History, 16.08.2020 01:01 orlandokojoasem1234

Explain the decreasing trend in HIV cases from the late 90's to 2016.

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The pace of decline in new HIV infectionsis faster both among adults and children in all regions. In 1995 new HIV infection has reached a peak level. There were 160,000 new infections annually, onwards this point of time, it progressively decline and the new infection dropped by 81% to 30,000 (19,000-41,000) cases in 2016. The incidence of HIV infection has declined by 0.1352 per 1000 population since 1990 and reached 0.33 per 1000 population among all ages in 2016. Even though there is remarkable decline in incidence of HIV in all regions; the rate of decline and the current incidence revealed huge disparities among regions in Ethiopia

Regional trends in the annual number of new HIV infections and the prevalence density among all ages have varied. The incidence rate of 2.39/1000 population in Gambela Region, 1.5/1000 population in Addis Ababa city administration and 1.36/1000 population in Hareri region are by far higher than the national incidence. Similarly the prevalence of HIV in Addis Ababa (4.8%), Gambela (4.5%), Hareri (3%) and Dire dawa (2.8%) are by far Higher than the national prevalence and the global cut off point for generalized epidemic declaration

After continuous decline for decades, since 2008; HIV incidence began to rise by 10% from the level of 0.3/1000 population to 0.33/1000 population in 2016. At the same time the number of new infection increased by 36% from the level of 22,000 to 30,000 among all ages and doubled from 12,000 to 26,000 among adults. The rate of increment is higher in Gambela, Addis Ababa and Hareri regions. However, the number of new infection has reduced by 60% among children at national level

Answer from: Quest

answer

$x = \pm 2\sqrt{2} \: \: or \: \: x = \pm 1$

explanation

the given equation is

${x}^{4} - 9 {x}^{2} + 8 = 0$

we can rewrite this as:

$({x}^{2})^{2} - 9 {x}^{2} + 8 = 0$

let

$u = {x}^{2}$

then the equation becomes:

${u}^{2} - 9 {u}^{2} + 8 = 0$

the factors of 8 that sum up to -9 are {-1,-8}

we split the middle term with these factors to obtain,

${u}^{2} - u - 8u+ 8 = 0$

we factor by grouping to obtain:

$u(u - 1) - 8(u - 1) = 0$

$(u - 8)(u - 1) = 0$

apply the zero product property to get:

$u - 8 = 0 \: \: or \: \: u - 1 = 0$

$u = 8 \: or \: u = 1$

but

$u = {x}^{2}$

${x}^{2} = 8 \: or \: {x}^{2} = 1$

$x = \pm \sqrt{8} \: \: or \: \: x = \pm \: \sqrt{ 1}$

$x = \pm 2\sqrt{2} \: \: or \: \: x = \pm 1$

Answer from: Quest

definitely a for sure. ill see and update if i get the answer

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Explain the decreasing trend in HIV cases from the late 90's to 2016....

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