Answer:
4. A) 
3. A) 
2. D) 
1. C) 
Step-by-step explanation:
4. Two families combined have <em>at</em><em> </em><em>least</em><em> </em>3 pets.
3. $1500 is your result, so everything else is straightforward simple, knowing that the keywords <em>at</em><em> </em><em>least</em><em> </em>$1500<em> </em>mean <em>greater</em><em> </em><em>than</em><em> </em><em>or</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em>$1500.
2. Find the correspondents, then determine what inequality to use. In this case, you will use <em>less than</em><em> </em><em>or equal</em><em> </em><em>to</em><em> </em>because overpaying is UNNECESSARY.
1. <em>Eight</em><em> </em><em>dollars</em><em> </em><em>each</em><em> </em><em>pizza</em><em> </em>and <em>twelve</em><em> </em><em>dollars</em><em> </em><em>each</em><em> </em><em>pizza</em><em> </em>is <em>12x</em><em> </em>and <em>8x</em>.<em> </em>The rest is straightforward simple because you now have your <em>initial</em><em> </em><em>costs</em><em> </em>of <em>$</em><em>45</em><em> </em>and <em>$</em><em>5</em>,<em> </em>and the exercise tells you that the total charge for Company A is <em>less than</em> the total charge for Company B, therefore you have your answer.
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Answer:
-95.78
Step-by-step explanation:
As the researcher decided to make the number of parties attended per week the explanatory variable, this would be variable x in the regression line, and of course, the variable y would be the number of text messages sent per day.
After constructing the linear regression equation, the researcher found that an approximate value
for the actual value of y could be represented by the line
Since this is an approximate value, it is not expected that it coincides with the actual value of y. We define then the residual for each value of x as the difference between the actual value of y and the approximation for the given x.
For the value x = 2 (the student attended 2 parties that week) the actual value of y is 20 (the student sent 20 text messages per day that week).
The approximate value of y would be according to the regression line
Hence, the residual value for x=2 would be
Answer:

Step-by-step explanation:
By applying <em>Pythagoras theorem</em> we can find hypotenuse of a right angle triangle.
Side of right angle triangle (a) = 5√2 cm
Side of right angle triangle (b) = 5√2 cm
Hypotenuse² = a² × b²

Answer:
the expected value of Xn , E(Xn) = 0 and the variance σ²(Xn) = n*(1-2n)
Step-by-step explanation:
If X1= number of tails when n fair coins are flipped , then X1 follows a binomial distribution with E(X1) = n*p , p=0,5 and the number of heads obtained is X2=n-X1
therefore
Xn =X1-X2 = X1- (n-X1) = 2X1-n
thus
E(Xn) =∑ (2*X1-n) p(X1) = 2*∑[X1 p(X1)] -n∑p(X1) = 2*E(X1)-n = 2*n*p--n= 2*n*1/2 -n = n-n =0
the variance will be
σ²(Xn) = ∑ [Xn - E(Xn)]² p(Xn) = ∑ [(2X1-n) - 0 ]² p(X1) = ∑ (4*X1²-4*X1*n+n²) p(X1) = = 4*∑ X1²p(X1) - 4n ∑X1 p(X1) - n²∑p(X1) = 2*E(X1²) -4n*E(X1)- n²
since
σ²(X1) = n*p*(1-p) = n*0,5*0,5=n/4
and
σ²(X1) = E(X1²) - [E(X1)]²
n/4 = E(X1²) - (n/2)²
E(X1²) = n(n+1)/4
therefore
σ²(Xn) = 4*E(X1²) -4n*E(X1)- n² = 4*n(n+1)/4 - 4*n*n/2 - n² = n(n+1) - 2n² - n²
= n - 2n² = n(1-2n)
σ²(Xn) = n(1-2n)
Work the information to set inequalities that represent each condition or restriction.
2) Name the
variables.
c: number of color copies
b: number of black-and-white copies
3)
Model each restriction:
i) <span>It
takes 3 minutes to print a color copy and 1 minute to print a
black-and-white copy.
</span><span>
</span><span>
3c + b</span><span>
</span><span>
</span><span>ii) He needs to print
at least 6 copies ⇒
c + b ≥ 6</span><span>
</span><span>
</span><span>iv) And must have
the copies completed in
no more than 12 minutes ⇒</span>
3c + b ≤ 12<span />
4) Additional restrictions are
c ≥ 0, and
b ≥ 0 (i.e.
only positive values for the number of each kind of copies are acceptable)
5) This is how you
graph that:
i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.
ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.
iii) since c ≥ 0 and b ≥ 0, the region is in the
first quadrant.
iv) The final region is the
intersection of the above mentioned shaded regions.v) You can see such graph in the attached figure.