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3 years ago

I WILL MARK BRAINLIEST!!!! I NEED HELP PLZZ EXTRAA 20 PONITS

Mathematics

2 answers:

Dafna11 [192]3 years ago

6 0

Answer: 3rd option

Step-by-step explanation:

Let’s look at the denominator. 2x-yep. (x-4) is also there.

There is 2 “x” denominators and 3 + 4 = 7. So it’ll look like 7x*2.

Then, there needs to be an + and - sign in the answer.

Finally, the third option is the only one that incorporates all three factors.

Hope this helps!

almond37 [142]3 years ago

4 0

The first answer because you have to write 3x/ x-4 as a denominator, then you have to keep simplifying

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An athlete trains on a circular track of radius 40 m and jogs 10 laps each day, 5 days a week. How far does he jog each week? Ro

mariarad [96]

Length of a circle=2πr
r=40 m
Length of the circular track=2*π*40m=80π m.

The athelete jogs each day 10 laps
1 lap-------------------80π m
10 laps------------- x
x=(10 laps * 80π m) / 1 lap=800π m

Calculate how far he jog eache week (5 days).
1 day------------------800π m
5 days---------------- x
x=(5 days*800π m)/1day=4000π m=4000m * 3.141592...≈12566,37 m

solution: 12566 m

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3 years ago

What is the average of the integers from 25 to 41?

Archy [21]

The average of the terms of a contiguous subsequence of any arithmetic progression is the average of the first and last terms. To see this, notice that if you remove 25 and 41 from your sequence, then the average of the remaining terms is still given by the average of the extreme terms 26+402=33 .

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3 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

5 employees works 6 4/15 hours pretty day how many hours in total do the 5 employees work pre day

Lubov Fominskaja [6]

You world need to multiply the 64/15 by 5

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3 years ago

You make an initial deposit of $2,500 in an account that offers 6% simple interest per year. How much will be in the account aft

Anarel [89]

Answer:

He will have 3100$ in 4 years.

Step-by-step explanation:

y= $

x= Years

2500 x 0.06 = 150$

y = ax + b

y . = 150x + 2500

y = 150(4) + 2500

y = 3100$

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3 years ago