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Zolol [24]
1 year ago
14

For each lettered part, a through c, examine the two given sets of numbers. without doing any calculations, decide which set has

the larger standard deviation and explain why. then check by finding the standard deviations by hand. set 1 set 2
a.4, 7, 7, 7, 10 4, 6, 7, 8, 10

b.100, 140, 150, 160, 200 10, 50, 60, 70, 110

c.10, 16, 18, 20, 22, 28 48, 56, 58, 60, 62, 70
Mathematics
1 answer:
romanna [79]1 year ago
5 0
<span>Standard deviation shows the distribution present in a set. Range can be used to approximate this property. Therefore, Set b (100, 140, 150, 160, 200 10, 50, 60, 70, 110) has the largest standard deviation because it has a range of 190 (i.e. 200 - 10)</span>
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Boden is making a prize wheel for the school fair. The ratio of winning spaces to losing spaces is shown in the diagram. The tab
IceJOKER [234]

The correct answers are Losing 12; Winning 15

Explanation:

The ratio of winning to losing is 5: 6 or 5/6. This means for every 5 winning spaces in the wheel there are 6 losing spaces. This ration should be used to complete the values of the table.

1. The first row shows there are 10 winning and you need to calculate the number of losing spaces. The process is shown below.

\frac{5}{6} = \frac{10}{x} - Express the ratios using fractions; use x to show the missing value

5x = 60 - Cross multiply to find the value of x

x = 60 / 5 - Solve the equation to find x

x = 12 - The number of losing is 12 if there are 10 winning spaces

2. The second row shows there are 18 losing spaces, and you need to calculate the number of winning spaces. Repeat the process.

\frac{5}{6} = \frac{x}{18}

6x = 90

x = 90 / 6

x =15 - The number of winning spaces is 15 if there are 18 losing spaces

6 0
2 years ago
Read 2 more answers
Find c, yenvelope(x,t), and ycarrier(x,t). express your answer in terms of a, k1, k2, x, t, ω1, and ω2. separate the three parts
steposvetlana [31]

Answer:

C,Y_{envelope}(x,t), Y_{carrier}(x,t)=2A, \cos ((k_1-k_2)x/2-(\omega_1 -\omega_2)t / 2 ) , \sin ((k_1+k_2)x / 2 - (\omega_1 +\omega_2)t / 2 )

Step-by-step explanation:

Given

Y_1(x,t)=A \sin(K_1x- \omega _1 t)\\\\Y_2(x,t)=A \sin(K_2x- \omega _2 t)

using a trigonometrical identity

sin p + sin q = 2 sin ( p+q/2) cos ( p-q/2)

and here the condition is

the choice is in between sinax and cosax

where a > b

so we get using above equation

C,Y_{envelope}(x,t), Y_{carrier}(x,t)=2A, \cos ((k_1-k_2)x/2-(\omega_1 -\omega_2)t / 2 ) , \sin ((k_1+k_2)x / 2 - (\omega_1 +\omega_2)t / 2 )

4 0
2 years ago
The Venn diagram shows the number of patients seen at a pediatrician’s office in one week for colds, C, ear infections, E, and a
MArishka [77]

Answer:

Correct answers: 1 question: How many p both? The Venn diagram shows the number of patients seen at a infections pediatrician's office in ... pediatrician's office in one week for colds, C, ear infections, E, and allergies, A.

1 answer

Step-by-step explanation:

4 0
2 years ago
Find A ∩ B if A = {2, 5, 8, 11, 14} and B = {1, 3, 5, 7}.
Gwar [14]

Answer:

<h2>{5}</h2>

Step-by-step explanation:

A=\{2,\ 5,\ 8,\ 11,\ 14\}\\\\B=\{1,\ 3,\ 5,\ 7\}\\\\A\ \cap\ B-\text{the intersection of two sets A and B,}\\\text{ is the set that contains all elements of A that also belong to B.}\\\\\text{Therefore:}\ A\ \cap\ B=\{5\}

8 0
2 years ago
Read 2 more answers
Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes
Ivan

Answer:

14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

Now, speed of boat 1 changes to 8+7 = 15 knots

At 5 pm:

Distance traveled by boat 1 = 33 + 2\times 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 \times 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled \triangle AOB.

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = \dfrac{Total\ distance}{Total\ time}

\Rightarrow \dfrac{74.28}{5} = 14.86\ knots

6 0
1 year ago
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