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3 years ago

Prime factorization of 1408

Mathematics

1 answer:

V125BC [204]3 years ago

8 0

Answer:

There are 7 copies of 2 and 1 copy of 11 in the product:

1408 = 2^7×11

Step-by-step explanation:

Factor the following integer:

1408

The last digit of 1408 is 8, which means it is even. Therefore 1408 is divisible by 2:

1408 = 2 704:

1408 = 2×704

The last digit of 704 is 4, which means it is even. Therefore 704 is divisible by 2:

704 = 2 352:

1408 = 2×2×352

The last digit of 352 is 2, which means it is even. Therefore 352 is divisible by 2:

352 = 2 176:

1408 = 2×2×2×176

The last digit of 176 is 6, which means it is even. Therefore 176 is divisible by 2:

176 = 2 88:

1408 = 2×2×2×2×88

The last digit of 88 is 8, which means it is even. Therefore 88 is divisible by 2:

88 = 2 44:

1408 = 2×2×2×2×2×44

The last digit of 44 is 4, which means it is even. Therefore 44 is divisible by 2:

44 = 2 22:

1408 = 2×2×2×2×2×2×22

The last digit of 22 is 2, which means it is even. Therefore 22 is divisible by 2:

22 = 2 11:

1408 = 2×2×2×2×2×2×2×11

11 is not divisible by 2 since 11 is odd and 2 is even:

1408 = 2×2×2×2×2×2×2×11 (11 is not divisible by 2)

The sum of the digits of 11 is 1 + 1 = 2, which is not divisible by 3. This means 11 is not divisible by 3:

1408 = 2×2×2×2×2×2×2×11 (11 is not divisible by 2 or 3)

The last digit of 11 is not 5 or 0, which means 11 is not divisible by 5:

1408 = 2×2×2×2×2×2×2×11 (11 is not divisible by 2, 3 or 5)

Divide 7 into 11:

| | 1 | (quotient)

7 | 1 | 1 |

- | | 7 |

| | 4 | (remainder)

11 is not divisible by 7:

1408 = 2×2×2×2×2×2×2×11 (11 is not divisible by 2, 3, 5 or 7)

No primes less than 11 divide into it. Therefore 11 is prime:

1408 = 2×2×2×2×2×2×2×11

There are 7 copies of 2 and 1 copy of 11 in the product:

Answer: 1408 = 2^7×11

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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

Someone please help! I'm confused with c and d.

Karo-lina-s [1.5K]

It would be 3 times the sum of x and seven plus 10. This is because first you would would add 7+x (7x) and multiply it by 3. And then add on 10.

5 0

3 years ago

Read 2 more answers

What does one eight equals

zvonat [6]

1/8 the fraction?

two eighths = 2/8=1/4