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Semenov [28]
3 years ago
8

Ali picked three numbers out of a hat. Use the information in the box to work out what numbers he picked. The lowest number was

11;the range was 13; the median was 17. Write All numbers in the boxes from lowest to highest
Mathematics
1 answer:
IgorC [24]3 years ago
7 0

Answer:

The numbers he picked are 11, 17 and 24

Step-by-step explanation:

To solve this, we will work with the information we are given, interpreting it properly.

From the information given , we can already decode that the first of the three numbers is 11. We can write that number down.

The next piece of information we are given is that the range is 13.

The range of a data set is the difference between the highest and lowest numbers in the data set.

Let us say that our highest number is x

Range =x-11

x= range + 11\\x=13+11=24

Now, we know that the highest number of the data set is 24

Finally, from the information given, we can see that the median number is 17. This is the middle number of the data set. we can simply write that down since the number of figures in our data set is an odd number.

Hence the numbers picked are

11, 17 and 24 arranging them in an ascending order

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Option B) y = (1.31)(2.8^{x}) is the correct option.

Step-by-step explanation:

<em>Exponential function is the one where, a constant is raised to the power of x, a independent variable.</em>

Here, only option B and D are such exponential functions, while option A is linear and option C is quadratic in nature.

Now, to verify which option is correct, we will input the values and check which has the least difference between the given values.

Or we can use the cost function,

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By applying the above results,we find Option B is correct.

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Prove set equality by showing that for any element x, x \in (A \backslash (B \cap C)) if and only if x \in ((A \backslash B) \cup (A \backslash C)).

Example:

A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace.

B = \lbrace0,\, 1 \rbrace.

C = \lbrace0,\, 2 \rbrace.

\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}.

\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}.

Step-by-step explanation:

Proof for [x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))] for any element x:

Assume that x \in (A \backslash (B \cap C)). Thus, x \in A and x \not \in (B \cap C).

Since x \not \in (B \cap C), either x \not \in B or x \not \in C (or both.)

  • If x \not \in B, then combined with x \in A, x \in (A \backslash B).
  • Similarly, if x \not \in C, then combined with x \in A, x \in (A \backslash C).

Thus, either x \in (A \backslash B) or x \in (A \backslash C) (or both.)

Therefore, x \in ((A \backslash B) \cup (A \backslash C)) as required.

Proof for [x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]:

Assume that x \in ((A \backslash B) \cup (A \backslash C)). Thus, either x \in (A \backslash B) or x \in (A \backslash C) (or both.)

  • If x \in (A \backslash B), then x \in A and x \not \in B. Notice that (x \not \in B) \implies (x \not \in (B \cap C)) since the contrapositive of that statement, (x \in (B \cap C)) \implies (x \in B), is true. Therefore, x \not \in (B \cap C) and thus x \in A \backslash (B \cap C).
  • Otherwise, if x \in A \backslash C, then x \in A and x \not \in C. Similarly, x \not \in C \! implies x \not \in (B \cap C). Therefore, x \in A \backslash (B \cap C).

Either way, x \in A \backslash (B \cap C).

Therefore, x \in ((A \backslash B) \cup (A \backslash C)) implies x \in A \backslash (B \cap C), as required.

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