The surface area is made up of six sides and the area is then just the sum of the areas of the six sides...

A=2xy+2xz+2yz

A=2(xy+xz+yz) in this case:

A=2(5*4.5+5*3.5+4.5*3.5)

A=2(22.5+17.5+15.75)

A=2(55.75)

A=111.5 in^2

3 0

3 years ago

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If Cody is draining his hot tub which holds 1,400 gallons of water at a rate of 27 gallons a minute, what impact will it have on

inessss [21]

Facts:

- tub has 1400 gallons

- draining at 27 gallons/minute

How much water will be left after 18 minutes?

We know that after 1 minute, there will be 27 gallons less.

So after 18 minutes, there must be 18 times less water.

27 x 18 = 486

This is how much water was lost in 18 min

So to find out how much he had left after <u>486 is subtracted.</u>

Therefore, the answer is A) -486 gallons

4 0

2 years ago

What is the exact solution to the equation 4^5x=3^(x-2)<br> list the steps to get the answer

scoray [572]

Answer: x = -0.377

Step-by-step explanation:

We have the equation:

4^(5*x) = 3^(x - 2)

Now we can use the fact that:

Ln(A^x) = x*Ln(A)

Then we can apply Ln(.) to both sides of the equation to get:

Ln(4^(5*x)) = Ln(3^(x - 2))

(5*x)*Ln(4) = (x - 2)*Ln(3)

(5*x)*Ln(4) - x*Ln(3) = -2*Ln(3)

x*(5*Ln(4) - Ln(3)) = -2*Ln(3)

x = -2*Ln(3)/(5*Ln(4) - Ln(3)) = -0.377

6 0

2 years ago

The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 manag

VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

We have to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a p-value of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{2050}{\sqrt{49}} = 574$

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

5 0

2 years ago