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2 years ago

Error Analysis Nolan began with $35 in his

Mathematics

2 answers:

gayaneshka [121]2 years ago

6 0

Answer:

Hi buddy

Step-by-step explanation:

Nolan begin with $35 spending $5 each day

X number of days and Y bank account balance

X=0 we can know that through the condition Y= $35

when X=1 Y=35-5=30

From these 2 conditions we can get the line model of the amount of money in her account bank is Y= -5x + 35

<h2># continue learning #</h2>

<h2>pls mark me as the brainliest</h2>

Irina-Kira [14]2 years ago

5 0

Answer:15

Step-by-step explanation:

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A group of people were asked which of three ice cream flavors they prefer. The results are shown in the table.

olchik [2.2K]

Using it's concept, it is found that the probability of a person being over 20 years old and preferring strawberry ice cream is of 12%.

<h3>What is a probability?</h3>

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem, there are 8 + 10 + 6 + 8 + 6 + 12 = 50 people, and of those, 6 are over 20 years old and prefer strawberry ice cream, hence the probability is given by:

p = 6/50 = 0.12 = 12%.

More can be learned about probabilities at brainly.com/question/14398287

#SPJ1

7 0

1 year ago

How do I do this? f(a)=4a+5; Find f(2)

Mariana [72]

F(2)=4(2)+5
f(2)=8+5
f(2)=13

6 0

2 years ago

Can u guy give me the AWNSER for number 3 or which ever on u know the Awnser to plz

FinnZ [79.3K]

3. Three and four tenths. 3+0.4=3.4
4. Two and fifty-one hundredths. 2+0.51=2.51
5. 8/10
6. 0.05
7. 46/100
8. 0.6
9. 9/10
10. 0.35

1 and 2 are correct btw

3 0

3 years ago

HELP PLS 20 POINTS! I WILL MARK BRAINLIEST

Alex777 [14]

Answer:

I believe it is 2 for $5.08

Step-by-step explanation:

1.27 times 2 is 2.54

2.54 times 5 is 12.70

7 0

3 years ago

Read 2 more answers

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