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Studentka2010 [4]
2 years ago
10

Is the statement 15=|-15| true

Mathematics
2 answers:
aleksandr82 [10.1K]2 years ago
7 0

Answer:

Yes! |-15| = 15

Step-by-step explanation:

Many people think of absolute value (the two bars symbol around the -15) as ALWAYS POSITIVE.

You can also think of absolute value as a distance. If you move 15 units, it doesn't matter in what direction you go. 15 units travelled is 15 units.

Lastly, absolute value has a V-shaped graph, that is because its always positive.

Paladinen [302]2 years ago
6 0
Yes, the lines on either side of the number -15 mean absolute value. Meaning that whatever number inside of the lines is positive.
You might be interested in
If f(x) = 3x + 2 and g(x) = x^2 + 1, which expression is equivalent to (f . g)(x)?
makvit [3.9K]

Answer:

d

Step-by-step explanation:

for this function you have to replace the function of g(x) where there's x in the function f(x) that's what the expression f.g(x) means and so this is going to be

3(x²+1)+2

I hope this helps

5 0
3 years ago
-7(v + 5) = 35 help
Airida [17]

Answer:

v=0

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLY PLEASE HELP!!!!!!!
marta [7]

Answer:

r² = 0.5652  < 0.7 therefore, the correlation between the variables does not imply causation

Step-by-step explanation:

The data points are;

X,          Y

0.7,       1.11

21.9,     3.69

18,         4

16.7,      3.21

18,         3.7

13.8,      1.42

18,          4

13.8,      1.42

15.5,      3.92

16.7,       3.21

The correlation between the values is given by the relation

Y =   b·X + a

b = \dfrac{N\sum XY - \left (\sum X  \right )\left (\sum Y  \right )}{N\sum X^{2} - \left (\sum X  \right )^{2}}

a = \dfrac{\sum Y - b\sum X}{N}

Where;

N = 10

∑XY = 499.354

∑X = 153.1

∑Y = 29.68

∑Y² = 100.546

∑X² = 2631.01

(∑ X)² = 23439.6

(∑ Y)² = 880.902

From which we have;

b = \dfrac{10 \times 499.354 -153.1 \times 29.68}{10 \times 2631.01 - 23439.6} = 0.1566

a = \dfrac{29.68 - 0.1566 \times 153.1}{10} = 0.5704

r = \dfrac{N\sum XY - \left (\sum X  \right )\left (\sum Y  \right )}{\sqrt{\left [N\sum X^{2} - \left (\sum X  \right )^{2} \right ]\times \left [N\sum Y^{2} - \left (\sum Y  \right )^{2} \right ]}}

r = \dfrac{10 \times 499.354 -153.1 \times 29.68}{\sqrt{\left (10 \times 2631.01 - 23439.6  \right )\times \left (10 \times 100.546- 880.902\right )}  } = 0.7518

r² = 0.5652  which is less than 0.7 therefore, there is a weak relationship between the variables, and it does not imply causation.

7 0
3 years ago
Write (5,-3) and (10,-18) in standard form
Lelu [443]

Answer:

y = - \frac{3}{4} x + 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c

Given

3x + 4y = 8 ( subtract 3x from both sides )

4y = - 3x + 8 ( divide all terms by 4 )

y = - \frac{3}{4} x + 2 ← in slope- intercept form

6 0
3 years ago
In a multiple choice quiz, there are 5 questions each with 4 answer choices (a, b, c, and d). Robin has not studied for the quiz
k0ka [10]

Answer:

A.  0.1035

B.  0.1406

C.  0.1025

Step-by-step explanation:

Given that:

the number of sample questions (n) = 5

The probability of choosing the correct choice (p) = 1/4 = 0.25

Suppose X represents the number of question that are guessed correctly.

Then, the required probability that she gets the majority of her question correctly is:

P(X>2) = P(X=3) + P(X =4) + P(X = 5)

P(X>2) =  [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4} + (^{5}C_{5}) \times (0.25)^5 (1-0.25)^{5-5}

P(X>2) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!}  \times (0.25)^4 (1-0.25)^{1} +\dfrac{5!}{5!(5-5)!}  \times (0.25)^5 (1-0.25)^{0} \Bigg ]

P(X>2) = [ 0.0879 + 0.0146 + 0.001 ]

P(X>2) = 0.1035

B.

Recall that

n = 5 and p = 0.25

The probability that the first Q. she gets right is the third question can be computed as:

P(X=x) = 0.25 ( 1- 025) ^{x-1}

Since, x = 3

P(X = 3) = 0.25 ( 1- 0.25 ) ^{3-1}

P(X =3) = 0.25 (0.75)^{3-1}

P(X =3) = 0.25 (0.75)^{2}

P(X=3) = 0.1406

C.

The probability she gets exactly 3 or exactly 4 questions right is as follows:

P(X. 3 or 4) = P(X =3) + P(X =4)

P(X=3 \ or \ 4) =  [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4}]

P(X=3 \ or \ 4) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!}  \times (0.25)^4 (1-0.25)^{1} \Bigg ]

P(X = 3 or 4) = [ 0.0879 + 0.0146 ]

P(X=3 or 4) = 0.1025

3 0
3 years ago
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