Answer:
We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.
Step-by-step explanation:
We have a sample of executives, of size n=160, and the proportion that prefer trucks is 26%.
We have to calculate a 95% confidence interval for the proportion.
The sample proportion is p=0.26.
The standard error of the proportion is:
The critical z-value for a 95% confidence interval is z=1.96.
The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 95% confidence interval for the population proportion is (0.192, 0.328).
We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.
24 students forget their pencil out of 120 students.
<h3>
<u>Answer:</u></h3>

<h3>
<u>Step-by-step explanation:</u></h3>
A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.
The inequality given to us is :-

Let's plot a graph to see its interval . Graph attached in attachment .
Now we can see that the Interval notation of would be ,
![\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5Corange%20%5Ctt%20%5Cpurple%7B%5Cleadsto%7Dy%20%5Cin%20%5B-2%2C-1%5D%20%7D%7D)
<h3>
<u>Hence</u><u> the</u><u> </u><u>standa</u><u>rd</u><u> </u><u>form</u><u> </u><u>of</u><u> </u><u>inequa</u><u>lity</u><u> </u><u>is</u><u> </u><u>y²</u><u>+</u><u>3y</u><u> </u><u>+</u><u>2</u><u> </u><u>≤</u><u> </u><u>0</u><u> </u><u>and</u><u> </u><u>the </u><u>Solution</u><u> </u><u>set</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ineq</u><u>uality</u><u> </u><u>is</u><u> </u><u>[</u><u> </u><u>-</u><u>2</u><u> </u><u>,</u><u> </u><u>-</u><u>1</u><u> </u><u>]</u><u> </u><u>.</u></h3>
Answer:
Number of families will have both radio and television = 44
Step-by-step explanation:
Given:
Number of family have radio = 794
Number of family have TV = 187
Number of family haven't both = 63
Find:
Number of families will have both radio and television
Computation:
Number of families will have Tv or radio = 1,000 - 63
Number of families will have Tv or radio = 937
Number of family have one of them = 794 + 187
Number of family have one of them = 981
Number of families will have both radio and television = 981 - 937
Number of families will have both radio and television = 44
Hi this is the answer and explaination
I'm sorry if it's incorrect
hope's it's help