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3 years ago

PLSSS HELP IF YOU TURLY KNOW THISS

Mathematics

2 answers:

Marrrta [24]3 years ago

4 0

Answer:

wouldnt X be 0.6?

Step-by-step explanation:

Elanso [62]3 years ago

4 0

<u>Answer:</u>

<u>Answer:2.0</u> is your answer.

I HOPE IT WILL HELP YOU.

in place of x u should write . 0

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Square root of 25 rounded to the nearest tenth

galina1969 [7]

5 is the answer
I used my phone

3 0

3 years ago

Read 2 more answers

9. GIVEN: y(x + 1) = 51; y = 3 PROVE: x = 16

ivann1987 [24]

Answer:

see explanation

Step-by-step explanation:

y(x + 1) = 51 ← substitute y = 3 into the equation

3(x + 1) = 51 ( divide both sides by 3 )

x + 1 = 17 ( subtract 1 from both sides )

x = 16

7 0

2 years ago

Read 2 more answers

A triangle has one side that is 6 units long and another side that is 3 units long.

kiruha [24]

The possible value of the third length is an illustration of Triangle inequality theorem

The possible third lengths are 4 units and 6 units

<h3>How to determine the possible length of the third side?</h3>

To determine the third length, we make use of the following Triangle inequality theorem.

a + b > c

Let the third side be x.

So, we have:

x + 6 > 3

x + 3 > 6

3 + 6 > x

Solve the inequalities

x > -3

x > 3

x < 9

Remove the negative inequality value.

So, we have:

x > 3 or x < 9

Rewrite as:

3 < x or x < 9

Combine the inequality

3 < x < 9

This means that the possible value of the third length is between 3 and 9 (exclusive)

Hence, the possible third lengths are 4 units and 6 units

Read more about Triangle inequality theorem at:

brainly.com/question/2403556

5 0

3 years ago

The critical value for a hypothesis test _______.

Jlenok [28]

Answer:

Option C) Critical value is based on the significance level and determines the boundary for the rejection region

Step-by-step explanation:

Critical Value:

In hypothesis testing, a critical value is a point that is compared to the test statistic
It is used to determine whether to reject the null hypothesis or accept the null hypothesis.
If the absolute value of your test statistic is greater than the critical value,we fail to accept the null hypothesis and reject it.
Critical value is affected by the significance level of the testing.
It is the value that a test statistic must exceed in order for the the null hypothesis to be rejected.

Thus, option C) is the correct interpretation of critical values.

Option C) Critical value is based on the significance level and determines the boundary for the rejection region

7 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

3 years ago