I believe it’s 5,832,000
5^3 (5 times 5= 25 times 5= 125)^6(750 times 6= 4,500 etc)
92% is 92 out of 100
therefore simply put 92 over 100 and simplify it.
Meaning divide top and bottom into 2. Then you get your answer, if needed simplify again.
Answer:
![r = \frac{3}{4\pi}V](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B3%7D%7B4%5Cpi%7DV)
Step-by-step explanation:
First, the formula for the volume of a sphere is:
![V = \frac{4}{3}\pi r^3](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3)
![\frac{3}{4}V = \pi r^3](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7DV%20%3D%20%5Cpi%20r%5E3)
![\frac{3}{4}V\frac{1}{\pi} = r^3](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7DV%5Cfrac%7B1%7D%7B%5Cpi%7D%20%3D%20r%5E3)
![r^3 = \frac{3}{4\pi}V](https://tex.z-dn.net/?f=r%5E3%20%3D%20%5Cfrac%7B3%7D%7B4%5Cpi%7DV)
![r = \sqrt[\leftroot{-2}\uproot{2}3]{\frac{3}{4\pi}V}](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B%5Cleftroot%7B-2%7D%5Cuproot%7B2%7D3%5D%7B%5Cfrac%7B3%7D%7B4%5Cpi%7DV%7D)
<u><em>If there is any steps you are unsure of, feel free to ask in the comments.</em></u>
Answer:
81.85%
Step-by-step explanation:
Given :
The average summer temperature in Anchorage is 69°F.
The daily temperature is normally distributed with a standard deviation of 7°F .
To Find:What percentage of the time would the temperature be between 55°F and 76°F?
Solution:
Mean = ![\mu = 69](https://tex.z-dn.net/?f=%5Cmu%20%3D%2069)
Standard deviation = ![\sigma = 7](https://tex.z-dn.net/?f=%5Csigma%20%3D%207)
Formula : ![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
Now At x = 55
![z=\frac{55-69}{7}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B55-69%7D%7B7%7D)
![z=-2](https://tex.z-dn.net/?f=z%3D-2)
At x = 76
![z=\frac{76-69}{7}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B76-69%7D%7B7%7D)
![z=1](https://tex.z-dn.net/?f=z%3D1)
Now to find P(55<z<76)
P(2<z<-1)=P(z<2)-P(z>-1)
Using z table :
P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185
Now percentage of the time would the temperature be between 55°F and 76°F = ![0.8185 \times 100 = 81.85\%](https://tex.z-dn.net/?f=0.8185%20%5Ctimes%20100%20%3D%2081.85%5C%25)
Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.
Answer:
0.46
Step-by-step explanation:bark brainliest plz