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STALIN [3.7K]

The answer would be 3.7 x 10 to the -6th power.

8 0

3 years ago

Find the fifth term and the nth term of the geometric sequence whose initial term, a, and common ratio, r, are given. a = 3 ; r

Anna [14]

Answer: 1875

Step-by-step explanation:

The nth term of a geometric progression is given as:

= ar^n-1

Given a = 3 ; r = 5, the nth term will be:

= ar^n-1

= 3 × 5^n-1

The fifth term will be:

= 3 × 5^n-1

= 3 × 5^5-1

= 3 × 5⁴

= 3 × 625

= 1875

The fifth term is 1875

6 0

3 years ago

F'(-7, 13) was dilated by a scale factor of 2 about (-2, -5), the resulting image is___

AveGali [126]

The resulting image of the point F(x, y) = (- 7, 13) is F'(x, y) = (- 12, 31).

<h3>What is the resulting image of a point by rigid transformation?</h3>

In this problem we have a point that has to be transformed by a kind of rigid transform known as dilation, which is defined by the following equation:

P'(x, y) = O(x, y) + k · [P(x, y) - O(x, y)] (1)

Where:

O(x, y) - Center of dilation
P(x, y) - Original point
P'(x, y) - Image
k - Scale factor

If we know that O(x, y) = (- 2, - 5), F(x, y) = (- 7, 13) and k = 2, then the resulting image is:

F'(x, y) = (- 2, - 5) + 2 · [(- 7, 13) - (- 2, - 5)]

F'(x, y) = (- 2, - 5) + 2 · (- 5, 18)

F'(x, y) = (- 2, - 5) + (- 10, 36)

F'(x, y) = (- 12, 31)

The resulting image of the point F(x, y) = (- 7, 13) is F'(x, y) = (- 12, 31).

To learn more on rigid transformations: brainly.com/question/28004150

#SPJ1

3 0

2 years ago

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

jeka94

Answer:

(a) The probability that the port handles less than 5 million tons of cargo per week is 0.7291.

(b) The probability that the port handles 3 or more million tons of cargo per week is 0.9664.

(c) The probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2373.

(d) The number of tons of cargo per week that will require the port to extend its operating hours is 5.35 million tons.

Step-by-step explanation:

Let X = amount of cargo the port handles per week.

The random variable X is Normally distributed with parameters,

μ = 4.5 million

σ = 0.82 million

(a)

Compute the probability that the port handles less than 5 million tons of cargo per week as follows:

$P(X$

*Use a z-table for the probability.

Thus, the probability that the port handles less than 5 million tons of cargo per week is 0.7291.

(b)

Compute the probability that the port handles 3 or more million tons of cargo per week as follows:

$P(X\geq 3)=P(\frac{X-\mu}{\sigma}\geq \frac{3-4.5}{0.82})\\=P(Z>-1.83)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that the port handles 3 or more million tons of cargo per week is 0.9664.

(c)

Compute the probability that the port handles between 3 million and 4 million tons of cargo per week as follows:

$P(3$

Thus, the probability that the port handles between 3 million and 4 million tons of cargo per week is 0.2373.

(d)

It is provided that P (X < x) = 0.85.

Then, P (Z < z) = 0.85.

The value of z for this probability is:

z = 1.04.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\1.04=\frac{x-4.5}{0.82}\\x=4.5-+1.04\times 0.82)\\x=5.3528\approx5.35$

Thus, the number of tons of cargo per week that will require the port to extend its operating hours is 5.35 million tons.

5 0

4 years ago

PLEASE HELP!!!!!! WILL MARK BRAINLIEST!!!!

JulijaS [17]

Answer:

23

Step-by-step explanation:

4 0

3 years ago

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