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Anna [14]
3 years ago
14

In a survey of 400 people randomly selected it was determined that 22 play soccer. What is the probability that a randomly selec

ted person play soccer?
Mathematics
1 answer:
galben [10]3 years ago
7 0
The answer to this problem is 0.055%
And you get that by doing 22/400
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If 16+4x is 10 more than 14, what is the value of 8x?
sattari [20]
16 + 4x = 10 + 14

16 + 4x = 24.

4x = 24 - 16

4x = 8

x = 8 ÷ 4

x = 2

8x = 2 × 8

8x = 16

Final answer = 16.
3 0
3 years ago
If a girl cuts the grass in 3 hours and a boy cuts it in 2 hours what would be the time if they cut it toghther.?
svet-max [94.6K]

Answer:

d

Step-by-step explanation:

the answer is d because if you do 3+2 it equals 5 so therefore it is b

4 0
3 years ago
Read 2 more answers
Math 7a using number sense to solve equations
Anika [276]
Simple...

you have: \frac{n}{3} =12


and

\frac{n}{3} =-12

To solve for....\frac{n}{3} =12

Multiply to remove the denominator-->>

3*  \frac{n}{3} =12*3

n=12*3

n=36

Then-->>>

\frac{n}{3} =-12

Do the same thing-->>

3* \frac{n}{3} =-12*3

n=-12*3

n=-36

Thus, your answer.



6 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
Identifying and Generating Equivalent Expressions - Item 31254
ioda

Answer:

2 · 7 · 4 + 2 · 10 · 7 + 2 · 4 · 10

2(10 · 4 + 10 · 7 + 4 · 7)

Step-by-step explanation:

Given dimensions of the square prism:

  • length = 7 units
  • width = 4 units
  • height = 10 units

In order to find the surface area, we must find the area of each face.

Area of a rectangle = width x length

The square prism has 3 pairs of faces:

SA of bases = 7 x 4

SA of side 1 = 4 x 10

SA of side 2 = 7 x 10

So the total surface area is

2(7 x 4) + 2(4 x 10) + 2(7 x 10) = 276 units squared

<u>Solutions</u>

2 · 7 · 4 + 2 · 10 · 7 + 2 · 4 · 10

2(10 · 4 + 10 · 7 + 4 · 7)

3 0
2 years ago
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