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wlad13 [49]
2 years ago
5

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, ho

w many skin cells were not burned? Must show work
Mathematics
1 answer:
lora16 [44]2 years ago
6 0

Answer:

The quantity of Jennie's skin cell were not burned is 119.601 × 10^{10}  .

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 × 10^{10}

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

Let Total quantity of whole skin cell = n

<u>Now, According to question</u>

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 × 10^{10} = n × 3 %

Or, 3.7 × 10^{10} = n × \dfrac{3}{100}

Or, n =  3.7 × 10^{10} × \dfrac{100}{3}

Or, n = \frac{3.7\times 10^{12}}{3}

i.e n = 1.233 × 10^{12}

Or, Total quantity of whole skin cell = n = 1.233 × 10^{12}

<u>Now, Again</u>

∵ 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 = \frac{97}{100} × n

Or, x =  \frac{97}{100} × 1.233 × 10^{12}

∴ x = 119.601 × 10^{10}

So, The quantity of Jennie's skin cell were not burned = x = 119.601 × 10^{10}

Hence,  The quantity of Jennie's skin cell were not burned is 119.601 × 10^{10}  . Answer

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a) We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:  

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And the region is on the figure attached

Step-by-step explanation:

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:  

Null hypothesis:\mu \geq 5.5  

Alternative hypothesis:\mu < 5.5  

Part b

For this case since we want to determine if the true mean is lower than an specified value is a one tailed test

Part c

For this case the significance level is given \alpha=0.01and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:

z_{cric}= -2.326

And the rejection zone would be: (\infty ,-2.326)

And the region is on the figure attached

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2 years ago
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