In Exercises 1-6, tell whether the ordered pair is a solution of the system of linear

Winnie Bracken opened a savings account at Dallas Trust Bank on March 1. It pays 4% interest compounded quarterly. She opened he

fiasKO [112]

The compounded interest is applied to the amount in the account at end

of a period specified in the rate of compounding.

The amount in the account at the end of 6 quarters is approximately <u>$16,767.2</u>

Reasons:

The interest paid on the account = 4% compounding

Amount with which she opened the account = $10,000

Amount she makes as deposit at the end of each quarter = $1,000

Therefore;

The interest per quarter = 4% ÷ 4 = 1%

Amount in the account after the end first quarter, A₁, is therefore;

A₁ = 10,000 × 0.01 + 10,000 + 1,000 = 11,100

The amount in the second quarter, A₂, is given as follows;

A₂ = 11,100 × 0.01 + 11,100 + 1,000 = 12211

A₃ = 12211 × 0.01 + 12211 + 1,000 = 13333.11

A₄ = 13333.11 × 0.01 + 13333.11 + 1,000 = 14466.4411

A₅ = 14466.4411 × 0.01 + 14466.4411 + 1,000 = 15611.105511

At the end of the 6th quarter, we have;

A₆ = 15611.105511 × 0.01 + 15611.105511 + 1,000 = 16767.2165991

The amount in the account at the end of 6 quarters, A₆ ≈ $16,767.2

Learn more about compounding interest rate here:

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3 0

2 years ago

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According to a recent survey, the population distribution of number of years of education for self-employed individuals in a c

creativ13 [48]

Answer:

a) X: number of years of education

b) Sample mean = 13.5, Sample standard deviation = 0.4

c) Sample mean = 13.5, Sample standard deviation = 0.2

d) Decrease the sample standard deviation

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 13.5 years

Standard deviation,σ = 2.8 years

a) random variable X

X: number of years of education

Central limit theorem:

If large random samples are drawn from population with mean $\mu$ and standard deviation $\sigma$ , then the distribution of sample mean will be normally distributed with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

b) mean and the standard for a random sample of size 49

$\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{49}} = 0.4$

c) mean and the standard for a random sample of size 196

$\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{196}} = 0.2$

d) Effect of increasing n

As the sample size increases, the standard error that is the sample standard deviation decreases. Thus, quadrupling sample size will half the standard deviation.

7 0

3 years ago

Find the sum of the first six terms of the sequence for which a2=0.7 and a3=0.49.

alina1380 [7]

A1 = 0.91
a6 = -0.14
sum = (0.91+-0.14) /2
sum = 2.31

8 0

3 years ago

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Makeeya only has $25 to spend on a custom T-shirt. It costs $10 for a plain T-shirt, and there is a charge of $$1.50 for each sq

kodGreya [7K]

Answer:

10 square inches

Step-by-step explanation:

cost of a plain t-shirt = $10

cost of 1 square inch of design addition = $1.50

let there be x square inches of design

then cost of x square inches of design = x*cost of 1 square inch of design

= 1.50*x = 1.5x

Total cost for t-shirt with x square inches of design = cost of a plain t-shirt + cost of x square inches of design = 10 + 1.5 x

It is given that Makeeya has only $25, then total cost which can be afforded by him will be $25 only

Thus,

10 + 1.5 x = 25

=> 1.5x = 25-10 = 15

=> x = 15/1.5 = 10

Thus, Makeeya can put 10 square inches of design on her T-shirt.

4 0

3 years ago

An algebra class has 21 students who all participate in a welcome activity introducing themselves to each other in pairs. One st

Nuetrik [128]

The handshake is an illustration of combination, and the values of n and r are 21 and 2, respectively

<h3>How to determine the values of n and r?</h3>

The number of students in the class is 21.

This represents the value of n

i.e. n = 21

Each student shakes hand with another student (i.e. 2 students per handshake)

This represents the value of r

i.e. r = 2

Hence, the values of n and r are 21 and 2, respectively