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3 years ago

5 The average rate of change of g(x) between x = 4 and x = 7 is Ē. Which statement must be true?

Mathematics

1 answer:

docker41 [41]3 years ago

8 0

Answer:

c. 5/6 = [ g(7) − g(4) ] / (7 − 4)

Step-by-step explanation:

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The state of New Jersey makes approximately $1.55 billion annually from tolls of the $1.55 billion about 3/4 comes from tolls on

Rama09 [41]

Percent of turn pike is 75% and percent of parkway is 25%.

Solution:

Given, The state of New Jersey makes approximately $1.55 billion annually from tolls of the $1.55 billion

About $\frac{3}{4}$ comes from tolls on the turnpike

Remaining $\frac{1}{4}$ comes from tolls on the parkway

To find the percent of the total revenue from turnpike:

$\begin{array}{l}{\text { Percent of turnpike }=\frac{\text { amount from parkway }}{\text { total revenue }} \times 100} \\\\ {=\frac{\frac{3}{4} \text { of total revenue }}{\text { tot al revenue }} \times 100} \\\\ {=\frac{\frac{3}{4} \times \text { total revenue }}{\text { total revenue }} \times 100} \\\\ {=\frac{3}{4} \times 100=3 \times 25=75 \%}\end{array}$

To find the percent of the total revenue from parkway:

$\begin{array}{l}{\text { Percent of parkway }=\frac{\text { amount from parkway }}{\text { total revenue }} \times 100} \\\\ {=\frac{\frac{1}{4} \text { of total revenue }}{\text { total revenue }} \times 100} \\\\ {=\frac{\frac{1}{4} \times \text { total revenue}}{\text { total revenue }} \times 100} \\\\ {=\frac{1}{4} \times 100=25 \%}\end{array}$

Summarizing the results:

Percent of turn pike is 75%

Percent of parkway is 25%

3 0

3 years ago

How much is 2000 × 36 =? help my please

PIT_PIT [208]

Answer:

72000 should be your answer

Step-by-step explanation:

Got it off of g o o g l e so it should be right!

4 0

2 years ago

(y-5)6+15–5y=5 What does the variable represent?

Strike441 [17]

Answer:

2.75

Step-by-step explanation:

(y-5)6+15–5y=5

(y-5+5)6-6+15-15–5y=5-6-15+5

y-5y=-11

-4y/-4=-11/-4

y=2.75

8 0

3 years ago

Please help me with these Algebra Questions!!

Bess [88]

Answer:

There is nothing here!

Step-by-step explanation:

I don't see anything to solve.

: )

3 0

3 years ago

A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade

Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

$P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}$

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

$P(x\geq1)=1-P(x$

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly n= 5

Number of blades in an assembly dull is M = 10.

The probability mass function is,

$P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}$

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

$P(x\geq 1)=1- P(x$

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly N = 5

Number of blades in an assembly dull is M = 10

From the information,

The probability that assembly is replaced (P) is 0.7069.

The probability that assembly is not replaced is (Q) is,

$q=1-p\\= 1-0.7069= 0.2931$

The geometric probability mass function is,

$P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)$

The probability that assembly is replaced no replaced until the third day of evaluation is,

$P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607$

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 2.

$P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005$

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 6.

$P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968$

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

$P(x\geq 1)=1- P(x$

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0

3 years ago