Answer:
In average, houses in the particular area use 119,6 therms of gas during the month of January.
Step-by-step explanation:
The μ formula is:
μ= ΣXi/N
ΣXi= is the sum of each xi. xi is each observation in the sample.
N= Total number of observations.
For this case:
ΣXi= 125+103+118+ 109+ 122+ 82+ 99+ 138+ 151+ 149
ΣXi= 1196
N= 10
μ= 1196/10
μ= 119,6
In average, houses in the particular area use 119,6 therms of gas during the month of January.
Answer:
The solutions are 
Step-by-step explanation:
To factor this cubic polynomial
you must:
- Group the polynomial into two sections

- Factor out -9 from


- Factor out
from 


- Factor out common term



- Factor




- Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

The solutions are

Answer:
a) c = (5/100)*d
b) c = 0.05*d
c) The ratio between what Lee recieves and what he sold.
d) d = 2000
Step-by-step explanation:
Since Lee earns 5% in commission for each sale then:
a) c = (5/100)*d
b) c = 0.05*d
c) In this context the constant of proportionality means the ratio of the value Lee sold that he'll earn as a comission.
d) Since he wants to earn $100, then c = 100 and we solve for d
c = 0.05*d
100 = 0.05*d
0.05*d = 100
d = 100/0.05 = $2000
He needs to sell $2000.
Let
x = the number of shorts bought
y = the number of t-shirts bought
A pair of shorts costs $16 and a t-shirt costs $10. Brandom has $100 to spend.
Therefore
16x + 10y ≤ 100
This may be written as
y ≤ - 1.6x + 10 (1)
Brandon wants at least 2 pairs of shorts. Therefore
x ≥ 2 (2)
Graph the equations y = -1.6x + 10 and x = 2.
The shaded region satisfies both inequalities.
Answer:
Two possible solutions are
(a) 3 pairs of shorts and 4 t-shirts,
(b) 4 pairs of shorts and 2 t-shirts.
Answer:
Step-by-step explanation:
Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?
Let p(x)=kpx+dp and q(x)=kqx+dq than
f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7
p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7
(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)
So you want:
−2kpk2q=0
and
kpkq=−1
and
2kpd2p−3kpdq+7=0
Now I amfraid this doesn’t work as −2kpk2q=0 that either kp or kq is zero but than their product can’t be anything but 0 not −1 .
Answer: there are no such linear functions.