the yearly increase of x% assumes is compounding yearly, so let's use that.

![95000=80000\left(1+\frac{~~ \frac{r}{100}~~}{1}\right)^{1\cdot 5}\implies \cfrac{95000}{80000}=\left( 1+\cfrac{r}{100} \right)^5 \\\\\\ \cfrac{19}{16}=\left( 1+\cfrac{r}{100} \right)^5\implies \sqrt[5]{\cfrac{19}{16}}=1+\cfrac{r}{100}\implies \sqrt[5]{\cfrac{19}{16}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[5]{\cfrac{19}{16}}=100+r\implies 100\sqrt[5]{\cfrac{19}{16}}-100=r\implies 3.5\approx r](https://tex.z-dn.net/?f=95000%3D80000%5Cleft%281%2B%5Cfrac%7B~~%20%5Cfrac%7Br%7D%7B100%7D~~%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%205%7D%5Cimplies%20%5Ccfrac%7B95000%7D%7B80000%7D%3D%5Cleft%28%201%2B%5Ccfrac%7Br%7D%7B100%7D%20%5Cright%29%5E5%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B19%7D%7B16%7D%3D%5Cleft%28%201%2B%5Ccfrac%7Br%7D%7B100%7D%20%5Cright%29%5E5%5Cimplies%20%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D%3D1%2B%5Ccfrac%7Br%7D%7B100%7D%5Cimplies%20%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D%3D%5Ccfrac%7B100%2Br%7D%7B100%7D%20%5C%5C%5C%5C%5C%5C%20100%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D%3D100%2Br%5Cimplies%20100%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D-100%3Dr%5Cimplies%203.5%5Capprox%20r)
Answer:
9+5x=t
Step-by-step explanation:
Answer:
The probability that the mean test score is greater than 290
P(X⁻ > 290 ) = 0.0217
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Mean of the Population (μ) = 281
Standard deviation of the Population = 34.4
Let 'X' be a random variable in Normal distribution
Given X = 290

<u><em>Step(ii):-</em></u>
<em> The probability that the mean test score is greater than 290</em>
P(X⁻ > 290 ) = P( Z > 2.027)
= 0.5 - A ( 2.027)
= 0.5 - 0.4783
= 0.0217
The probability that the mean test score is greater than 290
P(X⁻ > 290 ) = 0.0217
Answer:
The answer to this joke is a clockmaker.
Step-by-step explanation:
The math is just finding volume. You are asking for just the answer correct?
Answer:
Option C
Step-by-step explanation:
To find the corresponding graph substitute values of
.
The graph is:
. Substitute
.
We get:
. In the graphs only in Option C,
when
.