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LenKa [72]
3 years ago
8

Ms andrew made a line plots below to compare the quiz score for her first period math class and her second -period math class. s

he gave the same quiz to each class. what conclusion can ms andrew make about the performance to her first dn second period classes

Mathematics
2 answers:
lora16 [44]3 years ago
8 0

A. The first-period class had a higher median score than the second-period class.

B.The second-period class scores had a higher mean than the first-period class scores.

C.The first-period class scores had a greater range than the second-period

class scores.

D.The second-period class scores had a greater mean absolute deviation than the first-period class scores.

Answer : B. .The second-period class scores had a higher mean than the first-period class scores.

The score diagram is attached below

From the diagram we can see that the score in first period is more than the score in second period. Most of the score are greater in second period.

So ,The second-period class scores had a higher mean than the first-period class scores.

RoseWind [281]3 years ago
5 0
They should be about the same
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61 randomly selected students were asked the number of pairs of shoes they have. Let X represent the number of pairs of shoes. T
Molodets [167]

Complete question :

61 randomly selected students were asked the number of pairs of shoes they have. Let X represent the number of pairs of shoes. The results are as follows:

Pairs of Shoes4__5__6__7 __8 __9 __10 __11

Frequency : _ 8 _ 8 __5 _ 5 _ 9 __11 __7 ___8

Answer:

Mean = 7.64 ;

Median = 8

Q1 = 5

Q3 = 9

Atleast 10 pairs = 24.6

76% is equivalent to

Step-by-step explanation:

Round all your answers to 4 decimal places where possible.

10

The mean is:

Σfx /Σf

((8*4)+(5*8)+(6*5)+(7*5)+(8*9)+(9*11)+(10*7)+11*8)) ÷ (8+8+5+5+9+11+7+8) = 7.64

The median is:

0.5(n+1)th observation

n = frequency = 61

0.5(61 +1) = 1/2 * 62 = 31st observation

= 8

First quartile:

0.25(n+1)th observation

n = frequency = 61

0.25(61 +1) = 1/4 * 62 = 15.5

(15 + 16)th observation ÷ 2 = (5 + 5) / 2 = 5

The sample standard deviation is:

The third quartile is:

0.75(n+1)th observation

n = frequency = 61

0.75(61 +1) = 1/4 * 62 = 46.5

(46 + 47)th observation ÷ 2 = (5 + 5) / 2 = 9

What percent of the respondents have at least 10 pairs of Shoes? %

(7 + 8) / 61 = 15 / 61 = 0.246

76% of all respondents have fewer than how many pairs of Shoes?

(76 / 100) * 61

0.76 * 61

= 46.36

(46th + 47th)

(9 + 10) = 19 /2 = 9.5 = 10

6 0
3 years ago
Help!!<br> Giving brainliest!!
Elanso [62]
\dfrac{3}{5} (30x - 15) = 72

18x - 9 = 72    

18x = 81

x = 4.5

Answer: All that you checked are right, except the last one, x = 4.5 as well.
8 0
3 years ago
What is equivalent to 5x+2
svetlana [45]
Not sure I understand this but
5x+2=0
5x = -2
x = -2/5
Or
x = -.4
7 0
3 years ago
Need asap- will give brainliest-real answers only or you will be reported
Sav [38]

Answer: The answer is D.

Step-by-step explanation: In this case, you have to find the LCM. The LCM is 72 because both 24 and 36 are factors of 72, and that is the first product that both 24 and 36 are factors. Brainliest plz

6 0
3 years ago
Read 2 more answers
Find the area of the trapezoidal cross-section of the irrigation canal shown below. Your answer will be in terms of h, w, and θ.
miskamm [114]

[w + (h/tan θ)] × h

<h3>Further explanation</h3>

We aim is to find the area of the trapezoidal cross-section of the irrigation canal.

The formula for the area of the trapezoid is \boxed{\boxed{ \ Area = \frac{1}{2} \times (a + b) \times h \ }}

  • a & b = parallel sides
  • h = height

We assume the lower side is \boxed{ \ a = w \ } and the upper side is b = x + w + x or \boxed{ \ b = w + 2x \ }. See the attached picture.

In the right triangles located on the left and right of the trapezoid, we calculate the value of x based on trigonometric ratios namely tan of theta.

\boxed{ \ tan \ \theta = \frac{opposite}{adjacent} \ }

\boxed{ \ tan \ \theta = \frac{h}{x} \ }

\boxed{ \ x = \frac{h}{tan \ \theta} \ }

Substitute the equation of x to \boxed{ \ b = w + 2x \ }.

We get \boxed{ \ b = w + \frac{2h}{tan \ \theta} \ }

Finally all components are complete and can be substituted into a formula to calculate the area of a trapezoid.

\circ \ \boxed{a = w}\\ \circ \ \boxed{ \ b = w + \frac{2h}{tan \ \theta} \ }\\\circ \ and \ h

\boxed{ \ Area = \frac{1}{2} \times (w + w + \frac{2h}{tan \ \theta}) \times h \ }

\boxed{ \ Area = \frac{1}{2} \times (2w + \frac{2h}{tan \ \theta}) \times h \ }

\boxed{ \ Area = \frac{1}{2} \times 2(w + \frac{h}{tan \ \theta}) \times h \ }

Hence, the area of the trapezoidal cross-section of the irrigation canal is \boxed{\boxed{ \ Area = (w + \frac{h}{tan \ \theta}) \times h \ }}

<h3>Learn more</h3>
  1. Find out the area of parallelogram brainly.com/question/4459688
  2. The order of rotational symmetry of rhombus  brainly.com/question/4228574
  3. Find out the coordinates of the image of a point after the triangle is rotated 270° about the origin brainly.com/question/7437053

Keywords: the trapezoidal, cross-section of the irrigation canal, the area, the answer will be in terms of h, w, and θ, trigonometric ratios, tan, opposite, adjacent, height, sides

3 0
4 years ago
Read 2 more answers
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