3k+10=2k-21
move 2k to the other side
sign changes from +2k to -2k
3k-2k+10=2k-2k-21( combine like terms)
3k-2k+10= -21
k+10=-21
move +10 to the other side
k+10-10= -21-10
k= -21-10
answer:
k= -31
Answer:
30.17feet high
Step-by-step explanation:
Given the following
the distance from the base of the pole to the tip of the shadow = 49feet (hypotenuse)
angle of elevation = 38°
Required
Height of the flagpole (opposite)
Using the SOH CAH TOA identity
Sin theta = opposite/hypotenuse
Sin 38 = H/49
H = 49sin38
H = 49(0.6157)
H = 30.17feet
Hence the flagpole is 30.17feet high
Answer:
Step-by-step explanation:
FV =<u> p (1+i)^n -1</u>
i
pv = 700,000
i = .075/12 = .00625
n = (66 - 15)* 12 = 612
700,000 = P (( 1 + .00625)^ 612 -1 /.00625
4375 = P (1.00625)^612 -1)
P = $98.77
Answer:
For this case if we want to conclude that the sample does not come from a normally distributed population we need to satisfy the condition that the sample size would be large enough in order to use the central limit theoream and approximate the sample mean with the following distribution:

For this case the condition required in order to consider a sample size large is that n>30, then the best solution would be:
n>= 30
Step-by-step explanation:
For this case if we want to conclude that the sample does not come from a normally distributed population we need to satisfy the condition that the sample size would be large enough in order to use the central limit theoream and approximate the sample mean with the following distribution:

For this case the condition required in order to consider a sample size large is that n>30, then the best solution would be:
n>= 30
Answer:c(d) = 5 + (0.75*d)
Step-by-step :
In this equation the cost (c) of the distance traveled (d) has a base rate of 5, and increases by .75 for every mile traveled (d)