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pshichka [43]
2 years ago
15

What amount will an account have after 2 years if $2950 is invested at an annual rate of 8.4% compounded quarterly? Also, determ

ine the investment into an account at an annual rate of 7.8% compounded continuously for the same period of time to accumulate $3742.
Mathematics
1 answer:
ankoles [38]2 years ago
3 0

The amount if $2950 is invested at an annual rate of 8.4% compounded quarterly is $3486

<h3>Compound interest</h3>

Compound interest is given by:

A=P(1 + \frac{r}{n} )^{nt}

where t is the period, r is the rate, P is the initial amount, A is the final amount and n is the times compounded.

Given P = 2950, r = 0.084, n = 4, t = 2, hence:

A=2950(1+\frac{0.084}{4} )^{4*2}\\\\A=\$3486\\\\For \ r=0.078, A=3742,t=2,n=1:\\\\3742=P(1+\frac{0.078}{1} )^{1*2}\\\\P=\$3220

The amount if $2950 is invested at an annual rate of 8.4% compounded quarterly is $3486

Find out more on Compound interest at: brainly.com/question/24924853

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Twice x is 5” to an equation
Anna007 [38]
It is similar to an expresion just break it down x*2 is 5 nothing that is a whole number will complete this equation so the answer is 2.5

3 0
3 years ago
Read 2 more answers
Pls helppp What are the factors of the product represented below?​
Nookie1986 [14]
No sehhzvzhgVshsjsbskSksjkdisksvsjs
6 0
3 years ago
According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe
Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1 - p)}{n}}, as long as np \geq 10 and n(1 - p) \geq 10.

In this problem:

  • 22% of couples meet online, hence p = 0.22.
  • A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

\mu = p = 0.22

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem:

Z = \frac{X - \mu}{s}

Z = \frac{0.25 - 0.22}{0.0338}

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

Z = \frac{X - \mu}{s}

Z = \frac{0.2 - 0.22}{0.0338}

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

Z = \frac{X - \mu}{s}

Z = \frac{0.15 - 0.22}{0.0338}

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

4 0
2 years ago
Bob has two weekend jobs last weekend he made a total of $77 after working as a cashier for 5 hours and delivering newspapers fo
OlgaM077 [116]

Answer: $9 per hour at his job as a cashier and $8 per hour at his job delivering newspapers.

Step-by-step explanation:

1. Let's call the amount he got paid per hour at his job as a cashier: x.

Let's call the amount he got paid per hour at his job delivering newspapers: y.

2. Keeping on mind the information given in the problem above, you can make the following system of equations:

\left \{ {{5x+4y=77 \atop {6x+3y=78}} \right.

3. You can solve it by applying the Substitution method, as following:

- Solve for one of the variables from one of the equations and substitute it into the other equation to solve for the other variable and calculate its value.

- Substitute the value obtained into one of the original equations to solve for the other variable and calculate its value.

 4. Therefore, you have:

5x+4y=77\\5x=77-4y\\x=\frac{77-4y}{5}=15.4-0.8y

Then:

6(15.4-0.8y)+3y=78\\92.4-4.8y+3=78\\-1.8y=-14.4\\y=8

Finally:

5x+4(8)=77\\5x+32=77\\5x=45\\x=9

Therefore he got paid $9 per hour at his job as a cashier and $8 per hour at his job delivering newspapers.

5 0
3 years ago
ABCD is a rhombus, where m∠AED = 5x – 10. Use the properties of a rhombus to determine the value of x. Question 10 options: A) x
fomenos

Answer:

A) x = 20

Step-by-step explanation:

ABCD is a rhombus, AC and BD are diagonals which intersect each other at point E.

Since, diagonals of a rhombus are perpendicular bisector.

\therefore \: m \angle AED = 90 \degree \\  \therefore \: (5x - 10) \degree = 90 \degree \\ \therefore \: 5x - 10 = 90  \\ \therefore \: 5x  = 90 + 10  \\ \therefore \: 5x  = 100  \\ \\  \therefore \: x  = \frac{100}{5}  \\  \\ \huge \red{ \boxed{ \therefore \: x  =20}}

5 0
3 years ago
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