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2 years ago

Question 4 (1 point)

Mathematics

1 answer:

FromTheMoon [43]2 years ago

3 0

Answer:

Step-by-step explanation:

Inches to CM Conversion

1 Inch (in) is equal to 2.54 cm. To convert inches to cm, multiply the inch value by 2.54. For example, to find out how many cm is 12 inches, multiply 12 by 2.54, that makes 30.48 cm is 12 in.

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The next number in the series 13, 18, 16, 21, 19, is: 22 23 24 25 26 ...?

Lesechka [4]

I think the best option is 24.

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2 years ago

Identify the like term 9x, 10y, -9y, 10

quester [9]

the like terms are: 10y, and -9y because they are both of the same variable, y.

7 0

2 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

2 years ago

Which is the closest to the quotient 4367/0.004

VladimirAG [237]

Answer:

1091750

Step-by-step explanation:

if your dividing

8 0

2 years ago

A pair of Jordans sold at retail for $180 and now sell for $330 after market. A. Is this a percent INCREASE or DECREASE, what is

Elza [17]

Answer:

This is a percent INCREASE

The percent change is 83.33%

Step-by-step explanation:

Let us solve the question

∵ A pair of Jordans sold at retail for $180

∴ The initial price = 180

∵ Now it sells for $330 aftermarket

∴ The new price = 330

→ That means the new price is greater than the initial price

∵ The new price is greater than the initial price

∴ This is a percent INCREASE

∵ The percent of change = (amount of increase ÷ initial amount) × 100%

∵ The amount of increase = new price - initial price

∴ The amount of increase = 330 - 180 = 150

→ Substitute it in the rule of percent of increase above

∵ The percent change = $\frac{150}{180}$ × 100%

∴ The percent change = $\frac{5}{6}$ × 100%

∴ The percent change = 83.33%

4 0

2 years ago