Three percent of jennie’s skin cells were burned when she escaped from a fire. if 3.7×10’(of her skin cells were burned then, how many skin cells were not burned? must show work

We want to find the number of non-burned skin cells given that we know the percentage it represents.

1,196 cells were not burned.

Let's define N as the total number of cells.

3% were burned

then 97% were not burned.

We know that 3.7*10 cells were burned, and this is the 3% of N, then we have:

0.03*N = 3.7*10

N = (3.7*10)/0.03 = 1,233.33 in total.

The 97% of this is:

0.97*1,233.33 = 1,196.33

That we can round to the next whole number, 1,196

This means that 1,196 cells were not burned.

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The quantity of Jennie's skin cell were not burned is 119.601 ×  .

Step-by-step explanation:

Given as :

The percentage of Jennie's skin cell were burned = 3 %

The quantity of Jennie's skin cell were burned = 3.7 ×

Let The quantity of Jennie's skin cell were not burned = x

Let Total quantity of whole skin cell = n

Now, According to question

The quantity of Jennie's skin cell were burned = Total quantity of whole skin cell × percentage of Jennie's skin cell were burned

i.e 3.7 × = n × 3 %

Or, 3.7 × = n ×

Or, n =  3.7 × ×

Or, n =

i.e n = 1.233 ×

Or, Total quantity of whole skin cell = n = 1.233 ×

Now, Again

∵ percentage of Jennie's skin cell were burned = 3 %

So, percentage of Jennie's skin cell were not burned = 100 % - 3 % = 97 %

so , The quantity of Jennie's skin cell were not burned = 97 % of Total quantity of whole skin cell

Or, x = × n

Or, x =   × 1.233 ×

∴ x = 119.601 ×

So, The quantity of Jennie's skin cell were not burned = x = 119.601 ×

Hence,  The quantity of Jennie's skin cell were not burned is 119.601 ×  . Answer

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

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