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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?

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

You might be interested in

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