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Tanya [424]
3 years ago
8

Which one have is the coldest temperature negative numbers or positive numbers

Mathematics
1 answer:
aalyn [17]3 years ago
6 0

Answer:

Step-by-step explanat số âm vì số âm < số dương mà nhiệt đọ càng thấp thì càng lạnhion:

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Suppose a radio station desires to estimate the proportion of the market
In-s [12.5K]
About 3.4% large according to the coordinates
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3 years ago
Number of data values: 1007
kogti [31]

Answer:

The arithmetic mean is 10.9

Step-by-step explanation:

In this question, we are tasked with calculating the arithmetic mean using the information given.

The arithmetic mean or otherwise called the average is the sum of all the values in a given set divided by the number of elements or count of the elements in that particular data set

Mathematically, the arithmetic mean can be found by dividing the sum of the data values by the number of data values.

In equation form, we have Arithmetic -mean = \frac{sum of data values}{number of data values}

Inserting the values, we have Arithmetic mean = 10998/1007 = 10.92 which is 10.9 to the nearest tenth

5 0
3 years ago
What property was used to rewrite the polynomial expression (x+5)+3(a+(6x)y) to 5+(x+3a+(18x)y)
Umnica [9.8K]
<span>3(a+(6x)y) was clearly multiplied out as seen by the 3a and 18xy, so the distributive property was used there. In addition, the commutative and associative properties state that you can rearrange sums, so those were used too </span><span />
4 0
3 years ago
I need help with A,B,and C
Vinil7 [7]

Answer:

Answers!!!!!!

HOPE THIS HELPS!!!!!!!!!!

Step-by-step explanation:

8 0
3 years ago
Suppose that the length of a side of a cube X is uniformly distributed in the interval 9
Nastasia [14]

Answer:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

Step-by-step explanation:

Given

9 < x < 10 --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

v = x^3

For a uniform distribution, we have:

x \to U(a,b)

and

f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.

9 < x < 10 implies that:

(a,b) = (9,10)

So, we have:

f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Solve

f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Recall that:

v = x^3

Make x the subject

x = v^\frac{1}{3}

So, the cumulative density is:

F(x) = P(x < v^\frac{1}{3})

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right. becomes

f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.

The CDF is:

F(x) = \int\limits^{v^\frac{1}{3}}_9 1\  dx

Integrate

F(x) = [v]\limits^{v^\frac{1}{3}}_9

Expand

F(x) = v^\frac{1}{3} - 9

The density function of the volume F(v) is:

F(v) = F'(x)

Differentiate F(x) to give:

F(x) = v^\frac{1}{3} - 9

F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}

F'(x) = \frac{1}{3}v^{-\frac{2}{3}}

F(v) = \frac{1}{3}v^{-\frac{2}{3}}

So:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

8 0
3 years ago
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