Answer:
(c) H0 should be rejected
Step-by-step explanation:
Null hypothesis (H0): population mean is equal to 5
Alternate hypothesis (Ha): population mean is greater than 5
Z = (sample mean - population mean) ÷ (sd/√n)
sample mean = 5.3, population mean = 5, sd = 1, n = 500
Z = (5.3 - 5) ÷ (1/√500) = 0.3 ÷ 0.045 = 6.67
Using the normal distribution table, for a one tailed test at 0.01 significance level, the critical value is 2.326
Conclusion:
Since 6.67 is greater than 2.326, reject the null hypothesis (H0)
Answer:
The answer is B. (-7,-2)
Step-by-step explanation:
Multiply the first equation by 7,and multiply the second equation by -6.
7(−3x+6y=9)
−6(5x+7y=−49)
Becomes:
−21x+42y=63
−30x−42y=294
Add these equations to eliminate y:
−51x=357
Then solve−51x=357for x:
−51x=357 Divide both sides by -51
x=-7
Write down an original equation:
−3x+6y=9
Substitute−7forxin−3x+6y=9:
(−3)(−7)+6y=9
6y+21=9(Simplify both sides of the equation)
6y+21+−21=9+−21(Add -21 to both sides)
6y=−12
y=-2
Domain is all real numbers
range
hmm
we know that x^2=a positive number
then multiply it by that negative -3
therefor the range is going to be mostly negative
if we make the x-5 equal to zero, then there is no negative, so
wher x=5, then f(5)=0+4=4
the highest positive number you can get is f(5)=4
so therfor
domain=all real number
range is x≤4
I'd use synthetic div. here, with 2 as my divisor:
________________
2 / 1 -11 18
2 -18
----------------------------
1 -9 0
Note that the coefficients of the missing factor are given here and are 1 and -9. Thus, (x-9) is the missing factor.
Step-by-step answer:
Please refer to attached image.
1. Quad PQRS is cyclic (all vertices on the same circle), so opposite angles are supplementary, meaning
that
angles QPS and QRS are supplementary =>
QPS+QRS=180 =>
QRS = 180 - 74 = 106
2. Triangle RSQ is isosceles with RS=RQ =>
angles RSQ and RQS are congruent.
3. Angle RSQ = (180 - 106) /2 = 74 / 2 = 37
4. QP is a diameter => angle QSP is a right-angle.
5. From (3) and (4) above,
angle RSP = 37+90 = 127
6. Since PQRS is cyclic, angles RQP and RSP are supplementary =>
RQP+RSP = 180 =>
x + 127 = 180 =>
x = 180 - 127 = 53 degrees.
