Can someone help me find the median first quartile second quartile third quartile and the interquartile range
for the questions
4.
5.
6.
7.

Answer:

HI!!

Step-by-step explanation:

Answer:

They have 11 inches of rope left.

Step-by-step explanation:

First we do the 6 foot rope minus the 4 foot cut. That leaves us with 2 feet, or 24 inches.

24-13 = 11

Answer:

[tex] 93 \: {in}^{2} [/tex]

Step-by-step explanation:

Surface area of the figure

[tex] = \frac{1}{2} \times 6 \times 7 + 3 \times \frac{1}{2} \times 6 \times 8 \\ \\ = 3 \times 7 + 3 \times 3 \times 8 \\ \\ = 21 + 72 \\ \\ = 93 \: {in}^{2} [/tex]

Answer:

A) 60

Step-by-step explanation:

Find two points that intersect with the grid.

(0,0) and (1, 60)

Subtract the second y value from the first y value. Do the same to the X-values.

Write this as a fraction with the y values over the x-values.

60-0= 60

-------------

1 - 0 = 1

60/1 = 60

[tex]\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = \lim_{h\to0}\frac{(9+h)^4-9^4}h[/tex]

Carry out the binomial expansion in the numerator:

[tex](9+h)^4 = 9^4+4\times9^3h+6\times9^2h^2+4\times9h^3+h^4[/tex]

Then the 9⁴ terms cancel each other, so in the limit we have

[tex]\displaystyle \lim_{h\to0}\frac{4\times9^3h+6\times9^2h^2+4\times9h^3+h^4}h[/tex]

Since h is approaching 0, that means h ≠ 0, so we can cancel the common factor of h in both numerator and denominator:

[tex]\displaystyle \lim_{h\to0}(4\times9^3+6\times9^2h+4\times9h^2+h^3)[/tex]

Then when h converges to 0, each remaining term containing h goes to 0, leaving you with

[tex]\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = 4\times9^3 = \boxed{2916}[/tex]

or choice C.

Alternatively, you can recognize the given limit as the derivative of f(x) at x = 9:

[tex]f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h \implies f'(9) = \lim_{h\to0}\frac{f(9+h)-f(9)}h[/tex]

We have f(x) = x ⁴, so f '(x) = 4x ³, and evaluating this at x = 9 gives the same result, 2916.

Answer:

The degree of the polynomial is equal to 7.

Step-by-step explanation:

The given polynomial is :

7x⁴– 25x + 3x⁷​

We need to find the degree of the polynomial.

We know that,

Degree of the polynomial = The highest sum of exponents for anyone monomial.

In this case, the powers of x are 4,1 and 7.

Out of these three, the exponent of x is 7 which is highest.

Hence, the degree of the polynomial is equal to 7.

Answer:

Mike

Step-by-step explanation:

Mike is the traveling faster.

Answer:

10³ = l

Step-by-step explanation:

Answer:

6

Step-by-step explanation:

Volume of Rectangular Prism: (l)(w)(h)

8 x 4 x ? = 192

32 x ? = 192

/32        /32

height = 6

To check work:

(8)(4)(6) = 192

Answer:

36

Step-by-step explanation:

1×36

36×2

36×16

aaaaaaaaaaaaaaaaaaaaaa

Answer:

x= 17

y=62

Step-by-step explanation:

Answer:

!! , ' !!!

Step-by-step explanation:

ꜰɪʀꜱᴛ! ᴄᴏᴍʙɪɴᴇ ᴍᴜʟᴛɪᴘʟɪᴇᴅ ᴛᴇʀᴍꜱ ɪɴᴛᴏ ᴀ ꜱɪɴɢʟᴇ ꜰʀᴀᴄᴛɪᴏɴ:

        / = /

   / = /                 ᴊᴜꜱᴛ ʟɪᴋᴇ ᴛʜᴀᴛ!

ꜱᴇᴄᴏɴᴅ! ᴍᴜʟᴛɪᴘʟʏ ᴀʟʟ ᴛᴇʀᴍꜱ ʙʏ ᴛʜᴇ ꜱᴀᴍᴇ ᴠᴀʟᴜᴇ ᴛᴏ ᴇʟɪᴍɪɴᴀᴛᴇ ꜰʀᴀᴄᴛɪᴏɴ ᴅᴇɴᴏᴍɪɴᴀᴛᴏʀꜱ!! ᴇᴀꜱʏ ᴘᴇᴀꜱʏ:

                    / = /

                  9⋅/= 9⋅/     ꜱᴇᴇ ᴇᴀꜱʏ ᴘᴇᴀꜱʏ!!

! :

          9⋅/=!!!

                                   ᴀʟᴍᴏꜱᴛ ᴅᴏɴᴇ!!

ꜰᴏᴜʀᴛʜʟʏ!! ᴍᴜʟᴛɪᴘʟʏ ᴛʜᴇ ɴᴜᴍʙᴇʀꜱ (ᴊᴜꜱᴛ ᴛʜᴀᴛ ꜱɪᴍᴘʟᴇ!):

                9⋅/  

           ˜”*°•.˜”*°• Which Would equal 60=10x!! •°*”˜.•°*”˜

ꜰɪꜰᴛʜ!! ᴅɪᴠɪᴅᴇ ʙᴏᴛʜ ꜱɪᴅᴇꜱ ᴏꜰ ᴛʜᴇ ᴇQᴜᴀᴛɪᴏɴ ʙʏ ᴛʜᴇ ꜱᴀᴍᴇ ᴛᴇʀᴍ:

               60/10 = 10x/10

ʟᴀꜱᴛʟʏ!!! ꜱɪᴍᴘʟɪꜰʏʏ:

         =6

ᴛʜᴇɴ ʏᴏᴜ ɢᴇᴛ ʏᴏᴜʀ ᴀɴꜱᴡᴇʀʀ!!

              x=6  ʕ ˵• ₒ •˵ ʔ  

ᴀɴʏᴡᴀʏꜱ ᴛᴏᴏᴅʟᴇꜱ!!

    ʜᴀᴠᴇ ᴀɴ ᴀᴍᴀᴢɪɴɢ ɴɪɢʜᴛ ᴏʀ ᴅᴀʏ!

ꜰʀᴏᴍ ʏᴏᴜʀ ᴏɴᴇ ᴀɴᴅ ᴏɴʟʏ ᴘɪɴᴋʏ  ʕ ﾟ ● ﾟʔ

Answer:

x=2.83

Step-by-step explanation:

The solution to the equation, rounded to two decimal places, is x = 2.83.

The correct answer is option B: x = 2.83.

To solve the equation 5 + 6·log2(x) = 14 and round the answer to two decimal places, we can follow these steps:

1. Subtract 5 from both sides of the equation: 6·log2(x) = 9.

2. Divide both sides of the equation by 6: log2(x) = 9/6.

3. Simplify the right side: log2(x) = 1.5.

To find the value of x, we need to rewrite the equation in exponential form.

Remember that log2(x) = y is equivalent to [tex]2^y= x[/tex].

4. Rewrite the equation in exponential form: [tex]2^{1.5} = x[/tex].

5. Calculate the value of [tex]2^{1.5}[/tex] using a calculator or by approximating: x ≈ 2.83.

To learn more about the equation;

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Answer:

Step-by-step explanation:

yesssss

Answer:

D is the answer of this question

Answer:

155 are teenagers

Step-by-step explanation:

3100 -5%= 2945

2945-3100=155

Answer:

2/B

Step-by-step explanation:

The base is 9 and the height is 4.

Answer:

The answer is linear.

Step-by-step explanation:

The x int is being added by 1 while the y int is subtracting 1

Answer:

look at the picture i sent

In other words, choices 2 and 4 are false while everything else is true.

==========================================================

Explanation:

The boxplot, or box-and-whisker plot, is a visual way to display a five number summary. Recall that the five items of a five number summary are:

The min and max are exactly what they sound like: They're the smallest and largest values. Visually, this means they are the left and right most edges of the whiskers. This assumes that there are no outliers on either side.

The visual representation of Q1 and Q3 are the left and right edges of the box in that order. The gap between them is the IQR (interquartile range). The wider the box, the larger the IQR will be. The diagram shows the north cars have a larger IQR compared to the south ones.

Choice 1) is one of the answers since it is a true statement.

-------------------

The median is visually shown as a vertical line somewhere inside the box. It doesn't have to be at the exact midpoint of the left and right edges. We could have the median be closer to the left edge for instance, as the north cars boxplot shows.

Choice 2 is false because the diagram shows that the cars going south have a higher median speed, due to that lower vertical line being more to the right compared to the upper vertical line. In other words, the median speed of the cars going south is 64 mph while the median speed of the cars going north is 62 mph.

-------------------

Choice 3 is a bit tricky since the left edge of the blue box isn't lined up with a value on the number line. But it appears to be halfway between 58 and 60, so it seems like Q1 = 59 for the southbound cars. The value of Q3 for the southbound cars is Q3 = 68. The IQR is therefore,

IQR = Q3 - Q1 = 68-59 = 9

It's very likely that choice 3 is true

-------------------

For the northbound cars, we have

Q1 = 56

Q3 = 71 (assuming the right edge of the green box is exactly in the middle of the tickmarks of 70 and 72)

So,

IQR = Q3 - Q1 = 71 - 56 = 15

We don't get a result of 27, so choice 4 is not a true statement.

-------------------

Choice 5 is a true statement. The max speed of the northbound cars is larger than the max of the southbound cars. This is because the right tip of the north's whisker is further to the right compared to the south's corresponding whisker.

-------------------

The lower quartile is another name for "Q1" or "first quartile". So we'll be looking at the left edges of each box. The northbound cars have a smaller Q1 value because

Q1 = 56 for northbound cars

Q1 = 59 (or somewhere between 58 and 60) for southbound cars

Therefore, choice 6 is true.

Answer:

(x+2)2+(y-1)2=25

Step-by-step explanation:

x2+y2+4x-2y+4+1=20+4+1

Answer:b

Step-by-step explanation:

To analyze this data, you would use An ANOVA F test.

An example of a continuous probability distribution is the normal distribution, in which the majority of data points cluster in the middle of the range, while the remaining ones taper off symmetrically toward either extreme. The distribution's mean is another name for the center

The study involves comparing the weight loss of four groups of participants (weight-loss pill alone, weight-loss pill with a low-fat diet, placebo pill alone, or placebo pill with a low-fat diet) after six months of treatment. ANOVA (Analysis of Variance) is a statistical method used to analyze the differences among means of two or more groups.

In this case, ANOVA can be used to determine if there are significant differences in weight loss among the four treatment conditions.

A t-test or z-test would be appropriate if there were only two groups being compared.

A chi-square test is used to analyze categorical data, which is not applicable in this scenario where the data is continuous.

Therefore, an ANOVA F test would be used.

Learn more about normal distribution here:

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Answer:

C

Step-by-step explanation:

Answering your question so it will not get removed.

The correct option for the question is C.  [tex]S_{n}[/tex] = 15 [ 1 - [tex](3/4)^{n}[/tex] ]

If in a sequence of terms, each succeeding term is generated by multiplying each preceding term with a constant value, then the series is referred to as a geometric progression. (GP), while the constant value is called the common ratio.

Conclusion: The nth term from the end of the GP with the last term l and common ratio r = l/ [r(n – 1)]. The sum of infinite, i.e. the sum of a GP with infinite terms is S∞= a/(1 – r) such that 0 < r < 1.

[tex]a_{n}[/tex] = 5[tex](3/4)^{n}[/tex]

[tex]S_{n}[/tex]= Σ[tex]a_{n}[/tex]

[tex]S_{n}[/tex] = ∑5[tex](3/4)^{n}[/tex]

  = 5∑[tex](3/4)^{n}[/tex]

  = 5 [ 3/4 + [tex](3/4)^{2}[/tex] + ......+ [tex](3/4)^{n}[/tex]]

 =5 [ 3/4 * { 1- [tex](3/4)^{n}[/tex]} / 1 - 3/4 ]

[tex]S_{n}[/tex] = 15 [ 1 - [tex](3/4)^{n}[/tex] ]

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Answer:

180 units 2 ............

To start this is a ratio, and the goal is to get at least 1 to be a whole number so first lets get the number below 1 to 1.

0.75 : 2 5/16 ---> (x1.3333 on both sides) ---> 1 : 3 1/12.

Next, if needed for your answer, lets get the other number to be a whole number as well. We can do this my multiplying both sides by 12 since the fraction on 3 1/12 is 1/12.

1 : 3 1/12 --> (x12 on both sides) ---> 12 : 37

Your final answer will either be 1 : 3 1/12 or 12 : 37, depending on the formatting expected for your homework assignment.