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3 years ago

Use slopes and y-intercepts to determine if the lines y=6 and y=−5 are parallel.

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

7 0

Answer:lines are parallel

Step-by-step explanation:

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Hannah paddled the canoe for 3 hours at 2.2 miles per hour. How far did she paddle?

Grace [21]

6.6. 2.2*3= 6.6
i hope this makes since

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3 years ago

Read 2 more answers

In answering a question on a multiple-choice test, a student either knows the correct answer or guesses it. Let 0.7 be the proba

Dahasolnce [82]

Answer:

The probability is 0.9211

Step-by-step explanation:

Let's call K the event that the student know the answer, G the event that the student guess the answer and C the event that the answer is correct.

So, the probability P(K/C) that a student knows the answer to a question, given that she answered it correctly is:

P(K/C)=P(K∩C)/P(C)

Where P(C) = P(K∩C) + P(G∩C)

Then, the probability P(K∩C) that the student know the answer and it is correct is:

P(K∩C) = 0.7

On the other hand, the probability P(G∩C) that the student guess the answer and it is correct is:

P(G∩C) = 0.3*0.2 = 0.06

Because, 0.3 is the probability that the student guess the answer and 0.2 is the probability that the answer is correct given that the student guess the answer.

Therefore, The probability P(C) that the answer is correct is:

P(C) = 0.7 + 0.06 = 0.76

Finally, P(K/C) is:

P(K/C) = 0.7/0.76 = 0.9211

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3 years ago

Hii please help me!!!!!

inysia [295]

It should be eight if I remember correctly

3 0

2 years ago

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A. f(x) = l-2xl - 3 Domain: {1, 2, 3} Find the range

gizmo_the_mogwai [7]

Hello there!

We are given the function:

$\displaystyle \large{ f(x) = | - 2x| - 3} \\$

To find the range, we know that domain is the set of all x-values and also called 'input' while range is the set of all y-values and also called 'output'.

Basic Function - you add the input, you get the output. You add x-value in a function, you get y-value. You add domain, you get range.

So, we substitute x = 1,2 and 3 in the function.

x = 1

$\displaystyle \large{ f(1) = | - 2(1)| - 3} \\ \displaystyle \large{ f(1) = | - 2| - 3} \\$

Recall that any numbers in absolute value are always positive.

$\displaystyle \large{ f(1) = 2 - 3} \\ \displaystyle \large{ f(1) = - 1} \\$

x = 2

$\displaystyle \large{ f(2) = | - 2(2)| - 3} \\ \displaystyle \large{ f(2) = | - 4 | - 3} \\ \displaystyle \large{ f(2) = 4 - 3} \\ \displaystyle \large{ f(2) = 1}$

x = 3

$\displaystyle \large{ f(3) = | - 2(3)| - 3} \\ \displaystyle \large{ f(3) = | - 6| - 3} \\ \displaystyle \large{ f(3) = 6 - 3} \\ \displaystyle \large{ f(3) = 3}$