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Sergeu [11.5K]
3 years ago
12

. (06.02)

Mathematics
1 answer:
lora16 [44]3 years ago
7 0
Answers:

1.) C - Final exams: midterm = Q3 92 - Q1 80 = 12 and finals = Q3 85 - Q1 78 = 7, so finals exam has the smaller IQR.

2.) B - exam median is much higher than the class median: refer to the attached image of class and exam box and whisker graph

3.) A - <span>IQR is a better measure of spread for movies than it is for basketball games: because the data for movies are quite wide or apart from each other than the data in the basketball games

4.) A - </span><span>mean for April is higher than October's mean: mean in April is 67 and mean in October is 60

5.) A - </span><span>Neither data set has suspected outliers: meaning there's no data that is further apart from the group or set

6.) B - </span>There is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical 
<span>
7.) C - </span>college spread is best described by the IQR. The high school spread is best described by the standard deviation: this is because of the wide range of data in the college than the data in the high school.

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Effectus [21]

Answer:

Step-by-step explanation:

6 0
2 years ago
A circular cylindrical container, open at the top, and having a capacity of 24pi cubic inches, is to be manufactured. If the cos
otez555 [7]

Answer: radius r = 2 inches

height h = 6 inches

Step-by-step explanation:

Given;

Volume V = 24πin3

Volume of a cylinder is given by

V = πr^2h

h = V/πr^2. ....1

Where, h = height and r = radius of cylinder

For the surface area of the cylinder with open top. we have,

S = 2πrh + πr^2

For the cost of materials used, let k represent the cost of materials used for the body of the cylinder.

Then, for the bottom will be 3k

Total cost will be represented by C, which gives

C = 2πrhk + 3πr^2k. .....2

Substituting eqn 1 to 2, we have;

C = 2πrVk/πr^2 + 3πr^2k

C = 2Vk/r + 3πr^2k

The material cost is minimum at dC/dr = 0

dC/dr = -2Vk/r^2 + 6πrk =0

6πrk = 2Vk/r^2

r^3 = 2V/6π

r = (2×24π/6π)^-3

r = (8)^-3

r = 2

Substituting r = 2 into eqn1

h = 24π/π(2^2)

h = 24/4 = 6

h = 6

5 0
3 years ago
Find x. 12=6x what is x?
pickupchik [31]

Answer:

x=12/6

therefore x=2

7 0
3 years ago
Read 2 more answers
Which system of equations could be graphed to solve the equation below? log Subscript 0. 5 Baseline x = log Subscript 3 Baseline
Vanyuwa [196]

You can use the fact that two expressions in equality can be considered to be equal to a third variable(not used in given context).

The system of equations that could be graphed to solve the equation given is

  • y = \log_{0.5}(x)\\\\
  • y = \log_3(2+x)

<h3>How can we form a system of equations from an equation?</h3>

Suppose the equation be a = b\\

Let there is a symbol c such that we have a = b = c

It is because a and b are same measure (that is exactly what a = b means)

and we gave another name c to that measure.

Thus, we have

a = c\\b = c

in addition to  a = b\\

<h3>Using the above method to find the system of equations needed</h3>

Since the given equation is log_{0.5}(x) = log_2(2+x)

The 2d graphs are usually expressed as y = f(x) on X-Y plane.

Taking the equation's expressions equal to y, we get

log_{0.5}(x) = log_2(2+x) = y

or, we get system of equations as

y = log_{0.5}(x)\\\\y = log_3{(2 + x)}

Their graph is plotted below. The intersection point of both curves is the solution to the given equation as it satisfies both the equations of the system of equations formed from the given equation.

Thus,

The system of equations that could be graphed to solve the equation given is

  • y = \log_{0.5}(x)\\\\
  • y = \log_3(2+x)

Learn more about solutions to system of equations here:

brainly.com/question/14550337

4 0
3 years ago
Each of three coins has two sides, heads and tails. Represent the heads or tails status of each coin by a logical variable (A fo
jeka94

Answer:

(a) F(A, B, C) = AB'C' + A'BC' + A'B'C

(b) F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)

Step-by-step explanation:

(a) If F(A, B, C) = 1 iff exactly one of the coins is heads, then either

A is heads and the others are tails (AB'C')

B is heads and the others are tails (A'BC')

C is heads and the others are tails (A'B'C)

Hence, as a minterm expansion,

F(A, B, C) = AB'C' + A'BC' + A'B'C

(b) To get the corresponding maxterm expansion, we convert to binary.

F(A, B, C) = \sum (100, 010, 001) = \sum(4,2,1)

The maxterm is the product of the complements.

F(A, B, C) = \prod (0, 3, 5, 6, 7) = \prod(000, 011, 101, 110, 111)

Expanding,

F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)

3 0
3 years ago
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