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3 years ago

Ok ignore the thing that says do it the way i sowed you in class

Mathematics

1 answer:

Mamont248 [21]3 years ago

4 0

The way to do this question is : difference of the two sides < x < sum of the two sides. so for 1) 4 < x< 8
2) 8 < x< 16
3) 9 < x < 13
i’m pretty sure this is right! hope it helps!

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Which method could be used to factor x2 - 25?

Mandarinka [93]

Answer:

A. Perfect square Trinomial

Step-by-step explanation:

Just did its

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3 years ago

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4 Ayla s is paid $15 20 per hour for a 34 hour week. April has the same weekly pay but works 38

VashaNatasha [74]

Find Atlas weekly pay by multiplying rate by hours:

15.20 x 34 = 516.80 a week.

Now to find Apri’s rate divide the weekly pay by hours:

516.80 / 38 =$13.60 per hour

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3 years ago

Line m has the equation y = 1/2x - 4

Mandarinka [93]

Answer:

As the slopes of both lines 'm' and 'n' are the same.

Therefore, we conclude that the equation x-2y=4 represents the equation of the line 'n' if lines m and n are parallel to each other.

Step-by-step explanation:

We know that the slope-intercept of line equation is

$y=mx+b$

Where m is the slope and b is the y-intercept

Given the equation of the line m

y = 1/2x - 4

comparing with the slope-intercept form of the line equation

y = mx + b

Therefore,

The slope of line 'm' will be = 1/2

We know that parallel lines have the 'same slopes, thus the slope of the line 'n' must be also the same i.e. 1/2

Checking the equation of the line 'n'

$x-2y=4$

solving for y to writing the equation in the slope-intercept form and determining the slope

$x-2y=4$

Add -x to both sides.

$x - 2y + (-x) = 4+(-x)$

$-2y = 4 - x$

Divide both sides by -2

$\frac{-2y}{-2}=\frac{-x+4}{-2}$

$y=\frac{1}{2}x-2$

comparing ith the slope-intercept form of the line equation

Thus, the slope of the line 'n' will be: 1/2

As the slopes of both lines 'm' and 'n' are the same.

Therefore, we conclude that the equation x-2y=4 represents the equation of the line 'n' if lines m and n are parallel to each other.

3 0

3 years ago

Richard has just been given an l0-question multiple-choice quiz in his history class. Each question has five answers, of which o

myrzilka [38]

Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is $p = \frac{1}{5} = 0.2$

10 questions.

This means that $n = 10$

A) What is the probability that he will answer all questions correctly?

This is $P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024$

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so $P(X = 0)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

Since $P(X = 0) = 0.1074$ , from item b.

$P(X \geq 1) = 1 - 0.1074 = 0.8926$

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264$

$P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055$

$P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008$

$P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001$

$P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0$

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328$

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

8 0

3 years ago

Larry bought a used boat for $2400 and sold it for 2000 two years later what was the percent of decrease

guajiro [1.7K]

It pretty much would be 400

3 0

3 years ago