Classify the following data. Indicate whether the data is qualitative or quantitative, indicate whether the data is discrete, continuous, or neither, and indicate the level of measurement for the data.
The weights of a sample of 100 Maine lobsters.

Answer:

4x+ 4

Step-by-step explanation:

Answer:

I don't see any statements here.

Anyway, the answer is 8.

[tex]12 - 7 + 3 = 8[/tex]

Answer:

B

Step-by-step explanation:

Let me know if you need an explanation.

Answer:

B

Step-by-step explanation:

4x^3+x^2+5x+2

4x^3 cannot cancel with others= 4x^3

4x^2-3x^2= x^2

5x cannot cancel with others= 5x

-3+5= 2

4x^3+x^2+5x+2

Answer:  

a) h(x; 6, 9, 17).

b) P[X=2] = 0.2036

P[X ≤ 2] = 0.2466

P[X ≥ 2] = 0.9570.

c) Mean  = 3.176.

Variance = 1.028.

Standard deviation = 1.014.

Step-by-step explanation:

From the given details K=6, n=9, N=-17.

We conclude that it is the hypergeometric distribution:  

a) h(x; 6, 9, 17).

b)

[tex]P[X=2]=\frac{(^{g}C_{2})^{17-9}C_{6-2}}{^{17}C_{6}\textrm{}}[/tex]

P[X=2] = 0.2036

P[X ≤ 2] = P(x=0)+ P(x=1) + P(x=2)

P[X ≤ 2] = 0.2466

P[X ≥ 2] = 1-[P(x=0)+P(x=1)]

P[X ≥ 2] = 0.9570.

c)

Mean= [tex]n\frac{K}{N}[/tex]

            = 3.176.

Variance = [tex]n\frac{K}{N}( \frac{N-K}{N})(\frac{N-n}{n-1} )[/tex]

               = 2.824 x 0.6471 x 0.5625

               = 1.028.

Standard deviation = [tex]\sqrt{1.028}[/tex] = 1.014.

(a) Data with the eight day's measurement.

Raw data:      [60,58,64,64,68,50,57,82],  

Sorted data:  [50,57,58,60,64,64,68,82]

Sample size = 8 (even)

mean            = 62.875

median         = (60+64)/2 = 62

1st quartile   = (57+58)/2 = 57.5

3rd quartile  = (64+68)/2 =  66

IQR = 66 - 57.5 = 8.5

(b) Data without the eight day's measurement.

Raw data:      [60,58,64,64,68,50,57]

Sorted data:  [50,57,58,60,64,64,68]

Sample size = 7 (odd)

mean            = 60.143

median         = 60

1st quartile   = 57

3rd quartile = 64

IQR = 64 -57 = 7

Answers:

1. The average is the same with or without the 8th day's data.  FALSE

2. The median is the same with or without the 8th day's data.  FALSE

3. The IQR decreases when the 8th day is included.                  FALSE

4. The IQR increases when the 8th day is included.                   TRUE

5. The median is higher when the 8th day is included.              TRUE

Answer:

The Interquartile range increases when the 8th day is included.          

The median is higher when the 8th day is included.

Step-by-step explanation:

i got it right. Hope this helps.

Step-by-step explanation:

-27 okay 3^-3 its same as 3^3

Answer: A)

[tex]3^{-3}[/tex]

[tex]3^{-3}=\frac{1}{3^3}[/tex]

[tex]=\frac{1}{3^3}[/tex]

[tex]3^3=27[/tex]

[tex]=\frac{1}{27}[/tex]

OAmalOHopeO

Answer:

C. (-4099)^1/5

Step-by-step explanation:

[tex]x^{\frac{1}{2} } = \sqrt{x}[/tex]

you can not take roots (real roots) of a negative number if the exponent is

even ... A,B,D have even exponents (in the denominator of the exponent.. in other words the index of the radical is even)...

the only odd index is in "B" (the 5 in the 1/5)

Answer:

im pretty sure the degree is 4.

Step-by-step explanation:

Answer:

1+i

Step-by-step explanation:

I do believe i to be the imaginary unit.

Let's write out some partial sums from power=0 to power=7 or whatever we need to see a pattern.

i^0=1

i^0+i^1=1+i

i^0+i^1+i^2=1+i+-1=i

i^0+i^1+i^2+i^3=i+i^3=i+-i=0

i^0+i^1+i^2+i^3+i^4=0+i^4=0+1=1

Hmmm.... we might see 1+i, then i, then 0 again.... let's see.

i^0+i^1+i^2+i^3+i^4+i^5=1+i

Coolness so we should see a pattern

Sum from power=0 to power=multiples of 4 will give us 1.

Sum from power=0 to power=remainder of 1 when final power is divided by 4 gives us 1+i.

Sum from power=0 to power=remainder of 2 when final power is divided by 4 gives us i.

Sum from power=0 to power=remainder of 3 when final power is divided by 4 gives us 1

0.

So 2021 divided by 4....

Since 2020 is a multiple of 4, then 2021 has a remainder of 1 when divided by 4.

So the answer is 1+i.

Step-by-step explanation:

a) X is a discrete uniform distribution. As the number of outcomes is only 3.

b) sum is at least 4

X ≥ 4--------

i.e  (1,3) or (2,3)

probability of X ≥ 4 is 2/3

2/3= 0.667

66.7 % is the probability of the outcome to have  a sum at least 4.

c) The 3 likely outcome of X

(1,2) where X ; 1+2=3

(1,3) where X ; 1+3=4

(2,3) where X ;  2+3=5

Mean = 3+4+5/ 3

Mean = 4

Feel free to ask any uncleared step

Answer:

x=3

Step-by-step explanation:

7(x - 7) = -28

x - 7 = -4

x = 3

Answer:

x = 3

Step-by-step explanation:

Your goal is to isolate the x from the other numbers.

-28 = 7(x - 7)

Distribute the 7 to the (x - 7)

You will end up with:

-28 = 7x - 49

Add 49 to both sides of the equation to further isolate the x

21 = 7x

Finally, divide both sides by 7 so x is by itself

x = 3

Answer:

8×1000 milligrams

8000 milligrams

Answer: The image shown in your question as well as the one I provided is the correct answer

Step-by-step explanation:

a line with a slope of 2/3 must mean that the "m" is 2/3

y = mx + b

y = 2/3x + b

The question calls for the y-intercept to be equal to that of y=2/3x - 2

using the given equation, we easily figure out -2 is the y-intercept

so the line must go through (0,-2).

Answer:

The probability that 2 certain people will serve on that committee is 11.11%.

Step-by-step explanation:

Since to make a committee 4 men are chosen out of 6 candidates, to determine what is the probability that 2 certain people will serve on that committee the following calculation must be performed:

4/6 = 2/3

1/3 x 1/3 = X

0.333 x 0.333 = X

0.1111 = X

Therefore, the probability that 2 certain people will serve on that committee is 11.11%.

Answer:

[tex]\frac{2}{5}[/tex]

Step-by-step explanation:

6 groups, and 4 certain people

6

   C

        4

[tex]\frac{6!}{(6-2)!(2!)}[/tex]

1 × 2 × 3 × 4 × 5 × 6/1 × 2 × 3 × 4 × 1 × 2

1 × 2 × 3 × 4 × 5 × 6/1 × 2 × 3 × 4 × 1 × 2

5 × 6/ 1 × 2

30/2 = 15

15 possible combinations

4 people, and 2 specific ones

4

   C

        2

[tex]\frac{4!}{(4-2)!(2!)}[/tex]

1 × 2 × 3 × 4/1 × 2 × 1 × 2

1 × 2 × 3 × 4/1 × 2 × 1 × 2

12/2 = 6

[tex]\frac{6}{15}=\frac{\frac{6}{3} }{\frac{15}{3} } =\frac{2}{5}[/tex]

Given:

A line contains the points R(-1, 8), S(1, 4) and T(6, y).

To find:

The value of y.

Solution:

Three points are collinear if:

[tex]x_1(y_2-y_3)+x_2(y_3-y_2)+x_3(y_1-y_2)=0[/tex]

A line contains the points R(-1, 8), S(1, 4) and T(6, y). It means, these points are collinear.

[tex]-1(4-y)+1(y-8)+6(8-4)=0[/tex]

[tex]-4+y+y-8+48-24=0[/tex]

[tex]2y+12=0[/tex]

Subtract 12 from both sides.

[tex]2y=-12[/tex]

Divide both sides by 2.

[tex]y=\dfrac{-12}{2}[/tex]

[tex]y=-6[/tex]

Therefore, the value of y is -6.

Answer: The guy above me is right

Step-by-step explanation: I took the test and got it right

Answer:

Option D

Step-by-step explanation:

correct answer on edge :)

Answer:

D <3

Step-by-step explanation:

Answer:

23) No

24) No

25) Yes

Step-by-step explanation:

Question 23)

We want to determine if a zero exists between 1 and 2 for the function:

[tex]f(x)=x^2-4x-5[/tex]

Find the zeros of the function. We can factor:

[tex]\displaystyle 0 = (x-5)(x+1)[/tex]

Zero Product Property:

[tex]x-5=0\text{ or } x+1=0[/tex]

Solve for each case. Hence:

[tex]\displaystyle x = 5\text{ or } x=-1[/tex]

Therefore, our zeros are at x = 5 and x = -1.

In conclusion, a zero does not exist between 1 and 2.

Question 24)

We have the function:

[tex]f(x)=2x^2-7x+3[/tex]

And we want to determine if a zero exists between 1 and 2.

Factor. We want to find two numbers that multiply to (2)(3) = 6 and that add to -7.

-6 and -1 suffice. Hence:

[tex]\displaystyle \begin{aligned} 0 & = 2x^2-7x + 3 \\ & = 2x^2 -6x -x + 3 \\ &= 2x(x-3) - (x-3) \\ &= (2x-1)(x-3) \end{aligned}[/tex]

By the Zero Product Property:

[tex]2x-1=0\text{ or } x-3=0[/tex]

Solve for each case:

[tex]\displaystyle x=\frac{1}{2} \text{ or } x=3[/tex]

Therefore, our zeros are at x = 1/2 and x = 3.

In conclusion, a zero does not exist between 1 and 2.

Question 25)

We have the function:

[tex]f(x)=3x^2-2x-5[/tex]

And we want to determine if a zero exists between -2 and 3.

Factor. Again, we want to find two numbers that multiply to 3(-5) = -15 and that add to -2.

-5 and 3 works perfectly. Hence:

[tex]\displaystyle \begin{aligned} 0&= 3x^2 -2x -5 \\ &= 3x^2 +3x - 5x -5 \\ &= 3x(x+1)-5(x+1) \\ &= (3x-5)(x+1)\end{aligned}[/tex]

By the Zero Product Property:

[tex]\displaystyle 3x-5=0\text{ or } x+1=0[/tex]

Solve for each case:

[tex]\displaystyle x = \frac{5}{3}\text{ or } x=-1[/tex]

In conclusion, there indeed exists a zero between -2 and 3.

[tex] \sf \: d \: = 3.7m \\ \sf \: r \: = \frac{3.7}{2} = 1.85 \: m\\ \\ \sf \: c \: = \pi {r}^{2} \\ \\ \sf \: c \: = \pi ({1.85})^{2} \\ \sf c = 1.85 \times 1.85 \times \pi \\ \sf \: c = \boxed {\underline{ \bf a. \: 3.4225\pi \: m ^{2} }}[/tex]

Answer:

y + 1 = 7(x + 2).

Step-by-step explanation:

Point slope form:

y - y1 =m(x - x1).

Here m = 7 and (x1,  y1) = (-2,  -1)

So the answer is;

y - (-1) = 7(x - (-2))

Answer:

268.40

Step-by-step explanation:

We can write a ratio to solve

2.5 meters        22 meters

-----------------  = --------------

30.5 dollars       x dollars

Using cross products

2.5 * x = 30.5 * 22

2.5x =671

Divide each side by 2.5

2.5x / 2.5 = 671/2.5

x =268.4

Answer:

0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.

Step-by-step explanation:

We have the mean, which means that the Poisson distribution is used to solve this question.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

A mean of 3.5 customers arrive hourly at the drive-through window.

This means that [tex]\mu = 3.5[/tex]

What is the probability that, in any hour, more than 5 customers will arrive?

This is:

[tex]P(X > 5) = 1 - P(X \leq 5)[/tex]

In which

[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]

Then

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-3.5}*3.5^{0}}{(0)!} = 0.0302[/tex]

[tex]P(X = 1) = \frac{e^{-3.5}*3.5^{1}}{(1)!} = 0.1057[/tex]

[tex]P(X = 2) = \frac{e^{-3.5}*3.5^{2}}{(2)!} = 0.1850[/tex]

[tex]P(X = 3) = \frac{e^{-3.5}*3.5^{3}}{(3)!} = 0.2158[/tex]

[tex]P(X = 4) = \frac{e^{-3.5}*3.5^{4}}{(4)!} = 0.1888[/tex]

[tex]P(X = 5) = \frac{e^{-3.5}*3.5^{5}}{(5)!} = 0.1322[/tex]

Finally

[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0302 + 0.1057 + 0.1850 + 0.2158 + 0.1888 + 0.1322 = 0.8577[/tex]

[tex]P(X > 5) = 1 - P(X \leq 5) = 1 - 0.8577 = 0.1423[/tex]

0.1423 = 14.23% probability that, in any hour, more than 5 customers will arrive.

Answer:

7x10 ^-10

Step-by-step explanation:

Answer: distributive property

Step-by-step explanation: the 4 is multiplied by everting in the parenthesis

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

For each subset it can either contain or not contain an element. For each element, there are 2 possibilities. Multiplying these together we get 27 or 128 subsets. For generalisation the total number of subsets of a set containing n elements is 2 to the power n.

total subsets

Answer:

$170.40

Step-by-step explanation:

cost of policy=$8.52

Groups of $1000=20,00÷1000

=20

20,000 worth=$8.52×20

$170.40

Answer:

Line A

Step-by-step explanation:

graph it