Answer: 0.8461
Step-by-step explanation:
Let p be the proportion of residents are against the increase of taxes to fund alternatives to drug addiction treatment.
As per given , we have p=0.40
A random sample is taken with size : n= 400
Expecting sample proportion : 
Now , the probability that more than 150 of the residents surveyed will be against increasing taxes if a random sample of 400 residents are surveyed will be :
![=P(z>\dfrac{0.375-0.40}{\sqrt{\dfrac{0.40(0.60)}{400}}})\ \ [\because\ z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}]](https://tex.z-dn.net/?f=%3DP%28z%3E%5Cdfrac%7B0.375-0.40%7D%7B%5Csqrt%7B%5Cdfrac%7B0.40%280.60%29%7D%7B400%7D%7D%7D%29%5C%20%5C%20%5B%5Cbecause%5C%20z%3D%5Cdfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%5D)
[By z-table]
Hence, the approximate probability that more than 150 of the residents surveyed will be against increasing taxes if a random sample of 400 residents are surveyed is 0.8461 .
Answer:
1/3 =0.33
Step-by-step explanation:
<em>ju</em><em>st</em><em> </em><em>subst</em><em>itute</em><em> </em><em>the</em><em> </em><em>va</em><em>lues</em><em> </em><em>of</em><em> </em><em>x</em><em> </em><em>and </em><em>y</em><em> </em><em>int</em><em>o</em><em> </em><em>th</em><em>e</em><em> </em><em>exp</em><em>ression</em><em> </em><em>to</em><em> </em><em>get</em>
<em>4</em><em>(</em><em>2</em><em>)</em><em>+</em><em>1</em><em> </em><em>/</em><em>3</em><em>(</em><em>3</em><em>)</em><em>^</em><em>2</em>
<em>=</em><em>8</em><em>+</em><em>1</em><em> </em><em>/</em><em>3</em><em>×</em><em>9</em>
<em>=</em><em>9</em><em>/</em><em>2</em><em>7</em>
<em>=</em><em>1</em><em>/</em><em>3</em>
<em>=</em>0.33
The answer is f^-1(x)=8/3+x/3
