(Answer the given using PRB with the formula and pls give the full solution!!! )

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

The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2885 and standard deviation

aliina [53]

Answer:

a. X~N(2,885, 651)

b. 0.086291

c. 0.00058

d. 3213.10 calories

Step-by-step explanation:

a. -A normal distribution is expressed in the form X~N(mean, standard deviation).

-Let X a random variable denoting the number of calories consumed.

-X is a is a normally distributed random variable with mean 2885 and standard deviation 651.

-This distribution is expressed as X~N(2,885, 651)

b. The probability that less than 2000 calories are consumed is calculated using the formula:

$P(X$

#substitute the given values in the formula to solve for P:

$P(X$

Hence, the probability of consuming less than 2000 calories is 0.08691

c. The proportion of customers consuming more than 5000 calories is calculated as:

$P(X>x)=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(Z>\frac{5000-2885}{651})\\\\=P(z>3.2488)\\\\=1-0.99942\\\\=0.00058$

Hence, the proportion of customers consuming over 5000 calories is 0.00058

d. The least amount of calories to get the award is calculated as:

1% is equivalent to a z value of 0.50399.

-We equate this to the formula to solve for the mean consumption:

$0.01=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(z>\frac{\bar X-2885}{651})\\\\\1\%=0.50399 \\\\\frac{\bar X-2885}{651}=0.50399 \\\\\bar X=0.50399\times 651+2885\\\\=3213.09$

Hence, the least amount of calories consumed to qualify for the award is 3213.10 calories.

8 0

3 years ago

Point C divides segment AB so that AC:AB is 2:9. Which statement is NOT true?

jeka94

Answer:

A. Point C divides segment AB so that AC:CB is 1:3

or either c

5 0

3 years ago

The sea ice area around the North Pole fluctuates between about 6 million square kilometers in September to 14 million square ki

Aleksandr [31]

Answer:

there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

Step-by-step explanation:

Given the data in the question;

Let S(t) represent the amount sea ice around the North Pole in millions of square meters at a given time t,

t is the number of months since January.

Now, we use a cosine curve to model this scenario

Vertical shift will be;

D = ( 6 + 14 ) / 2 = 20 / 2

D = 10

Next is the Amplitude;

|A| = ( 6 - 14 ) / 2

|A| = 4

Now, the horizontal stretch factor will be;

B = 2π / 12

B = π/6

Hence;

S(t) = 4cos( π/6 × t ) + 10 ----------- let this be equation 1

Now we find when there will be less than 9 million square meters of sea ice;

S(t) = 9

so we have

9 = 4cos( π/6 × (t-2) ) + 10

9 - 10 = 4cos( π/6 × (t-2) )

-1 = 4cos( π/6 × (t-2) )

-1/4 = cos( π/6 × (t-2) )

so we have;

cos⁻¹( -1/4 ) = π/6 × (t₁-2) -------- let this be equation 2

2π - cos⁻¹( -1/4 ) = π/6 × (t₂-2) -------- let this be equation 3

so we solve equation 2 and 3

we have'

t₁ - t₂ = 6/π × ( 2π - cos⁻¹( -1/4 ) - cos⁻¹( -1/4 ) )

t₁ - t₂ = 6/π × ( 2π - 2cos⁻¹( -1/4 )

t₁ - t₂ = 6/π × ( π - cos⁻¹( -1/4 )

t₁ - t₂ = 6/π × ( π - 104.4775 )

t₁ - t₂ = 5.035

therefore, there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

6 0

3 years ago

Help me please brainly

3241004551 [841]

Step-by-step explanation:

1)

3x+3

2)

10+2x

3)

8x+24

4)

8a+8b+8c

6 0

3 years ago