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3 years ago

What’s the answer I don’t no if it’s right

Mathematics

2 answers:

sukhopar [10]3 years ago

7 0

Hello!

We could connect these two points making a right triangle with leg lengths 4 and 3. Let's use the Pythagorean Theorem to find the hypotenuse.

16+9=25

√25=5

Therefore, our answer is D) 5 miles

I hope this helps!

Elena L [17]3 years ago

4 0

You got it right!

Hope this helped

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All rectangles are parallelograms.<br> True<br> False

navik [9.2K]

Answer:

false

Step-by-step explanation:

Not all parallelograms are rectangles, only some of them are.

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3 years ago

Mechanics paper/Math (please helpp!!!)

sp2606 [1]

Answer:

(i) 15 m, 6 m/s

(ii) 90 m

Step-by-step explanation:

(i) For some acceleration (a) from rest, the distance covered (d) in time t is ...

d = (1/2)at^2

The distance covered by Ben in the 5 seconds he is accelerating is ...

d = (1/2)(1.2 m/s²)(5 s)² = 15 m

Of course, Ben's speed at that point is ...

s = (1.2 m/s²)(5 s) = 6 m/s

(ii) When Ben has been walking 5 s, Alan has been walking 10 s, so Alan has covered (10 s)(4 m/s) = 40 m. Their distance difference of 40 -15 = 25 m is being made up at the rate of their speed differences: (6 m/s) -(4 m/s) = 2 m/s.

It will take (25 m)/(2 m/s) = 12.5 s additional time for Ben to catch Alan. In the 22.5 s that Alan has been walking before they meet, he will have walked ...

(22.5 s)(4 m/s) = 90 m . . . the distance OP

4 0

3 years ago

Please help with this

GalinKa [24]

The answer is 20

Reason: so the W is 4 and the L is 6 so you do 4+4=8 then you do 6+6=12 then you add the two together and get 20

7 0

3 years ago

Which unit would you use to measure the amount of water in a bathtub?

zepelin [54]

Liters the rest are to small and kg is for solids

7 0

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

3 years ago