Ask question

Ask question

All categories

2 years ago

12

Two restaurants each surveyed diners to find out the ages of people eating at their restaurants. The box plots show the results

of the surveys.
Restaurant A

E

Restaurant B

+

10

+

15

+

50

20

45

55

60

25 30 35 40

Age of Diners (years)

Move options to the blanks to complete the statement, using data from the box plots.

than Restaurant B, because the

of data for Restaurant A is greater than

Restaurant A

that of Restaurant B.

served a greater variety of ages

served a greater number of diners

maximum

median

range

1 answer:

ipn [44]2 years ago

3 0

Answer:

Restaurant A served a greater variety of ages than Restaurant B, because the range of date for Restaurant A is greater than that of Restaurant B.

Step-by-step explanation:

You might be interested in

What's geometric mean?

Black_prince [1.1K]

<span>It means that it is pertaining to a type of math or in this case Geometry, for the use of its method.

</span>

3 0

2 years ago

Read 2 more answers

For f(x)-4x+1 and g(x)-x^2-5, find (g/f) (x)

antoniya [11.8K]

Answer:

A.

Step-by-step explanation:

let me know if you want an explanation :))

8 0

2 years ago

Read 2 more answers

In AJKL, the measure of ZL=90°, JL = 55, LK = 48, and KJ = 73. What

Juli2301 [7.4K]

Together I think it’s the same way that you can get the one to the one you have on your computer problems you have no proof that I have it on my computer problems I have lost all the same information I I don’t have any proof that you are in my account so so much you have to pay for it and I’m sorry you don’t know where I can get you I know I

3 0

2 years ago

I need help with this

andre [41]

Answer:

$f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}$

Step-by-step explanation:

Each vertical asymptote corresponds to a zero in the denominator. When the function does not change sign from one side of the asymptote to the other, the factor has even degree. The vertical asymptote at x=-4 corresponds to a denominator factor of (x+4). The one at x=2 corresponds to a denominator factor of (x-2)², because the function does not change sign there.

__

Each zero corresponds to a numerator factor that is zero at that point. Again, if the sign doesn't change either side of that zero, then the factor has even multiplicity. The zero at x=1 corresponds to a numerator factor of (x-1)².

__

Each "hole" in the function corresponds to numerator and denominator factors that are equal and both zero at that point. The hole at x=-3 corresponds to numerator and denominator factors of (x-3).

__

Taken altogether, these factors give us the function ...

$\boxed{f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}}$

8 0

3 years ago

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

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.

8 0

2 years ago