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bekas [8.4K]
4 years ago
12

Which equation supports the given conjecture?

Mathematics
1 answer:
Arada [10]4 years ago
6 0
It should've d because both numbers are odd and =even so the answer is D
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Please help. 25 points!
Misha Larkins [42]

Step-by-step explanation:

1-

(9 - 2) ^{2}  - (2 ^{2}  - 3 ^{2} ) + 2

calculate the powers behind the brackets from left to right

2-

(49) ^  - (4 - 9 ) + 2

Calculate the powers (a power squared means the number times itself so 2 squared equals 4)

3-

49 - - 5+2

Calculate the sum ( minus minus equals plus so for example 4- -5= 4+5=9)

4- 49+5+2= 56

8 0
3 years ago
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Three lines intersect to form angles. If <2 is equivalent to (2x-8)°, what is the value of x?
timofeeve [1]

Answer:

Step-by-step explanation:

3 0
3 years ago
The graph represents a functional relationship.
baherus [9]

Answer:

The value of 4 is an input of the function

Step-by-step explanation:

Define the variables

Let

x ----> the independent variable or input value

y ----> the dependent variable or output value

Looking at the graph

The domain of the graph (all possible values of x) is equal to the interval

[1,∞)

x\geq 1

All real numbers greater than or equal to 1

The range of the function (all possible values of y) is equal to the interval

(-∞,3]

y\leq 3

All real numbers less than or equal to 3

therefore

The value of 4 is an input of the function

4 0
3 years ago
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Is the following statement a good definition? Why?
Natasha_Volkova [10]
Yes it is. The statement is precise and reversible.
3 0
4 years ago
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Suppose the probability of an athlete taking a certain illegal steroid is 10%. A test has been developed to detect this type of
Travka [436]

Answer:

93.25% probability that they have taken this steroid

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

P(B|A) = \frac{P(B)*P(A|B)}{P(A)}

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Positive test

Event B: Taking the steroid.

Suppose the probability of an athlete taking a certain illegal steroid is 10%.

This means that P(B) = 0.1

Given that the athlete has taken this steroid, the probability of a positive test result is 0.995.

This means that P(A|B) = 0.995

Positive test:

99.5% of 10%(If the athlete has taken).

100-99.2 = 0.8% of 100-10 = 90%(Athlete has not taken)

Then

P(B) = 0.995*0.1 + 0.008*0.9 = 0.1067

Given that a positive test result has been observed for an athlete, what is the probability that they have taken this steroid

P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.1*0.995}{0.1067} = 0.9325

93.25% probability that they have taken this steroid

4 0
3 years ago
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