Answer:
![0.625kg](https://tex.z-dn.net/?f=0.625kg)
Step-by-step explanation:
we are given half-life of PO-210 and the initial mass
we want to figure out the remaining mass <u>after</u><u> </u><u>4</u><u>2</u><u>0</u><u> </u><u>days</u><u> </u>
in order to solve so we can consider the half-life formula given by
![\displaystyle f(t) = a {0.5}^{t/T}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20f%28t%29%20%3D%20a%20%7B0.5%7D%5E%7Bt%2FT%7D%20)
where:
- f(t) is the remaining quantity of a substance after time t has elapsed.
- a is the initial quantity of this substance.
- T is the half-life
since it halves every 140 days our T is 140 and t is 420. as the initial mass of the sample is 5 our a is 5
thus substitute:
![\displaystyle f(420)=5\cdot{0.5}^{420/140}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28420%29%3D5%5Ccdot%7B0.5%7D%5E%7B420%2F140%7D)
reduce fraction:
![\displaystyle f(420)=5\cdot{0.5}^{3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28420%29%3D5%5Ccdot%7B0.5%7D%5E%7B3%7D)
By using calculator we acquire:
![\displaystyle f(420)=0.625](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28420%29%3D0.625)
hence, the remaining sample after 420 days is 0.625 kg
Answer:
There is not evidence that the population mean waiting time is different from 3.9 minutes.
Step-by-step explanation:
From the question we know that the size (n), the mean (x) and the standard deviation (s) of the sample are 81, 4.05 minutes and 0.9 minutes. Additionally, we are going to decide if the waiting time is different or not form 3.9 minutes, so the null and alternative hypotheses are:
H0: m=3.9
H1: m≠3.9
Where m is the mean of the population.
Then, <em>we don't need to be concerned about the shape of the population distribution because the value of the n is bigger than 30</em> and we can use the statistic z as:
![z=\frac{x-m}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-m%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
So, replacing the values, the test statistic is:
![z=\frac{4.05-3.9}{\frac{0.9}{\sqrt{81}}}=1.50](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B4.05-3.9%7D%7B%5Cfrac%7B0.9%7D%7B%5Csqrt%7B81%7D%7D%7D%3D1.50)
On the other hand, the p value for this test is calculated as:
p value = 2P(z>1.50) = 2(0.0668) = 0.134
Taking into account that the p value is bigger than the level of significance 0.01, the null hypothesis is not reject, and there is not evidence that the population mean waiting time is different from 3.9 minutes.
It’s D 3,5 bc when you reflect over y-axis instead of -3 its 3 and 5 just stays 5
Hello there!
![4v - 1 = -5v - 10](https://tex.z-dn.net/?f=4v%20-%201%20%3D%20-5v%20-%2010)
Explanation:
↓↓↓↓↓↓↓↓↓↓↓
First you had to add by 1 from both sides of the equation.
![4v-1+1=-5v-10+1](https://tex.z-dn.net/?f=4v-1%2B1%3D-5v-10%2B1)
Simplify
![4v=-5v-9](https://tex.z-dn.net/?f=4v%3D-5v-9)
Then you add by 5v from both sides of the equation.
![4v+5v=-5v-9+5v](https://tex.z-dn.net/?f=4v%2B5v%3D-5v-9%2B5v)
Simplify
![9v=-9](https://tex.z-dn.net/?f=9v%3D-9)
Divide by 9 from both sides of the equation.
![\frac{9v}{9}=\frac{-9}{9}](https://tex.z-dn.net/?f=%5Cfrac%7B9v%7D%7B9%7D%3D%5Cfrac%7B-9%7D%7B9%7D)
Simplify it should be the correct answer.
![v=-1](https://tex.z-dn.net/?f=v%3D-1)
<em><u>Answer⇒⇒⇒⇒v=-1</u></em>
Hope this helps!
Thank you for posting your question at here on Brainly.
Have a great day!
-Charlie
So the first one, 1.2 >11/5. and the second one 1/3> 1/2