-(-(-)(-)(-10x))=-5 solve for x

Is (8,2) a solution for the system 2y=x-12

erica [24]

I hope this helps you

4 0

3 years ago

Read 2 more answers

| x+5 | > -1 | x-4 | < 0

Daniel [21]

Answer:

|x+5|>-1

-5 -5

|x| > -6

All Real Numbers

|x-4|<0

+4 +4

|x|<4

4 0

3 years ago

An investor bought 40 shares of ABC corporation's stock at $80 a share. Two weeks later, the investor receives notice that the c

Eva8 [605]

The correct answer is C: 80 shares of the stock and the price of each share is $40.

8 0

3 years ago

If a car travels 23 miles on 1.0 gal of gas, how many liters of gasoline are needed for a 135 mile trip?

cluponka [151]

Solution can be gained by using "Product of means = Product of extremes" concept. So the information we have is 23 miles on 1 gal and 135 mile on 'x' gal. 23 : 1 :: 135 : x As per Product of means = Product of extremes, 23 multiplied x = 135 multiplied by 1. And x will be 135 divided by 23 which is 5.86 gal Now, we have to convert gal to liters, 1 gal = 3.78 liters, so 5.86 gal = 5.86 multiplied by 3.78, which is 22.15 liters. Hence, 22.15 liters of gasoline are needed for a 135 mile trip

3 0

3 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