The answer would be disagree because it you reflected it over the x axis then it would be in the third quadrant
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The answer is D: -1
If you plug it in, it’s the only one that works
A digit is a number in one of the places, so for example the number 54 has two digits; a tens place digit (5) and a ones place digit (4).
Say the mystery number is a two digit number = xy
* that's not x times y but two side by side digits.
Info given:
<span>the sum of the digits of a two-digit number is 6
x + y = 6 </span>
<span>if the digits are reversed, yx the difference between the new number and the original number is 18.
**To obtain the number from digits you must multiply by the place and add the digits up. (Example: 54 = 10(5) + 1(4))
Original number = 10x + y
Reversed/New number = 10y + x
Difference:
10y + x - (10x + y) = 18
9y - 9x = 18
9(y - x) = 18
y - x = 18/9
y - x = 2
Now we have two equations in two variables
</span>y - x = 2
<span>x + y = 6
Re-write one in terms of one variable for substitution.
y = 2 + x
sub in to the other equation to combine them.
x + (2 + x) = 6
2x + 2 = 6
2x = 6 - 2
2x = 4
x = 2
That's the tens digit for the original number. Plug this value into either of the equations to obtain y, the ones digit.
2 + y = 6
y = 4
number "xy" = 24
</span>
First, remove the parentheses
-7+2x+7=-7
Cancel the equal terms.
2x+7=0
Subtract 7
2x= -7
Divide both sides by 2
Final answer is:
X= -3.5
Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.
<h3>What is the binomial distribution formula?</h3>
The formula is:
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 0.54% of the calls receive a busy signal, hence p = 0.0054.
- A sample of 1300 callers is taken, hence n = 1300.
The probability that at least 5 received a busy signal is given by:
In which:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).
Then:
Then:
P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.
0.8295 = 82.95% probability that at least 5 received a busy signal.
More can be learned about the binomial distribution at brainly.com/question/24863377
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