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3 years ago

15

Jewel wanted to know if there was a correlation between the weight of a dog and its age. She collected data and created the scat

terplot shown:
scatter plot with points distributed all over quadrant 1

Determine the closest correlation coefficient (r value) for these data.

0

1

−1

5

2 answers:

sukhopar [10]3 years ago

8 0

Answer:

A. 0

Step-by-step explanation:

Because there is no correlation between any of the points.

Aleks [24]3 years ago

5 0

Answer:

0

Step-by-step explanation:

You might be interested in

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

Lets talk games whar are your favorie pc or console

lina2011 [118]

Answer:

my favorite is console because i play on a PS4

Step-by-step explanation:

6 0

2 years ago

Given g(x)=-5x+2,find g(4)

kakasveta [241]

Answer: -18

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A zoo in oklahoma charged $22 for adults and $15 for children. During a summer day, 732 zoo tickets were sold, and the total rec

Vlad [161]

Answer: Cost of tickets for adults = $22

Cost of tickets for children = $15

Total number of tickets sold = 732

Total receipts = $12,912

Let x be number of adult ticket's sold and y be the number of child ticket's sold.

Therefore, we have:

$x+y=732$ --------------- equation 1

$x=732-y$

Also, we have:

$22 x+15y=12,912$

Now, let's substitute the value of x:

$22 (732-y)+15y=12912$

$16104-22y+15y=12912$

$16104-7y=12912$

$7y=16104-12912$

$7y=3192$

$y=\frac{3192}{7} =456$

Now, let's substitute the value of y=456 in equation 1, we have:

$x+456=732$

$x=732-456$

$x=276$

Therefore, the number of child ticket's sold is 456 and number of adult ticket's sold is 276

6 0

4 years ago

IVE BEEN STUCK ON THIS FOR AN HOUR PLEASE HELP WHAT IS 9.85714285714 X 7

dsp73

Answer:

69

Step-by-step explanation:

lol

4 0

3 years ago