Help me please ill give brainlest

Write the expression as a square of a monomial. a 81x^4

poizon [28]

Answer:

(9x²)²

Step-by-step explanation:

Given the expression 81x⁴, to write the expression as a square of a monomial, first we will assign a variable to the expression.

y = 81x⁴

Then we take the square root of both sides of the expression

√y = √81x⁴

y^½ = √81 × √x⁴

y^½ = 9x²

Squaring both sides of the resulting equation to get y back

(y^½)² = (9x²)²

y = (9x²)²

The expression as a square of a monomial is (9x²)²

7 0

3 years ago

Read 2 more answers

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

Factor the expression 20k + 50

Yuliya22 [10]

Answer is 10(2k+5)

-----------------------------------

Work Shown:

20k+50 = 10*2k + 10*5
20k+50 = 10(2k+5)

Note how we can distribute the 10 back in to check our work
10(2k+5) = 10(2k)+10(5) = 20k+50
so that confirms we have the right answer

Another thing to notice is that 10 is the largest factor that we can pull out of 20k and 50. The value 10 is the GCF (greatest common factor) of 20 and 50.

4 0

2 years ago

8+2(5 - x) = Need help

nikdorinn [45]

We can distribute
2(5)+2(-x)=10-2x

we get
8+10-2x=?
18-2x=?

whatever that ? is, we don't know
multiply both sides by -1
2x-18=-?
add 18 to both sides
2x=18-?
divide bth sides by 2
x=9-(?/2)
whatever the left side it, just divide it by 2 and subtract it from 9

7 0

3 years ago

Solving Exponential and Logarithmic Equations In Exercise, solve for x. 100(1.21)x = 110

SVETLANKA909090 [29]

Answer:X=110/121

Step-by-step explanation:

Multiply 100 by 1.21 then by x

121x=110

Divide both side by 121

X=110/121

X=0.91

8 0

3 years ago