Which statements about the graph of the function y = (StartFraction 1 Over 3 EndFraction) Superscript x are true?

A lightbulb manufacturer ships large consignments of lightbulbs to big industrial users. When the production process is function

Mashutka [201]

Answer:

a) P[ Pf | Ptd ] = 0,428

b) P[ Pf | Nd ] = 0,692

Step-by-step explanation:

We will call:

Pf Probability of functioning correctly 60 % or 0.6

Pfn Probability of malfunctioning 40 % or 0,4

PNd Probability of nondefective bulb

D₁ Probability of defective bulb when the process is correctly working

D₂ Probability of defective bulb when the process is in malfunction condition

Ptd Total probability of defective bulbs

Then applying theorem´s Bayes have

P [ A | B ] = P(A) * P [ B | A ] / P(B)

a) Probabilty of the process is functioning correctly given that we pick up a defective bulb

The total probability of defective bulbs is equal to, probabilty of defective bulbs when the process is Ok, plus probability of defective bulbs when the process is in malfunctioning condition, therefore

Ptd = (0,6)*(0,25) + (0,40)*(0,50) = 0,15 * 0,20 = 0,35

P[ Pf | Ptd ] = Pf * P [ Pd | Pf ] / Ptd = 0,6 * 0,25 / 0,35

P[ Pf | Ptd ] = 0,428

b) P [ Pf | Nd ]

P[ Pf | Nd ] = Pf * P [Nd | Pf ] / PNd

PNd = 1 - Ptd = 1 - 0,35 = 0,65

P[ Pf | Nd ] = 0,6 * 0,75 / 0,65 = 0,692

P[ Pf | Nd ] = 0,692

6 0

2 years ago

There were 200 members at a skating club. After some new members went in, there were 256 members altogether at the skating club.

mojhsa [17]

If I think I’m understanding it, the new members out of the 200 original would be adding 28% to the mix

4 0

2 years ago

If an intravenous solution containing 123 mg of a drug substance in each 250-mL bottle is to be administered at the rate of 200

12345 [234]

Answer:

24.39mL of the solution would be given per hour.

Step-by-step explanation:

This problem can be solved by direct rule of three, in which there are a direct relationship between the measures, which means that the rule of three is a cross multiplication.

The first step to solve this problem is to see how many mg of the solution is administered per hour.

Each minute, 200 ug are administered. 1mg has 1000ug, so

1mg - 1000 ug

xmg - 200 ug

$1000x = 200$

$x = \frac{200}{1000}$

$x = 0.2mg$

In each minute, 0.2 mg are administered. Each hour has 60 minutes. How many mg are administered in 60 minutes?

1 minute - 0.2 mg

60 minutes - x mg

$x = 60*0.2$

$x = 12mg$

In an hour, 12 mg of the drug is administered. In 250 mL, there is 123 mg of the drug. How many ml are there in 12 mg of the drug.

123mg - 250mL

12 mg - xmL

$123x = 250*12$

$x = \frac{250*12}{123}$

$x = 24.39$ mL

24.39mL of the solution would be given per hour.

7 0

2 years ago

ANSWER ALL 5 PARTS.

N76 [4]

A function $f$ from a set A to a set B is defined as a relation that assings to each element $x$ in the set A exactly one element $y$ in the set B. The set A is called the domain of the function while the set B is the range. So we have five statements and need to find some functions. Melissa decides to reserve a patch in her vegetable garden for growing bell peppers. If each side of the tomato patch is $x$ feet, then we have a square patch as shown in the Figure below.

1.a) Write the function Wa(x) representing the width of the bell pepper patch.

We know that she wants its width to be half the width of the tomato patch. Let $x$ be the width of the tomato patch, then the function that matches this statement is:

$\boxed{Wa(x)=\frac{x}{2}}$

1.b) Write the function La(x) representing the length of the bell pepper patch.

In this case Melissa wants its length to exceed the length of the tomato patch by 2 feet. To do this we enlarge the length of the tomato patch 2 feet. Therefore the function is the following:

 $\boxed{La(x)=x+2}$

2. Area of the bell pepper patch in terms of x.

Given that the bell pepper patch is a rectangle, then the area of a rectangle is the product of the length and width. So:

 $A=(\frac{x}{2})(x+2) \\ \\ \therefore \boxed{A=\frac{x(x+2)}{2}}$

3. Combined area of the tomato patch and the bell pepper patch.

This function is the sum of both the area of the tomato patch and the bell pepper patch. So:

 $Aar(x)=x^{2}+\frac{x(x+2)}{2} \rightarrow Aar(x)=x^{2}+ \frac{x^{2}}{2}+x \rightarrow Aar(x)=\frac{3x^{2}}{2}+x \\ \\ \therefore \boxed{Aar(x)=\frac{x(3x+2)}{2}}$

4. Write the function Aa(x) for the remaining planting area in the garden.

The remaining planting area in the garden are the rectangles in red. So we need to subtract the width of the bell pepper patch from the width of the tomato patch and multiply it by 2. In mathematical language this is given by:

 $Aa(x)=2(x-\frac{x}{2}) \rightarrow Aa(x)=x$

5. Find the area of the remaining space in the garden after planting tomatoes and bell peppers.

Given that Melissa wants the area of the bell pepper patch to be 31.5 square feet, then it is true that:

 $31.5=\frac{x(x+2)}{2} \rightarrow x^{2}+2x-64=0 \\ \\ solving \ for \ x: \\ x=7.06$

Therefore the area of the remaining space is:

 $\boxed{Aa(7.06)=7.06ft^{2}}$

6 0

2 years ago

Joey wants to build a rectangular garden. He plans to use a side of a river for one side of the garden, he will not place fencin

jonny [76]

We let x and y be the measures of the sides of the rectangular garden. The perimeter subtracted with the other side should be equal to 92.

2x + y = 92

The value of y in terms of x is equal to,

y = 92 – 2x

The area is the product of the two sides,

A = xy

Substituting,

A = x (92 – 2x) = 92x – 2x2

Solving for the derivative and equating to zero,

0 = 92 – 4x ; x = 23

Therefore, the area of the garden is,

A = 23(92 – 2(23)) = 1058 yard2

4 0

2 years ago

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