All categories

2 years ago

What is the value of the fourth term in a geometric sequence for which a1=10 and r =0.5

Mathematics

1 answer:

Lena [83]2 years ago

4 0

Answer: 1.25

Step-by-step explanation:

The explicit formula for the sequence is $a_{n}=10(0.5)^{n-1}$

Substituting in n=4,

$a_{4}=10(0.5)^{4-1}=\boxed{1.25}$

You might be interested in

A company that has 350000 shares declares an annual gross profit of $2450665, pays 18.5% of this in tax, and reinvests 25% of th

podryga [215]

Gross profit, G = $2450665

Tax, t=18.5%

reinvestment, r = 25%

Total dividends

= G(1-t)(1-r)

=2450665*(1-0.185)(1-0.25)

=1497968.98

Dividend per share

=1497968.98/350000

=4.280

Earnings per share

EPS = Net profit / number of shares

= 2450665(1-0.185)/350000

=5.7065

Current price = 43.36

P/E ratio

= Current price/EPS

= 43.36/5.7065

= 7.598

=7.6 to one decimal place.

8 0

3 years ago

X squared minus 7x plus 10 equals 0

joja [24]

X² - 7x + 10 = 0

Let's factor out the equation first.

x² - 7x + 10

x -2
x -5

This factoring of (x-2)(x-5) fit the equation.

(x-2)(x-5) = 0

One of the factors must equal 0 as to equal zero.

Either x - 2 = 0 or x - 5 = 0

x - 2 = 0
x = 2

x - 5 = 0
x = 5

The two answers are x = 2 or x = 5. Hope this helps!

3 0

3 years ago

What’s the exact volume?

MAXImum [283]

I think the answer is option 3.

8 0

3 years ago

Read 2 more answers

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

$P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)$

Calculation the value from standard normal z table, we have,

$P(x < 5) =0.7287= 72.87\%$

b) P( port handles 3 or more million tons of cargo per week)

$P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)$

Calculating the value from the standard normal table we have,

$1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%$

c)P( port handles between 3 million and 4 million tons of cargo per week)

$P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%$

$P(3 \leq x \leq 4) = 23.7\%$

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85$

$= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85$

$=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15$

Calculation the value from standard normal z table, we have,

$P( z \leq -1.036) = 0.15$

$\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65$

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0

3 years ago

I nee help. Can anyone help please.

Radda [10]

You have to take 60 degrees and add it to 90 degrees because that’s how much the box in the corner is worth

Then add those 2 together which is 150 after that’s subtract it by 270 which is 120

Answer: 120 degrees or 120

7 0

3 years ago