Marie has a building that is 40 feet wide, 70 feet long, and 16 feet high. How many cubic feet of

Assume the rate of inflation is 8% per year for the next 2 years. What will be the cost of goods 2 years from now, adjusted fo

Paha777 [63]

<h3>Answer: $326.59</h3>

==============================================

Work Shown:

F = future value

P = present value = 280

r = rate of inflation in decimal form = 0.08

t = elapsed time in years = 2

---------

F = P*(1+r)^t

F = 280*(1+0.08)^2

F = 326.592

F = 326.59

8 0

3 years ago

Simplify the expression below.

QveST [7]

Answer:

(c)

Step-by-step explanation:

x^3+3 = x^6

hope it helped :)

5 0

2 years ago

In a lottery game, a player picks six numbers from 1 to 27. If the player matches all six numbers, they win 40,000 dollars. Othe

Alexus [3.1K]

Answer:

We conclude that expected value of this game is -0.865$.

Step-by-step explanation:

We know that in a lottery game, a player picks six numbers from 1 to 27.

We know that

$C_6^{27}=296010$

As there is only one advantageous combination, we conclude that the number of non-winning combinations is 296009.

He can win 40,000 dollars.

We calculate:

$E(X)=\frac{1}{296010}\cdot 40000\$- \frac{296009}{296010}\cdot 1\$\\\\E(X)=\frac{40000-296009}{296010}\, \$\\\\E(X)=-0.865\, \$$

We conclude that expected value of this game is -0.865$.

7 0

3 years ago

Any suggestions or answers

Sati [7]

Answer:

$x=81\degree$

$y=99\degree$

$z=129\degree$

Step-by-step explanation:

The sum of the interior angles of a triangle is 180 degrees.

$\implies 51\degree+48\degree+x=180\degree$

$\implies 99\degree+x=180\degree$

$\implies x=180\degree-99\degree$

$\implies x=81\degree$

Use the exterior angles theorem to find y.

The exterior angle y is the sum of the two remote interior angles:

$y=51\degree+48\degree=99\degree$

The sum of angles on a straight line is 180 degrees.

$\implies z+51\degree=180\degree$

$\implies z=180\degree-51\degree$

$\implies z=129\degree$

7 0

3 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

IgorLugansk [536]

Answer:

(a) 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

Step-by-step explanation:

We are given that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.75.

(a) Also, the average porosity for 20 specimens from the seam was 4.85.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.85

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 20

$\mu$ = true average porosity

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.85-1.96 \times {\frac{0.75}{\sqrt{20} } }$ , $4.85+1.96 \times {\frac{0.75}{\sqrt{20} } }$ ]

= [4.52 , 5.18]

Therefore, 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) Now, there is another seam based on 16 specimens with a sample average porosity of 4.56.

The pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.56

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 16

$\mu$ = true average porosity

Here for constructing 98% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 98% confidence interval for the true mean, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98 {As the critical value of z at 1% level

of significance are -2.3263 & 2.3263}

P(-2.3263 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.3263$ ) = 0.98

P( $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.56-2.3263 \times {\frac{0.75}{\sqrt{16} } }$ , $4.56+2.3263 \times {\frac{0.75}{\sqrt{16} } }$ ]

= [4.12 , 4.99]

Therefore, 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

7 0

3 years ago