Write 5,129.347 in word

Cynthia is planning a party. for entertainment, she has designed a game that involves spinning two spinners. if the sum of the n

vesna_86 [32]

The correct statement of the problem is attached

Spinner A: 4 parts, 3/4 is 1, 1/4 is 3.----------1 1 1 3
Spinner B: 3 parts, 1/3 is 5, 1/3 is 7, 1/3 is 9--------5 7 9

The combinations of the two spinners are
1 and 5
1 and 7
1 and 9
3 and 5
3 and 7
3 and 9

cases get you thrown in the pool
1 and 5
1 and 7
3 and 5

calculation of the probabilities
1 and 5--------3/4*1/3 = 1/4
1 and 7--------3/4*1/3 = 1/4
3 and 5--------1/4*1/3 = 1/12
1/4+1/4+1/12=7/12-----------58.33%

probabilities that the guests will be thrown in the pool--------58.33%

5 0

3 years ago

Question 2 pater romuli remique erat ______. question 2 options: 1) mars 2) numitor 3) amulius 4) pastor save question 3 which i

Mamont248 [21]

"Pater romuli remique erat _______" translates as "The father of Romulus and Remus is____". The answer is 1) Mars. According to myth Rhea Silvia, who is the mother of the two brothers, claimed that the father of her children was the god Mars.

3 0

4 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

4 years ago

Use the picture below to find the length of x. Round your answer to the nearest hundredth.

valentinak56 [21]

Answer: 6.71

=======================================

Work Shown:

The longest horizontal portion of length 6 breaks up into two equal pieces of length 3 each. Focus on the smaller right triangle on the right hand side. This right triangle has legs of 3 and 6. The hypotenuse is x.

Use the pythagorean theorem with a = 3, b = 6, c = x to find the value of x

a^2 + b^2 = c^2

3^2 + 6^2 = x^2

9 + 36 = x^2

45 = x^2

x^2 = 45

x = sqrt(45)

x = 6.7082039

x = 6.71

4 0

3 years ago

WILL MARK BRAINLIEST!! PLZ HELP!!

olasank [31]

Answer:

The answer is D.

Step-by-step explanation:

When x=0, then y=300, so it has to be more than that.

When x=25, y=300, so it has to be less than that.

8 0

3 years ago