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densk [106]
3 years ago
15

I NEED THIS ASAP, PLEASE HELP!!!

Mathematics
1 answer:
WINSTONCH [101]3 years ago
7 0

Answer: 22.5 ; 5 ; 14

Step-by-step explanation:

Given the dataset:

{20,22,23,24,26,26,28,29,30}

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 9

Q1 = 1/4 (9 + 1)th term

Q1 = 1/4(10) = 2.5

We average the 2nd and 3rd term:

(22 + 23) / 2

45 / 2 = 22.5

B) The interquartile range(IQR) of the dataset :

{62,63,64,65,67,68,68,68,69,74}

IQR = Q3 - Q1

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 10

Q1 = 1/4 (10 + 1)th term

Q1 = 1/4(11) = 2.75 term

We take the average of the 2nd and 3rd term:

(63 + 64) / 2

45 / 2 = 63.5

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 10

Q3 = 3/4 (10 + 1)th term

Q3 = 3/4(11) = 8.25 term

We take the average of the 8th and 9th term:

(68 + 69) / 2

137 / 2 = 68.5

IQR = Q3 - Q1

IQR = 68.5 - 63.5

IQR = 5

C) give the dataset :

{7,8,8,9,10,12,13,15,16}

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 9

Q3 = 3/4 (9 + 1)th term

Q3 = 3/4(10) = 7.5 term

We take the average of the 7th and 8th term:

(13 + 15) / 2

28 / 2 = 14

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If the radius of a cylinder was shrunk down to a quarter of its orignal size and the height was reduced to a third of its origin
svetoff [14.1K]

The modified area is (1/48) (2πr(4h+3r))

<u>Step-by-step explanation:</u>

Let the radius be 'r' and height be 'h'.

Area of cylinder= 2π r(h+r)

The radius is shrunk down to quarter of its original radius

 r = r/4

The height is reduced to a third of its original height

h = h/3

New Area = 2π(r/4) [(h/3) +(r/4) ]

= (1/4)2πr[(4h+3r) /12]

= (1/48) (2πr(4h+3r))

6 0
3 years ago
I NEED HELP PLEASE, THANKS! :) Determine the zeros for and the end behavior of f(x) = x(x – 4)(x + 2)^4.
Amiraneli [1.4K]

Answer:  a) zeros: x = {0, 4, -2}

               b) as x → ∞,  y → ∞

                   as x → -∞,  y → ∞

<u>Step-by-step explanation:</u>

I think you mean (a) find the zeros and (b) describe the end behavior

(a) Find the zeros by setting each factor equal to zero and solving for x:

      x (x - 4) (x + 2)⁴ = 0

  • x = 0   Multiplicity of 1  --> odd multiplicity so it crosses the x-axis
  • x = 4  Multiplicity of 1  --> odd multiplicity so it crosses the x-axis
  • x = -2 <u>Multiplicity of 4 </u> --> even multiplicity so it touches the x-axis

                       Degree  =  6

(b) End behavior is determined by the following two criteria:

  1. Sign of Leading Coefficient (Right side): Positive is ↑, Negative is ↓
  2. Degree (Left side): Even is same direction as right side, Odd is opposite direction of right side

Sign of the leading coefficient is Positive so right side goes UP

                                                                        as x → ∞,  y → ∞

Degree of 6 is Even so Left side is the same direction as right (UP)

                                                                        as x → -∞,  y → ∞

7 0
3 years ago
At a farmers' market Saturday, 5/8 of the vegetables sold were green. Of these, 3/4 were string beans. What fraction of the vege
Arte-miy333 [17]

Answer:

47%

Step-by-step explanation:

3/4 of 5/8

3/4 • 5/8

15/32 = 0.46875 or 47%

7 0
3 years ago
When the sum of a number and 3 is subtracted from 10 the result is 5. What is the integer
Rashid [163]

In algebraic terms this is:

10-(x+3)=5\\10-x-3=5\\10-x=5+3\\10-x=8\\-x=8-10\\-x=-2\\x=2

The integer is 2.

4 0
3 years ago
During the mayoral election,two debates were held between the candidates. The first debate lasted 1 2/3 hours.The second one was
butalik [34]

Answer:

2nd debate was 3 hours long.

Step-by-step explanation:

We have been given that during the mayoral election,two debates were held between the candidates. The first debate lasted 1 2/3 hours. The second one was 1 4/5 times as long as the first one.

Let us find the estimate of time spent on 2nd debate.

1 2/3 hours would be approximately 2 hours. 1 4/5 times would be equal to 2 times.

2*2=4

Therefore, the estimated time is less 4 hours.

To find the time spent on 2nd debate, we will multiply 1 2/3 by 1 4/5.

First of all, we will convert mixed fractions into improper fractions as:

1\frac{2}{3}=\frac{5}{3}

1\frac{4}{5}=\frac{9}{5}

Now, we will multiply both fractions as:

\frac{5}{3}\times \frac{9}{5}

\frac{1}{3}\times \frac{9}{1}

\frac{9}{3}\Rightarrow 3

Therefore, the 2nd debate was 3 hours long.

5 0
3 years ago
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