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2 years ago

No random answers or links, please.

Mathematics

2 answers:

kondaur [170]2 years ago

6 0

Answer:

the ans is 1 hope ut may help u

Step-by-step explanation:

x=-2

x = -(-2) -1

x=2-1

x=+1

alina1380 [7]2 years ago

4 0

Answer:

Step-by-step explanation:

By question it's given that ,

=> h(x) = -x - 1

For finding h(-2) , substitute x = -2 .

=> h(-2) = -(-2)-1

=> h(-2) = 2 - 1

=> h(-2) = 1

<h3>Hence the required answer is 1.</h3>

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Ainat [17]

Answer:

Option D, Yes, Devon should have evaluated f(3)

Step-by-step explanation:

f(x) = x^3 + x^2 − 10x + 8

Step 1: Evaluate using F(3) NOT f(-3)

f(3) = 3^3 + 3^2 - 10(3) + 8

f(3) = 27 + 9 - 30 + 8

f(3) = 14

Since the factor is (x - 3) the x value is x - 3 + 3 = 0 + 3 -> x = 3 not x = -3

Answer: Option D, Yes, Devon should have evaluated f(3)

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3 years ago

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Musya8 [376]

The answer is £287.98 is the correct answer

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2 years ago

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irakobra [83]

Pretty sure it’s the second one

3 0

3 years ago

35+14 distributive property

Lera25 [3.4K]

Answer:

7(2)+7(5) {using distributive property}

14+35 {7 multiplied by 2 gives 14 and 7 multiplied by 5 gives 35}

49 {therefore 35+14 gives 49}

Step-by-step explanation:

Your answer is 49

Hope this helps :)

5 0

3 years ago

Read 2 more answers

Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose fur

stealth61 [152]

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.
$P(A \cap B)$ is the probability of both A and B happening.
P(A) is the probability of A happening.

In this problem:

Event A: Three heads.
Event B: Fair coin.

The probability associated with 3 heads are:

$0.5^3 = 0.125$ out of 0.5(fair coin).
1 out of 0.5(biased).

Hence:

$P(A) = 0.125 + 0.5 = 0.625$

The probability of 3 heads and the fair coin is:

$P(A \cap B) = 0.5(0.125) = 0.0625$

Then, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0625}{0.625} = 0.1$

0.1 = 10% probability that the chosen coin was the fair coin.

A similar problem is given at brainly.com/question/14398287

4 0

2 years ago