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9966 [12]
4 years ago
9

Ethan is conducting an experiment to determine whether a new medication is effective in reducing sneezing. He finds 1,500 volunt

eers with sneezing issues and divides them into 2 groups. The control group does not receive any medication; the treatment group receives the medication. The patients in the treatment group show reduced signs of sneezing. What can Ethan conclude from this experiment?
Mathematics
2 answers:
andre [41]4 years ago
6 0
The medicine helps but not very much
Nuetrik [128]4 years ago
5 0

Answer:

The medicine is effective in reducing sneezing.

Step-by-step explanation:

In this example, Ethan wants to perform an experiment that will determine whether a medication is effective in reducing sneezing.

In order to make sure that the resulting effect is actually a result of the medicine, Ethan needs a control group. As the control group and the treatment group are equal in every aspect except the use of the medicine, he can assume that any differences in outcome will be a result of the medicine. Therefore, when the treatment group show reduced signs of sneezing, he knows that this is an effect of the treatment, and that the medicine works.

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In 1955 an antique car that originally cost $3,668 is valued today at $62,125 if in excellent condition, which is 1 3/4 times as
chubhunter [2.5K]

Answer:

The value of the car in very nice condition will be $35500.

Step-by-step explanation:

Let the value of car in a very nice condition be = x

The value $62125 is 1\frac{3}{4} or \frac{7}{4} or 1.75 times of x.

Now, we can calculate for x:

1.75x=62125

x=\frac{62125}{1.75}

x = 35500

Hence, the value of the car in very nice condition will be $35500.

4 0
4 years ago
Which of the following statements is true for all sets A,B and C? Give a proof
ELEN [110]

Answer:

(a) and (b) are not true in general. Refer to the explanations below for counterexamples.

It can be shown that (c) is indeed true.

Step-by-step explanation:

This explanation will use a lot of empty sets \phi just to keep the counterexamples simple.

<h3>(a)</h3>

Note that A \cap B can well be smaller than A. It should be alarming that the question is claiming A\! to be a subset of something that can be smaller than \! A. Here's a counterexample that dramatize this observation:

Consider:

  • A = \left\lbrace 1 \right\rbrace.
  • B = \phi (an empty set, same as \left\lbrace \right\rbrace.)
  • C = \phi (another empty set.)

The intersection of an empty set with another set should still be an empty set:

A \cap B = \left\lbrace 1\right\rbrace \cap \left\lbrace\right\rbrace = \left\lbrace\right\rbrace.

The union of two empty sets should also be an empty set:

((A \cap B) \cup C) = \left\lbrace\right\rbrace \cup \left\lbrace\right\rbrace = \left\lbrace\right\rbrace.

Apparently, the one-element set A = \left\lbrace 1 \right\rbrace isn't a subset of an empty set. A \not \subseteq ((A\cap B) \cup C). Contradiction.

<h3>(b)</h3>

Consider the same counterexample

  • A = \left\lbrace 1 \right\rbrace.
  • B = \phi (an empty set, same as \left\lbrace \right\rbrace.)
  • C = \left\lbrace 2 \right\rbrace (another empty set.)

Left-hand side:

(A \cup B) \cap C = \left(\left\lbrace 1 \right\rbrace \cup \left\lbrace \right\rbrace\right) \cap \left\lbrace 2 \right\rbrace\right = \left\lbrace 1 \right\rbrace \cap \left\lbrace 2 \right\rbrace = \left\lbrace \right\rbrace.

Right-hand side:

(A \cap B) \cup C = \left(\left\lbrace 1 \right\rbrace \cap \left\lbrace \right\rbrace\right) \cup \left\lbrace 2 \right\rbrace\right = \left\lbrace \right\rbrace \cup \left\lbrace 2 \right\rbrace = \left\lbrace 2 \right\rbrace.

Apparently, the empty set on the left-hand side \left\lbrace \right\rbrace is not the same as the \left\lbrace 2 \right\rbrace on the right-hand side. Contradiction.

<h3>(c)</h3>

Part one: show that left-hand side is a subset of the right-hand side.

Let x be a member of the set on the left-hand side.

x \in (A \backslash B) \cap C.

\implies x\in A \backslash B and x \in C (the right arrow here reads "implies".)

\implies x \in A and x \not\in B and x \in C.

\implies (x \in A\cap C) and x \not\in B \cap C.

\implies x \in (A \cap C) \backslash (B \cap C).

Note that x \in (A \backslash B) \cap C (set on the left-hand side) implies that x \in (A \cap C) \backslash (B \cap C) (set on the right-hand side.)

Therefore:

(A \backslash B) \cap C \subseteq (A \cap C) \backslash (B \cap C).

Part two: show that the right-hand side is a subset of the left-hand side. This part is slightly more involved than the first part.

Let x be a member of the set on the right-hand side.

x \in (A \cap C) \backslash (B \cap C).

\implies x \in A \cap C and x \not\in B \cap C.

Note that x \not\in B \cap C is equivalent to:

  • x \not \in B, OR
  • x \not\in C, OR
  • both x \not\in B AND x \not \in C.

However, x \in A \cap C implies that x \in A AND x \in C.

The fact that x \in C means that the only possibility that x \not\in B \cap C is x \not \in B.

To reiterate: if x \not \in C, then the assumption that x \in A \cap C would not be true any more. Therefore, the only possibility is that x \not \in B.

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

In other words, x \in (A \cap C) \backslash (B \cap C) \implies x \in (A \backslash B)\cap C.

(A \cap C) \backslash (B \cap C) \subseteq (A \backslash B)\cap C.

Combine these two parts to obtain: (A \backslash B) \cap C = (A \cap C) \backslash (B \cap C).

8 0
4 years ago
The fourth term of a sequence is 108. Each term after the first is 3 times the previous term. Write an explicit expression that
marishachu [46]

Answer:

f(n)=4(3)^{n-1}

Step-by-step explanation:

Let the first term be x, then the terms of the sequence are:

x,3x,9x,27x,81x,....

Since the fourth term is 108, we have

27x=108

\implies x=4

Hence the sequence becomes:

4,12,36,108,324,....

The explicit expression for a geometric expression is given by:

f(n)=ar^{n-1}

where a=4 and r=3

The required formula is:

f(n)=4(3)^{n-1}

8 0
3 years ago
A−9=6 addition equation.
blondinia [14]

Answer:

15

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Evaluate 3z- 3 when z=7<br><br> a 34<br> b 21<br> c 18
OLga [1]

Answer: C (18)

Step-by-step explanation: If we plug in 7 for z, 3 times 7 is 21. 21-3=18.

6 0
3 years ago
Read 2 more answers
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