PLEASE HELP!! how do you solve - 3/4 + 0.25= PLEASE show work if u can?

If the required reserve ratio is 10 percent, currency in circulation is $1,200 billion, checkable deposits are $1,600 billion, and excess reserves total $2,500 billion

Given that

Reserve ratio is 10 percent

Currency in circulation is $1,200 billion

Checkable deposits are $1,600 billion

Excess reserves total $2,500 billion

Here we are calculating M1 measure of money supply

C: Currency in circulation

D: Checkable Deposits

M1=C+D = 1200+1600= 2800

C: Currency in circulation

D: Checkable Deposits

Currency Deposit Ratio= C/D = 1200/1600 = 0.75

ER: Excess Reserves

CD: Checkable Deposits

Excess Reserve Ratio= ER/CD= 2500/1600 = 1.56

The formula for the M1 money multiplier is:

M1 money multiplier= M1/MB

M1 money multiplier= C+D / R+C

M1 money multiplier= C+D / q∗D+ER+C

M1 money multiplier= (1,200+1,600) / (0.1∗1,600+2,500+1,200)

M1 money multiplier= 0.73

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Answer:

C

Step-by-step explanation:

We know that secant is the same as \(\frac{1}{cos}\) , so knowing that we can immediately find cosΘ by flipping the given secant value to be \(\frac{12}{-13}\) . Since only C has this for cosΘ, this is your answer. However, if you wanted to check further using a triangle. Using the knowledge that secant is \(\frac{1}{cos}\) and cosine is adjacent over hypotenuse (SOH CAH TOA), we can find that our hypotenuse is -13 and the adjacent side is 12. From there we can find the remaining functions using 5 as your opposite (given in the solutions) and SOH CAH TOA for each sine, cosine and tangent, flipping each for cosecant and cotangent.

1. The correct exponential function for each graph is

a.  f(x) = 2ˣ is graph (III)     b. f(x) = 2⁻ˣ is graph (I)   c. f(x)= -2ˣ graph (II)  D. f(x) = -2⁻ˣ is graph (IV)

2. The evaluated exponential functions are;

a. f(1) = 1  b. f(-1) = 0.1  c. f(0) = 0.3

3. The exponential function for the graph  is f(x) = \(\frac{1}{2}^x\)

4. The model for the rate population is N(t) = 320 × \(2^{t}\) . After 8 years, the population will be 81,920.

5. The compounded interest after 2 years: 2628.17. After 3 years 2694.7. After 6 years 2904.57

6. The compounded interest after a) 3 years = 4221.81  b) 6 years = 5092.47  c) 9 years = 6142.69

7. At time t = 0, the mass is 13 kg. After 45 days, the mass is approximately 6.62 kg

1. For the exponential function,  

a. f(x) = 2ˣ has a coordinate (0,1).  f(1) = 2⁰ = 1 It matches (III)

b.  f(x) = 2⁻ˣ When x = -1, f(x) = 2,  x = 0, f(x) = 1, x = 1, f(x) = 0.5; x = 2, f(x) = 0.25. If you plot this coordinates, you will see the graph falls from left to right

c. f(x)= -2ˣ it should pass through  (0, -1),  x = -1, when f(x) = -0, x = 0, f(x) = -1, x = 1, f(x) = -2

d. f(x) = -2⁻ˣ will pass through the point (0, -1). When x = -1, f(x) = -2; x = 0, f(x) = -1; x = 1, f(x) = -0.5. The graph should rise from left to right.

2. If the exponential function is f(x) = 3ˣ⁻¹ then we evaluate

f(1) = 3¹⁻¹ = 3⁰ = 1

f(-1) = 3⁻¹⁻¹ = 3⁻² = 1/3² or 0.1

f(0) = 3⁰⁻¹ = 3⁻¹ = 1/3 or 0.3

3. For the graph whos exponential function should be in the form f(x) = aˣ, we can find the actual function by looking at the coordinated (-3, 8)

a⁻³ = 8

a = \(8^1/-3\)

a = 1/2

f(x) = (1/2)ˣ

4. The mouse population can be modeled as

N(t) = N₀ × (1 + \(r)^{t}\) ⇒ N(t) = 320 × \(2^{t}\)

After 8 years ⇒ N(8) = 320 × 2⁸

N(8) = 320 × 256

N(8) = 81920

5. We use the formula A = P(1 + \(r/n)^{nt}\)

$2500 invested at an annual interest rate of 2.5% compounded daily,

After 2 years (t = 2):                          After 3 years

A = 2500(1 + 0.025/365)⁽³⁶⁵ˣ²⁾         A = A = 2500(1 + 0.025/365)⁽³⁶⁵ˣ³⁾

A = 2628.17                                       A = 2694.7

After 6 years

A = 2500(1 + 0.025/365)⁽³⁶⁵ˣ⁶⁾

A = 2904.57

6. For a  continuous compounded interest, we use the formula;

A = P × \(e^{rt}\)

the principal is $3500, the interest rate is 6.25%, or 0.0625 as a decimal.

a) After 3 years:                 After 6 years:             After 9 years

A = 3500 × \(e^{0.0625 * 3}\)          A = 3500 × \(e^{0.0625 * 6}\)     A = 3500× \(e^{0.0625 * 9}\)

A = 4221.81                       A = 5092.47                A = 6142.69

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Answer:

\(x = 34m\)

Step-by-step explanation:

Given

\(Height = 68\ metres\)

Required

Determine the halfway

Represent the halfway with x

\(x = \frac{1}{2} * Height\)

Substitute 68 for Height

\(x = \frac{1}{2} * 68m\)

\(x = \frac{68m}{2}\)

\(x = 34m\)

The sequence is decreasing as n increases and sequence converges to the value 0.

The given sequence is defined as aₙ = 1 / (7n + 3).

To determine if the sequence converges or diverges, we need to analyze its behavior as n approaches infinity.

As n increases, the denominator 7n + 3 also increases which means that the values of aₙ will get smaller and smaller, approaching zero as n becomes larger.

The sequence converges to the value 0.

The sequence is decreasing as n increases.

The sequence converges to the value 0.

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Answer:

The purchase value at the 30th percentile=34.24

Step-by-step explanation:

We are given that

Mean,\(\mu=41.34\)

Standard deviation,\(\sigma=13.54\)

We have to find the purchase value at the 30th percentile.

\(xth percentile =\mu+Z\times \sigma\)

Where Z is the critical value of x% confidence interval

x=30

Critical value of Z at 30% confidence interval=-0.5244

Using the formula

30th percentile=\(41.34+(-0.5244)(13.54)\)

30th percentile=\(41.34-7.100376\)

30th percentile\(\approx 34.24\)

Hence,  the purchase value at the 30th percentile=34.24

Given that 16 families went on a trip and the cost of the trip was Rs. 2,16,352.The amount paid by each family is to be determined by unitary method Hence each family paid Rs.13522

Now, let's solve this by using the method of unitary method. To find the cost of 1 family trip, we will divide the total cost of the trip by the number of families.2,16,352 / 16 = 13,522 So, the cost of the trip per family is Rs. 13,522.Hence, each family paid Rs. 13,522 for the trip.

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Answer:

Step-by-step explanation

1. The total cost of the trip for all 16 families is Rs 2,16,352.
2. To find out how much each family paid, we need to divide the total cost by the number of families: Rs 2,16,352 ÷ 16.
3. When we do the division, we get the result: Rs 13,522.

Now let's check if this result is correct:

1. If each family paid Rs 13,522 for the trip, then the total cost for all 16 families would be: 16 × Rs 13,522 = Rs 2,16,352.
2. This is exactly the same as the total cost given in the problem statement.

So we have shown that each family paid **Rs 13,522** for the trip

Answer:

1.45cm

Step-by-step explanation:

I just added

A square of the length of the sides are:

The upper bound is  7.6.

The lower bound is 7.4

The area of the square:

The value of upper bound is :

\(7.6^2 = 54.76 cm^2\)

The value of lower bound is:

\(7.4^2= 57.76cm^2\)

We have the following information from the question is:

A square has sides of 7.5 cm, measured correctly to 1 d. p.

We have to calculate the Upper and lower bounds for the length of the sides.

and, the lower and upper bounds for the area of the square.

Now, According to the question:

1 d. p. = 1 decimal place.

A square has side is 7.5 cm

The upper bound is going to be 7.5 plus . 1, which would be 7.6.

The lower bound is going to be 7.5 minus . 1, which is going to be 7.4

Now, The lower and upper bounds for the area of the square is:

We know the

area of square = side^2

So, the value of upper bound is 7.6

\(7.6^2 = 54.76 cm^2\)

Similarly, The value of lower bound is:

\(7.4^2= 57.76cm^2\)

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Answer:

Step-by-step explanation:

If f(x)=x²+4 then (fof⁻¹)(x)=(f⁻¹of)(x)=x is proved

A relation is a function if it has only One y-value for each x-value.

Composite functions are when the output of one function is used as the input of another.

If f(x)=x²+4

Then we have to prove f⁻¹of(x)=f⁻¹(f(x)

Let us consider fof⁻¹(x)

f(f⁻¹(x))

x

So (fof⁻¹(x))=x

Now f⁻¹of(x)=f⁻¹(f(x)

=x

Hence, If f(x)=x²+4 then (fof⁻¹)(x)=(f⁻¹of)(x)=x is proved

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Answer:

  46, 48

Step-by-step explanation:

The average of the two integers will be the odd integer that lies between them. The average of the two integers is half their sum.

The average is ...

  n = 94/2 = 47

The two even integers are (n-1) = 46, and (n+1) = 48.

Answer:

60 contain 6 tens

then

60+10

contain

7 tens

the

correct answer

is

7 tens

Answer:

6 tens + 1 ten = 7 tens

Step-by-step explanation:

6 tens = 60

1 ten = 10

7 tens = 70

Answer:

I am pretty sure the answer is 27.02.

Step-by-step explanation:

Hope this helps!

The number of students who participated in the survey, based on the histogram showing the number of hours students watch television on the weekend, is 125.

A histogram is a pictorial or graphical representation of categorical data, proportional to the frequency of a variable and whose width is equal to the class interval.

The number of students who watched between 0 - 4 hours = 20

The number of students who watched between 5 - 9 hours = 40

The number of students who watched between 10 - 14 hours = 30

The number of students who watched between 15 - 19 hours = 20

The number of students who watched between 20 - 24 hours = 15

The total number of students who participated = 125

Thus, using the histogram, the total number of students who participated in the survey was 125.

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Answer:

The answer is measles

Step-by-step explanation:

Answer:

17.89 miles long.

Step-by-step explanation:

The pedestrian route that runs diagonally across the rectangular park is the hypotenuse of a right triangle whose legs are the length and width of the park. We can use the Pythagorean theorem to find the length of this diagonal route.

According to the Pythagorean theorem, the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of its legs. In this case, one leg of the right triangle is 16 miles long and the other leg is 8 miles long. So, we have:

d² = 16² + 8²

d² = 256 + 64

d² = 320

d = √320

d ≈ 17.89

So, the pedestrian route that runs diagonally across the park is approximately 17.89 miles long.

Answer:

Sorry I could only find the answer to the second part.

y=56x

Answer:

y=3x-3

Step-by-step explanation:

Answer:

-7

Step-by-step explanation:

8-15 = -7 but since it's an absolute value, it is 7 and then you add the negative sign in front since it is not in the absolute value.

Answer:

56

Step-by-step explanation:

7 times 8=56

Answer:

1 1/8

Step-by-step explanation:

Answer:

1 1/8

Step-by-step explanation:

hope this is right

The expected number of investigators who will report that 3, 4, or 5 people are consumers of rice is 57 investigators.

The binomial distribution B(n, p) describes the behavior of a count X of successes in n independent trials, each with its own boolean-valued outcome: success (with probability p) or failure (with probability q = 1-p).

We want to find the probability that an investigator reports 3, 4, or 5 people are consumers of rice. The probability P(X=k) for a binomial distribution is given by the formula:

P(X=k) = C(n, k) * p^k * q^(n-k)

So we want to find P(X=3) + P(X=4) + P(X=5) :

P(X=3) = C(10, 3) * (0.5)^3 * (0.5)^(10-3)

= 0.117

P(X=4) = C(10, 4) * (0.5)^4 * (0.5)^(10-4)

= 0.205

P(X=5) = C(10, 5) * (0.5)^5 * (0.5)^(10-5)

= 0.246

The expected number of investigators who will report three or less of the people interviewed consumed is :

= P(X=3) + P(X=4) + P(X=5)

= 0.117 + 0.205 + 0.246

= 0.568

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Full question is:

Suppose that half the population of a town are consumers of rice. Each of 100 investigators interviews 10 individualas to see wheather they are consumers of rice. How many investigators do you expert to report that three or more but less than six people are consumers of rice?

Answer:

Step-by-step explanation:

44/8

= 5 whole 4 by 8

= 5 4/8

Answer:

5/2 = 2.5

Step-by-step explanation:

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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The statement A and statement B are supported by the data in the table. The solution has been obtained by studying the bar graph.

What is a bar graph?

Bar graphs or bar charts are visual depictions of groups of data that are made up of vertical or horizontal rectangular bars with lengths that are equal to the data's measure.

We are given a graph showing grasshopper numbers before and after blue jays preyed on them.

From the graph, we can observe that there is no change in the number of green grasshoppers. This means that blue jays did not prey them.

Also, there is a great fall in the brown grasshoppers which means that they are most visible to the blue jays.

Hence, the statement A and statement B are supported by the data in the table.

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The complete question has been attached below

Answer:

i think it's -1 / -1

nut i'm not sure so if its wrong then sorry , and don't thank's me...

Length is 6 area is 000 it most be answer you please write in your copy.

Answer:

The statements that can be used to compare the characteristics of the functions are:

1. g(x) has the greatest maximum value.

2. All three functions share the same domain.

Explanation:

- The table shows the values of three functions - f(x), g(x), and h(x) - evaluated at different values of x.

- We cannot determine the domain of f(x) or h(x) from the given table but we can see that g(x) has a domain of all real numbers.

- We can see that g(x) has the highest maximum value among the three functions, which is 8.

- We cannot determine the range of f(x) or g(x) from the given table but we can see that h(x) has a range of all negative numbers.

- We cannot say anything about the domain or range of f(x) based on the given table.

- Therefore, the two statements that can be used to compare the characteristics of the functions are: g(x) has the greatest maximum value and all three functions share the same domain.

Answer: 45%

Step-by-step explanation: