the table below show the number of studnts in the school choir.

school choir

Students Number
Girls 48
Boys 64

the choir teacher plans to arrange the students in equal rows. Only girls and boys will be in each row. What is the greatest number of students that could be in each row.​

Answer:

-11

I went to Starbucks today knowing that i had -3 dollars in my bank account, but i wanted this Frappe so bad and it was 8 dollars. Long story short i ended up leaving with -11 dollars in my account.

Answer:

307°

Step-by-step explanation:

The sum of all the angles of a hexagon is equal to 900°.

So,

112+x+120+100+128+133 = 900

593+x = 900

x = 900-593

x = 307°

So, the value of x is equal to 307°.

Answer:

\(\huge\boxed{Answer\hookleftarrow}\)

\(_____ x 14.32 = 554.76\)

Let's take '____' as x

\(x \times 14.32 = 554.76 \\ x = \frac{554.76}{14.32} \\ x = \frac{554.76 \times 100}{14.32 \times 100} \\ x = \frac{55476}{1432} \\ x = 38.74\)

⎆ The answer of ____ x 14.32 = 554.76 in 2 decimal places will be ⇻38.74

NOTE:

We multiplied the fraction × 100 to remove the decimal points so that it'll be easier to calculate.

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

# ꧁❣ RainbowSalt2²2² ࿐

Answer:

Hundred thousands place.

Answer:

In this case the number 7 would be located in the hundreds of thousands.

---

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Answer: 80

Step-by-step explanation:

8 times more than ten is 80

The solution by substituting 15 for b in the Original equation. The answer is true.

The steps needed to solve the equation b + 6 = 21 are:

Subtract 6 from both sides of the equation: b + 6 - 6 = 21 - 6, which simplifies to b = 15.

Check the solution by substituting 15 for b in the original equation: 15 + 6 = 21, which is true.

Therefore, the correct steps to solve the equation are:

Subtract 6 from both sides of the equation. The answer is b = 15

the solution by substituting 15 for b in the original equation. The answer is true.

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Answer: the Nswer for this is 30 people

Step-by-step explanation:

Answer:

10n=40    n=4

Step-by-step explanation:

10n=40  divide each side by 10

n=40/10   simplify

n=4

The given relationship is called "segment addition postulate".

The points lying on the same segment are called collinear points.

Given that, For any two points A and B on a line, there is another point C between A and B if and only if A, B, and C are collinear and AC + CB = AB

The relationship shown here is "segment addition postulate".

Segment addition postulate:-

According to segment addition postulate,

The length of the whole segment is equal to the sum of the lengths of all the parts of the segment.

Hence, This relationship is called the segment addition postulate.

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\(x^2-3x+2=0\\x^2-x-2x+2=0\\x(x-1)-2(x-1)=0\\(x-2)(x-1)=0\\x=2 \vee x=1\\\\x^2=4 \vee x^2=1\)

Answer:

Step-by-step explanation:

First we try to factor x²-3x+2.

We have to look for two numbers that multiply to 2 and add -3.

The two numbers are -1 and -2.

(x-1)(x-2) = 0

x-1 = 0 -> x = 1

x-2 = -> x = 2

Now we find x^2.

(1)^2 = 1

(2)^2 =4

The final matrix will be as follows   \(\rm D-C=\left[\begin{array}{ccc}12&-2&3\\3&1&-2\\\end{array}\right]\)

A matrix is a rectangular array or table of numbers, symbols, or expressions arranged in rows and columns in mathematics.

The calculation of the given matrix is as follows:-

\(\rm D-C= \left[\begin{array}{ccc}13&2&-5\\6&-4&0\\\end{array}\right] - \left[\begin{array}{ccc}1&4&-8\\3&-5&2\\\end{array}\right] \\\)

\(\rm D-C= \left[\begin{array}{ccc}(13-1)&(2-4)&(-5+8)\\(6-3)&(-4+5)&(0-2)\\\end{array}\right]\)

\(\rm D-C=\left[\begin{array}{ccc}12&-2&3\\3&1&-2\\\end{array}\right]\)

Hence the final matrix will be as follows   \(\rm D-C=\left[\begin{array}{ccc}12&-2&3\\3&1&-2\\\end{array}\right]\)

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The LCD of 1/(x+2)^2, 3/(x-2)^2, and 4/(x^2-4) is (x+2)^2(x-2)^2(x+2)(x-2) = (x+2)^3(x-2)^3.

To find the lowest common denominator (LCD) of three fractions, we need to identify the factors of each denominator and select those factors that are common to all denominators.The given fractions are: 1/ (x+2)^2, 3/ (x-2)^2, and 4/ (x^2-4).

The first denominator (x+2)^2 can be factored using the formula for the difference of two squares. The denominator of 3/(x-2)^2 can also be factored using the formula for the difference of two squares.

Finally, the denominator of 4/(x^2-4) can be factored using the formula for the difference of two squares and the formula for the sum of two squares.

So, 1/(x+2)^2, 3/(x-2)^2, and 4/(x^2-4) can be rewritten as follows:1/ (x+2)^2 = A/(x+2) + B/(x+2)^2. Multiplying both sides by (x+2)^2 gives:1 = A(x+2) + B ... (1)3/ (x-2)^2 = C/(x-2) + D/(x-2)^2.

Multiplying both sides by (x-2)^2 gives:3 = C(x-2) + D ... (2)4/ (x^2-4) = E/(x-2) + F/(x+2). Multiplying both sides by (x^2-4) gives:4 = E(x-2)(x+2) + F(x+2)(x-2) ... (3)

Now, we need to solve these equations for A, B, C, D, E, and F. First, we will solve equation (1) for A and B by substituting x = -2, which gives A = -1/4. Then we will substitute x = -2 again and solve for B, which gives B = 1/4. Next, we will solve equation (2) for C and D by substituting x = 2, which gives C = 3/4.

Then we will substitute x = 2 again and solve for D, which gives D = -3/4. Finally, we will solve equation (3) for E and F by substituting x = 2 and x = -2, which gives E = 1/2 and F = 1/2.

Now, we have A = -1/4, B = 1/4, C = 3/4, D = -3/4, E = 1/2, and F = 1/2. The LCD is the product of all the factors that appear in the denominators of the fractions, each raised to the highest power that appears.

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The simplified value of the given expression (2x^3 + 3x + 7) / (x^2 + x + 10) is: the quotient will be 2x - 2 and the remainder will be -15x + 27.

Let us solve the given algebraic expression by the long division method:

We are given the expression:

(2x^3 + 3x + 7) / (x^2 + x + 10)

We need to perform the division of the given algebraic expression:

x^2 + x + 10  ) 2x^3 + 3x + 7 ( 2x - 2

                       2x^3 + 2x^2 + 20x

                     -         -            -        

                         -2x^2 - 17x + 7

                         -2x^2 - 2x  - 20

                         +        +       +          

                                  -15x + 27

So,

quotient = 2x - 2

remainder =  -15x + 27

Thus, the simplified value of the given expression (2x^3 + 3x + 7) / (x^2 + x + 10) is: the quotient will be 2x - 2 and the remainder will be -15x + 27.

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Answer:

(x + 4)^2 + (y + 7)^2 = 36

Step-by-step explanation:

The given information describes a circle with its center at (-4, -7) and tangent to the vertical line x = 2. To determine the radius of the circle, we need to find the distance between the center and the tangent line.

The distance between a point (x1, y1) and a line Ax + By + C = 0 is given by:

d = |Ax1 + By1 + C| / sqrt(A^2 + B^2)

In this case, the equation of the line is x = 2, which can be written as 1x + 0y - 2 = 0. Therefore, A = 1, B = 0, and C = -2. The center of the circle is (-4, -7), so x1 = -4 and y1 = -7. Substituting these values into the formula, we get:

d = |1*(-4) + 0*(-7) - 2| / sqrt(1^2 + 0^2)

d = |-6| / sqrt(1)

d = 6

Therefore, the radius of the circle is 6 units. The equation of a circle with center (h,k) and radius r is given by:

(x - h)^2 + (y - k)^2 = r^2

Substituting the values we have found, we get:

(x + 4)^2 + (y + 7)^2 = 36

This is the equation of the circle that satisfies the given conditions.

Answer:

1-23.5

2-2

3-5.59

4-0.278

The probability that a sample of size n = 62 is randomly selected with a mean between 189.7 and 213 is approximately 0.9702.

To find the probability that a single randomly selected value is between 189.7 and 213, we can use the standard normal distribution.

Step 1: Calculate the z-scores for the given values using the formula:

  z = (x - μ) / σ

  For 189.7:

  z1 = (189.7 - 189.7) / 96.7 = 0

  For 213:

  z2 = (213 - 189.7) / 96.7 ≈ 0.2417

Step 2: Utilize a standard typical conveyance table or number cruncher to find the probabilities comparing to the z-scores.

  P(189.7 < X < 213) = P(0 < Z < 0.2417) ≈ 0.0939

Therefore, the probability that a single randomly selected value is between 189.7 and 213 is approximately 0.0939.

To find the probability that a sample of size n = 62 is randomly selected with a mean between 189.7 and 213, we use the central limit theorem. Under specific circumstances, the testing dispersion of the example mean methodologies a typical conveyance

Step 1: Calculate the standard error of the mean (σ_m) using the formula:

  σ_m = σ / sqrt(n)

  σ_m = 96.7 / sqrt(62) ≈ 12.2878

Step 2: Convert the given qualities to z-scores utilizing the equation:

  z = (x - μ) / σ_m

  For 189.7:

  z1 = (189.7 - 189.7) / 12.2878 = 0

  For 213:

  z2 = (213 - 189.7) / 12.2878 ≈ 1.8967

Step 3: Utilize a standard typical conveyance table or mini-computer to find the probabilities relating to the z-scores.

  P(189.7 < M < 213) = P(0 < Z < 1.8967) ≈ 0.9702

Therefore, the probability that a sample of size n = 62 is randomly selected with a mean between 189.7 and 213 is approximately 0.9702.

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Answer:

lol no thanks for the points

Step-by-step explanation:

dwadawdwadawdssadws

Answer:

  the sum of the prices divided by 7

Step-by-step explanation:

The mean is the sum of the elements of a set, divided by the number of elements in the set. This set has 7 members, so its mean is ...

  the sum of the prices divided by 7

Answer:

−4304

Step-by-step explanation:

1. The given determinant is :

\(\begin{vmatrix}7 &31 \\ 142& 14\end{vmatrix}\)

We need to find its determinant . It can be solved as follows :

\(\begin{vmatrix}7 &31 \\ 142& 14\end{vmatrix}=7(14)-142(31)\\\\=-4304\)

So, the value of determinant is equal to −4304.

Answer:

A= -4269

B= 1768

C= 647.36

Step-by-step explanation:

Answer:

AB= 7.45

Anwer C)

Step-by-step explanation:

Cos (angle) = Nearest side / Huypothenuse

Cos(20)      = 7 / AB

Cos(20) * AB = (7 /AB) * AB

Cos (20) * AB = 7

(Cos(20) *AB) / Cos(20) = 7 / Cos(20)

AB = 7 / cos(20)

AB=    7.45

Answer:

The number of potential members predicted in next year's estimate becomes 193. The further explanation is given below.

Step-by-step explanation:

The slope of the model,

m = 15.345

c = 24

After comparing with the equation,

⇒  \(y=mx+c\)

we get,

⇒     \(=15.345x + 24\)

The slope seems to be a positive value which also means that another always improves this as component decreases. Which ensures that perhaps the number of contributors rising as time went by.

So that the above is the right answer.

Answer:

------------------

Simplify in below steps:

Answer:

i think a.... sorry if im wrong

Step-by-step explanation:

Answer:

11$ per 2 hours

Step-by-step explanation:

Answer:

First - y=-2x+3

Second - y=-7+3

Step-by-step explanation:

Answer:

Y= -2 x + 3

Y= 3 x + -7

Step-by-step explanation:

Answer:

Samantha can make 16 necklaces.

Step-by-step explanation:

If she has 8 1/4 yards of string and uses 1/2yd for each necklace, then our equation is (8 1/4)/(1/2) or \(\frac{8\frac{ 1}{4} }{\frac{1}{2} }\)  

8 1/4 as an improper fraction is 33/4.

Our new equation is (33/4)/(1/2)

To divide fractions, flip the divisor into a reciprocal and multiply:

\(\frac{33}{4}x\frac{2}{1}\)

Simplify:

\(\frac{33}{2}x\frac{1}{1}\)

33/2

Simplify to a mixed number:

16 1/2

Samantha can make 16.5 necklaces, but you can't have half a necklace, so she can make 16 necklaces.

Answer: The answer is 16

Step-by-step explanation: I have my ways ;>

1.Bricks and timber purchased for construction. (Debit: Bricks - Rs 60,000, Debit: Timber - Rs 35,000, Credit: Bank - Rs 95,000)

2.Transfer of Rs 13,000 to fixed deposit account. (Debit: Fixed Deposit - Rs 13,000, Credit: Current Account - Rs 13,000)

3.Appointment of Mr. S.N. Rao as Accountant. (Debit: Salary Expense - Rs 300, Debit: Security Deposit - Rs 1,000, Credit: Accountant - Rs 300)

4.Goods sold to Shruti with discounts. (Debit: Accounts Receivable - Shruti - Rs 80,000, Credit: Sales - Rs 80,000)

5.Goods purchased from Richa with discounts. (Debit: Purchases - Rs 60,000, Credit: Accounts Payable - Richa - Rs 60,000)

6.Goods sold to Shilpa with discounts and received payment. (Debit: Accounts Receivable - Shilpa - Rs 50,000, Credit: Sales - Rs 50,000)

7.Goods sold to Garima with discounts and received partial payment. (Debit: Accounts Receivable - Garima - Rs 1,00,000, Credit: Sales - Rs 1,00,000)

8.Purchase of land with additional charges. (Debit: Land - Rs 2,00,000, Debit: Brokerage Expense - Rs 2,000, Debit: Registration Charges - Rs 15,000, Credit: Bank - Rs 2,17,000)

9.Proprietor took goods and cash for personal use. (Debit: Proprietor's Drawings - Rs 65,000, Credit: Goods - Rs 25,000, Credit: Cash - Rs 40,000)

10.Goods sold to Charu with profit and discount. (Debit: Accounts Receivable - Charu - Rs 1,20,000, Credit: Sales - Rs 1,20,000)

11.Rent paid for the building. (Debit: Rent Expense - Rs 60,000, Credit: Bank - Rs 60,000)

12.Goods sold to Sunil with profit and discount. (Debit: Accounts Receivable - Sunil - Rs 24,000, Credit: Sales - Rs 24,000)

13.Purchased goods from Nanda and supplied to Helen. (Debit: Purchases - Rs 1,000, Debit: Accounts Payable - Nanda - Rs 1,000, Credit: Accounts Receivable - Helen - Rs 1,300, Credit: Sales - Rs 1,300)

14.Purchased goods from Rohit and Sons and supplied to Madan. (Debit: Purchases - Rs 2,700, Credit: Accounts Payable - Rohit and Sons - Rs 2,700, Debit: Accounts Receivable - Madan - Rs 3,000, Credit: Sales - Rs 3,000)

Here are the journal entries for the given transactions:

1. Bricks and timber purchased for construction:

Debit: Bricks (Asset) - Rs 60,000

Debit: Timber (Asset) - Rs 35,000

Credit: Bank (Liability) - Rs 95,000

2. Transfer to fixed deposit account:

Debit: Fixed Deposit (Asset) - Rs 13,000

Credit: Current Account (Asset) - Rs 13,000

3. Appointment of Mr. S.N. Rao as Accountant:

Debit: Salary Expense (Expense) - Rs 300

Debit: Security Deposit (Asset) - Rs 1,000

Credit: Accountant (Liability) - Rs 300

4. Goods sold to Shruti:

Debit: Accounts Receivable - Shruti (Asset) - Rs 80,000

Credit: Sales (Income) - Rs 80,000

5. Goods purchased from Richa:

Debit: Purchases (Expense) - Rs 60,000

Credit: Accounts Payable - Richa (Liability) - Rs 60,000

6. Goods sold to Shilpa:

Debit: Accounts Receivable - Shilpa (Asset) - Rs 50,000

Credit: Sales (Income) - Rs 50,000

7. Goods sold to Garima:

Debit: Accounts Receivable - Garima (Asset) - Rs 1,00,000

Credit: Sales (Income) - Rs 1,00,000

8.Purchase of land:

Debit: Land (Asset) - Rs 2,00,000

Debit: Brokerage Expense (Expense) - Rs 2,000

Debit: Registration Charges (Expense) - Rs 15,000

Credit: Bank (Liability) - Rs 2,17,000

9. Goods and cash taken away by proprietor:

Debit: Proprietor's Drawings (Equity) - Rs 65,000

Credit: Goods (Asset) - Rs 25,000

Credit: Cash (Asset) - Rs 40,000

10. Goods sold to Charu:

Debit: Accounts Receivable - Charu (Asset) - Rs 1,20,000

Credit: Sales (Income) - Rs 1,20,000

Credit: Cost of Goods Sold (Expense) - Rs 80,000

Credit: Profit on Sales (Income) - Rs 40,000

11. Rent paid for the building:

Debit: Rent Expense (Expense) - Rs 60,000

Credit: Bank (Liability) - Rs 60,000

12. Goods sold to Sunil:

Debit: Accounts Receivable - Sunil (Asset) - Rs 24,000

Credit: Sales (Income) - Rs 24,000

Credit: Cost of Goods Sold (Expense) - Rs 20,000

Credit: Profit on Sales (Income) - Rs 4,000

13. Goods purchased from Nanda and supplied to Helen:

Debit: Purchases (Expense) - Rs 1,000

Debit: Accounts Payable - Nanda (Liability) - Rs 1,000

Credit: Accounts Receivable - Helen (Asset) - Rs 1,300

Credit: Sales (Income) - Rs 1,300

14. Goods received from Rohit and Sons and supplied to Madan:

Debit: Purchases (Expense) - Rs 2,700 (after 10% trade discount)

Credit: Accounts Payable - Rohit and Sons (Liability) - Rs 2,700

Debit: Accounts Receivable - Madan (Asset) - Rs 3,000

Credit: Sales (Income) - Rs 3,000

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