. **At least one pet:**

  - Boys: 18

  - Girls: 39

  - Total: 57

2. **No pets:**

  - Boys: 27

  - Girls: 26

  - Total: 53

Now, let's check each statement:

a. 27% of the boys surveyed have no pets.

  - Percentage of boys with no pets = (Number of boys with no pets / Total number of boys surveyed) * 100%

  - = (27 / 45) * 100% ≈ 60%

  - This statement is false.

b. 40% of the boys surveyed have at least one pet.

  - Percentage of boys with at least one pet = (Number of boys with at least one pet / Total number of boys surveyed) * 100%

  - = (18 / 45) * 100% = 40%

  - This statement is true.

c. 49% of the girls surveyed have no pets.

  - Percentage of girls with no pets = (Number of girls with no pets / Total number of girls surveyed) * 100%

  - = (26 / 65) * 100% ≈ 40%

  - This statement is false.

d. 57% of the students surveyed have at least one pet.

  - Percentage of students with at least one pet = (Total number of students with at least one pet / Total number of students surveyed) * 100%

  - = (57 / 110) * 100% ≈ 52%

  - This statement is false.

So, the only true statement is b. 40% of the boys surveyed have at least one pet.

The probable question may be:

110 students are surveyed about their pets. The results are shown in the table. Which statement is true?

  Boys | Girls | Total

At least one pet | 18 | 39 | 57

No pets | 27 | 26 | 53

Total | 45 | 65 | 110

a. 27% of the boys surveyed have no pets.

b. 40% of the boys surveyed have at least one pet.

c. 49% of the girls surveyed have no pets.

d. 57% of the students surveyed have at least one pet.

Answer: 6 sq. units

Step-by-step explanation:

The formula to find the circumference of a circle is given by :-

[tex]C=2\pi r[/tex], where r is the radius of the circle.

Given : Circumference = 10 units

Then , [tex]10=2\pi r[/tex]

[tex]\\\\\Rightarrow\ r=\dfrac{10}{2\pi}\approx1.59[/tex]

The area of a circle is given by :-

[tex]A=\pi r^2=\pi(1.59)^2=7.9422603875\approx7.94\text{ sq. units}[/tex]

Now, the 75 % of the area is given by :-

[tex]0.75\times7.94=5.955\approx6\text{ sq. units}[/tex]

Hence, 75% of the area of a circle with a circumference of 10 units = 6 sq. units

Answer:

Option 3 - 45 tens

Step-by-step explanation:

Given : Expression 45 ones times 10.

To find : What is the expression?

Solution :

We know that,

"one" means "  1

We  have 45 ones means multiply 45 by 1.

[tex]45\text{ ones}=45\times 1=45[/tex]

Now, 45 ones times 10 means

[tex]45\times 10=450[/tex]

450 means 45 tens as tens is 10.

So, The correct choice is 45 tens.

Therefore, Option 3 is correct.

Final answer:

To find the binary equivalent of the decimal number 97, one divides it by 2 repeatedly and records the remainders in reverse order, resulting in the binary number 1100001.

Explanation:

The binary equivalent of the decimal value 97 can be found using a process of dividing by 2 and keeping track of the remainders. Firstly, divide 97 by 2, which gives a quotient of 48 and a remainder of 1. We write down the remainder. Continuing this process:

After collecting all the remainders in reverse order, the binary equivalent of decimal 97 is 1100001.

Final answer:

The student needs to solve a quadratic equation to find the base and height of a triangle using the area formula and the given relationship between height and base. The solution involves substitution, expansion, and application of the quadratic formula or factoring.

Explanation:

The problem involves finding the base and height of a triangular front of an A-frame cabin based on its given area and a relationship between the height and base. It's a typical quadratic equation problem found in the high school mathematics curriculum when dealing with geometry and algebra.

To find the base (b) and height (h) of the triangle, we first use the area formula of a triangle A = 1/2 × base × height. We know that the area (A) is 189 ft² and that the height (h) is 1 foot less than twice the base, so h = 2b - 1. Substituting h into the area formula, we get 189 = 1/2 × b × (2b - 1). Solving this quadratic equation, we find the values for the base (b) and substitute back to find the height (h).

The process entails expanding the equation, moving all terms to one side to set the equation to zero, and then using the quadratic formula or factoring to find the value of b. Once the base is found, we use the relationship h = 2b - 1 to determine the height.

Jeremiah needs additional information such as the location of the foci, the other vertex on the major axis, a covertex along the minor axis, or the length of the minor axis to write the equation of the ellipse that is options A, B, C, D and G are correct.

Jeremiah is asked to write the equation of an ellipse given one vertex along the major axis and the location of the center.

He realizes he does not have enough information. He could ask for the following additional pieces of information to help him write the equation:

With any of these pieces of information, he could determine the necessary parameters to complete the equation of the ellipse.

Jeremiah could ask for the other information that is the location of the other vertex along the major axis, the location of one covertex along the minor axis, the length of the minor axis and the location of the focus nearest the given vertex.

To write the equation of an ellipse, Jeremiah needs more information. Given the center and one vertex along the major axis, he can ask for:

With any of this additional information, Jeremiah can confidently determine the parameters required to write the equation of the ellipse.

Answer:

[tex]s=4[/tex].

Step-by-step explanation:

We have been given an equation [tex]3r=10+5s[/tex]. We are asked to find the value of 's', when [tex]r=10[/tex].

To find value of 's', we will substitute [tex]r=10[/tex] in our given equation as shown below:

[tex]3(10)=10+5s[/tex]

[tex]30=10+5s[/tex]

Upon subtracting 10 from both sides of our given equation, we will get:

[tex]30-10=10-10+5s[/tex]

[tex]20=5s[/tex]

Now, we will divide both sides of our equation by 5.

[tex]\frac{20}{5}=\frac{5s}{5}[/tex]

[tex]4=s[/tex]

Therefore, the value of 's' is 4, when [tex]r=10[/tex].

(a) [tex]\( P(X = 5) \approx 0.234 \)[/tex]

(b) [tex]\[ P(X \geq 6) \approx 0.892 \][/tex]

(c) [tex]\[ P(X < 4) \approx 0.020 \][/tex]

To solve this problem, we can use the binomial probability formula since each man's response (considering themselves a baseball fan or not) is independent and there are only two possible outcomes (success or failure).

Given:

- Probability of success (considering themselves a baseball fan) [tex]\( p = 0.57 \)[/tex]

- Probability of failure (not considering themselves a baseball fan) [tex]\( q = 1 - p = 1 - 0.57 = 0.43 \)[/tex]

- Number of trials [tex]\( n = 10 \)[/tex]

We'll calculate the probabilities for each case:

(a) To find the probability that exactly five men consider themselves baseball fans:

[tex]\[ P(X = 5) = \binom{10}{5} \times (0.57)^5 \times (0.43)^{10 - 5} \][/tex]

(b) To find the probability that at least six men consider themselves baseball fans, we can find the probability of six, seven, eight, nine, and ten men being baseball fans, and then sum them up.

(c) To find the probability that less than four men consider themselves baseball fans, we need to find the probabilities of zero, one, two, and three men being baseball fans, and then sum them up.

Let's calculate each probability:

(a) [tex]\[ P(X = 5) = \binom{10}{5} \times (0.57)^5 \times (0.43)^{5} \][/tex]

(b) To find the probability of at least six men being baseball fans:

[tex]\[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \][/tex]

(c) To find the probability of less than four men being baseball fans:

[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]

We'll use these formulas to find the probabilities for each case. Let me do the calculations.

(a) To find the probability that exactly five men consider themselves baseball fans:

[tex]\[ P(X = 5) = \binom{10}{5} \times (0.57)^5 \times (0.43)^{5} \][/tex]

Using the binomial coefficient formula [tex]\(\binom{n}{k} = \frac{n!}{k!(n - k)!}\)[/tex], where [tex]\(n = 10\)[/tex] and [tex]\(k = 5\)[/tex]:

[tex]\[ \binom{10}{5} = \frac{10!}{5!(10 - 5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \][/tex]

Now, we plug in the values:

[tex]\[ P(X = 5) = 252 \times (0.57)^5 \times (0.43)^{5} \][/tex]

[tex]\[ P(X = 5) \approx 0.234 \][/tex]

(b) To find the probability of at least six men being baseball fans:

[tex]\[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \][/tex]

For each [tex]\(k = 6, 7, 8, 9, 10\)[/tex], we calculate [tex]\(P(X = k)\)[/tex] using the binomial formula and sum them up.

(c) To find the probability of less than four men being baseball fans:

[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]

Similarly, for each [tex]\(k = 0, 1, 2, 3\)[/tex], we calculate [tex]\(P(X = k)\)[/tex] using the binomial formula and sum them up. Let me do the calculations.

(a) [tex]\( P(X = 5) \approx 0.234 \)[/tex]

(b) To find the probability of at least six men being baseball fans:

[tex]\[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \][/tex]

Using the binomial formula for each [tex]\( k = 6, 7, 8, 9, 10 \)[/tex] and summing the probabilities:

[tex]\[ P(X \geq 6) \approx 0.892 \][/tex]

(c) To find the probability of less than four men being baseball fans:

[tex]\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]

Using the binomial formula for each [tex]\( k = 0, 1, 2, 3 \)[/tex] and summing the probabilities:

[tex]\[ P(X < 4) \approx 0.020 \][/tex]

The irrational number among the options is root 3 + 8.486. option C.

An irrational number is a number that cannot be expressed as a fraction of two integers and has a non-repeating, non-terminating decimal expansion.

Let's examine each option:

A. [tex]\(7.51\ldots \times -4\)[/tex]

This is a rational number because it's the product of a rational number [tex](\(7.51\ldots\)) and \(-4\),[/tex] which is also rational. So, option A is not irrational.

B. [tex]\(\sqrt{16} + \frac{3}{4}\)[/tex]

[tex]\(\sqrt{16} = 4\)[/tex], so this expression simplifies to [tex]\(4 + \frac{3}{4}\)[/tex], which is a rational number. So, option B is not irrational.

C. [tex]\(\sqrt{3} + 8.486\)[/tex]

If [tex]\(\sqrt{3}\)[/tex] is not exactly equal to 8.486 , then this expression is irrational because it's the sum of an irrational number and a rational number. However, if [tex]\(\sqrt{3}\)[/tex] does happen to be exactly 8.486, then this expression would be rational. To determine if [tex]\(\sqrt{3}\)[/tex] is exactly 8.486, we need to compute [tex]\(\sqrt{3}\).[/tex] Since [tex]\(\sqrt{3}\)[/tex] is irrational (it's not a perfect square), 8.486 is not [tex]\(\sqrt{3}\)[/tex], so this expression is irrational. Therefore, option C is the correct answer.

D.[tex]\(8 \frac{2}{3} \times 17.75\)[/tex]

This is a rational number because it's the product of a rational number [tex](\(8 \frac{2}{3}\)) and \(17.75\),[/tex] which is also rational. So, option D is not irrational.

Complete question: Which of the following is irrational?

A. 7.51... x -4

B. root 16 + 3/4

C. root 3 + 8.486

D. 8 2/3 x 17.75

there are 4 Aces and 4 8's for a total of 8 cards

52 cards in a deck

 so probability of picking an Ace or 8 = 8/52

Final answer:

The probability of being dealt an ace or an 8 from a standard 52-card deck is 2/13, which is approximately 15.38%.

Explanation:

The question asks for the probability of being dealt an ace or an 8 from a standard 52-card deck. There are 4 aces and 4 eights in a 52-card deck. To calculate the probability of getting either an ace or an 8, we add the probability of getting an ace to the probability of getting an 8. The probability of drawing an ace (P(Ace)) is 4/52 and the probability of drawing an 8 (P(8)) is also 4/52. Since these are mutually exclusive events (you cannot draw an ace and an 8 simultaneously with one card), we can simply add these probabilities together:

P(Ace or 8) = P(Ace) + P(8)
= (4/52) + (4/52)
= 8/52
= 2/13

Therefore, the probability of being dealt an ace or an 8 is 2/13, which is approximately 0.1538 or 15.38%.

Hence, it will take 4.71 minutes  for all the liquid in the funnel to pass through the nozzle.

A cone shaped funnel has a radius(r) of 3 inches and a height(h) of 7 inches.

Now, the volume(V) of the cone is given as:

[tex]V=\dfrac{1}{3}\times (\pi r^2h)[/tex]

Hence, on putting the value of r and h in the formula of volume we obtain the volume of cone funnel as:

[tex]V=\dfrac{1}{3}\times (3.14\times (3)^2\times 7)\\\\\\V=\dfrac{1}{3}\times (197.82)\\\\V=65.94 \ in^3[/tex]

If the liquid drips at the rate of 14 cubic inches per minute.

i.e. for 14 cubic inches it takes 1 minutes.

Now for 1 cubic inches it will take:

[tex]\dfrac{1}{14} \ min.[/tex]

Hence, for all the liquid ( i.e. 65.94 cubic inches) to pass the nozzle is the time taken is:

[tex]\dfrac{65.94}{14}\ min.\\\\=4.71\ min.[/tex]

Hence, it will take 4.71 minutes  for all the liquid in the funnel to pass through the nozzle.

The phrase "oh heck another hour of algebra" can help a student recall the trigonometric ratios by associating key words in the phrase with math concepts. The phrase contains words related to math, such as "algebra" and "hour," as well as a word similar to "angle." By linking these words to the trigonometric ratios, a student can better remember and understand them.

The phrase "oh heck another hour of algebra" can help a student recall the trigonometric ratios by focusing on the key words within the phrase. The phrase contains the words "algebra" and "hour," which are related to math, and the word "heck," which is similar to the word "angle." By associating these words with the phrase, a student can remember the trigonometric ratios, which involve angles and algebraic calculations. For example, the phrase can remind a student that the sine ratio involves the ratio of opposite and hypotenuse sides, similar to finding lengths in algebraic equations.

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