Answer: I found Optimum Internet advertisement effective.
Explanation: I found optimum internet effective because not only does it cost less but it's a good deal.
I really don't know if this is will help but that's what I find effective.
The growth-share matrix defines four types of sbus: Cash cows are low-growth, high-share businesses or products.
Each of the four quadrants represents a particular combination of relative market share, and growth: Low Growth, High Share High Growth, High Share. Stars are high-growth, high –share businesses or products.
They often need heavy investments to finance their zoom. The market rate varies from industry to industry but usually shows a cut-off point of 10% – growth rates more than 10% are considered high, while growth rates below 10% are considered low.
Low market share business is a smaller amount than half the industry leader's share, and successful companies are those whose five-year average return on equity surpasses the industry median.
Growth-share business matrix may be a business tool, which uses relative market share and industry rate of growth factors to guage the potential of business brand portfolio and suggest further investment strategies.
The BCG matrix relies on Industry rate and relative market share. BCG matrix may be a framework created by Boston Consulting Group to guage the strategic position of the business brand portfolio and its potential.
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I think the rate of interest is 14 hope this helps
Because if you divide 880 by 63 you would just keep 14 without adding
Answer:
Conversion Cost Equivalent units FIFO 39, 125
Explanation:
Beginning WIP 5,000 30% completed
transferred units 39,500
ending WIP 4,500 25% completed
<u>The equivalent units will be:</u>
the transferred units
- complete portion for the beginning WIP
+ complete portion of the ending WIP
transferred out 39,500
work in previous period
5,000 x 30% = (1,500)
worked but not complete
4,500 x 25% = <u> 1, 125 </u>
Equivalent units FIFO 39, 125
Answer: V=7.43m/s
d =2.82m
Explanation:
a) For the first part, the initial velocity immediately after ejection, by using momentum conservation
before ejection, the momentum of the squid/water system is zero
there are no external forces acting on the system at the moment of ejection, so we can find the speed of the squid by noting
momentum before ejection = momentum after ejection
0 = M1U + M2V
0=-0.26 kg x 20 m/s + 0.7kg x V
where the speed of the water is taken as the negative sign, and V is the speed of the squid right after ejection, solving for V we get
V=7.43m/s
B. we use the equation vf^²=v0^²+2ad
where vf=final velocity = 0 since velocity is zero at motion's apex
v0=initial velocity = 7.43m/s
a = acceleration = -9.8m/s/s
d=height (to be found)
Therefore,
0=7.43^²+2(-9.8)d
Mathematically, it becomes
d=7.43^²/2(9.8)= 2.82m
d = 2.82m
