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2 years ago

Can someone do this for me

Mathematics

2 answers:

Dafna11 [192]2 years ago

6 0

Angle 8 and Angle 3 are supplementary same side angles

cestrela7 [59]2 years ago

4 0

Answer:

angle 2 and angle 3

Step-by-step explanation:

Supplementary angles means that they add up to 180 degrees. 2 and 3 are on the same straight line that are side by side with each other. This is a hint that they are supplementary angles.

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If R is between P and Q, find the value of x and PR. PQ=17, PR=5x+2, RQ=x^2-1

vodka [1.7K]

The value of x will be2.217 and PR will be 13.085 units.

<h3>What is a line segment?</h3>

A line section that can connect two places is referred to as a segment.

In other words, a line segment is just part of a big line that is straight and going unlimited in both directions.

The line is here! It extends endlessly in both directions and has no beginning or conclusion.

Given that,

Line PQ consists of two segments PR and RQ

So,

PQ = PR + RQ

17 = 5x+2 + x² - 1

x² + 5x - 16 = 0

The roots of the equation will be

x = [-5 + √(25 + 64) ] ÷ 2

x = 2.217

Now,

PR = 5(2.217) + 2 = 13.085

Hence "The value of x will be2.217 and PR will be 13.085 units".

For more about line segment,

brainly.com/question/25727583

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1 year ago

How would I find the tangent of this angle?

Wittaler [7]

Answer:

hoal

Step-by-step explanation:

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2 years ago

Sally went 3.5 miles per hour on a treadmill for 5 hours. She was sore and decided to take 400 mg of Advil. The half portion sta

xenn [34]

Answer:

8 hours

Step-by-step explanation:

it says: The half portion stays in her body for 8 hours.

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3 years ago

Each Cube inside the rectangle Prisma has a edge length of 3/4 inches which is the value of the Prisma

Sphinxa [80]

Answer:

Step-by-step explanation:

6 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