What is the inverse of p(m) = m/5 + 8

navik [9.2K]

Find the total cost of producing 5 widgets. Widget Wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x. The company also discovered that it costs $15.50 to produce 3 widgets, $23.50 to produce 7 widgets, and $56 to produce 12 widgets.
OK...so we have
a(7)^2 + b(7) + c = 23.50 → 49a + 7b + c = 23.50 (1)
a(3)^2 + b(3) + c = 15.50 → 9a + 3b + c = 15.50 subtracting the second equation from the first, we have
40a + 4b = 8 → 10a + b = 2 (2)
Also
a(12)^2 + b(12) + c = 56 → 144a + 12b + c = 56 and subtracting (1) from this gives us
95a + 5b = 32.50
And using(2) we have
95a + 5b = 32.50 (3)
10a + b = 2.00 multiplying the second equation by -5 and adding this to (3) ,we have
45a = 22.50 divide both sides by 45 and a = 1/2 and using (2) to find b, we have
10(1/2) + b = 2
5 + b = 2 b = -3
And we can use 9a + 3b + c = 15.50 to find "c"
9(1/2) + 3(-3) + c = 15.50
9/2 - 9 + c = 15.50
-4.5 + c = 15.50
c = 20
So our function is
c(x) = (1/2)x^2 - (3)x + 20
And the cost to produce 5 widgets is = $17.50

7 0

3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

3 years ago

Sample Response: First, I wrote each ratio as a fraction. Then I found a common denominator of 20. The fraction 4/10 is 8/20. Co

lesantik [10]

In the case above, the option that one need to include in your answer is to Find a common denominator.

<h3>What is Fraction in lowest terms?</h3>

When the numerator and denominator of a fraction are known to not have common factor other than 1, then a person can say that the fraction in its simple form or say its in its lowest term.

Since there is a common numerator and the common denominator will be between 10 and 20. Since 10 is less that than 20, the common denominator will be 20 as both can divide it.

Hence, In the case above, the option that one need to include in your answer is to Find a common denominator.

Learn more about Fraction from

brainly.com/question/78672

#SPJ1

8 0

2 years ago

Hello! Please help me! I'm not good at subtracting fractions!

mafiozo [28]

The answer in an improper fraction is
47/36

As a mixed number, it is 1 11/36

Hope you become an expert at this soon :)

3 0

3 years ago

Read 2 more answers