Each medal is different and is handed out depending on the outcome of the race/order of the racers, so we're counting the number of permutations. 4 medals are given to 4 runners, so there are
![4!=4\cdot3\cdot2\cdot1=24](https://tex.z-dn.net/?f=4%21%3D4%5Ccdot3%5Ccdot2%5Ccdot1%3D24)
possible ways to do it.
Answer:
6 titles and 16 acts were elaborated
Step-by-step explanation:
Let the number of titles be t and the number of acts be a
Since the total is 22 operations:
a + t = 22 •••••••(i)
Total amount from titles = 3500 * t = 3500t
Total amount from acts = 2000 * a = 2000a
2000a + 3500t = 53,000 ••••••(ii)
From i, a = 22-t
Insert this in ii
2000(22-t) + 3500t = 53,000
44,000 - 2000t + 3500t = 53,000
1,500t = 53000-44000
1500t = 9000
t = 9000/1500
t = 6
But a = 22 -
= 22-6 = 16
Answer:
130/l in where l is the length of the rectangle
Step-by-step explanation:
The formula for area of a rectangle is length by width where the width is the height of the rectangle
Given that area=130 in²
Lets assume the length of this rectangle is =l inches
Then ;
A=l*w
130 =l*w
130/l = w
130/l in = height
Answer:
1.345*10^7
Step-by-step explanation:
Here we need to rewrite 13,450,000 in scientific notation, which looks like x.xx*10^n.
To move the decimal point in 13,450,000 to its proper place between the 3 and the 4, a move of 7 digits, we need to multiply 1.345 by 10^7.
13,450,000 in scientific notation is 1.345*10^7.
Answer: (0.8115, 0.8645)
Step-by-step explanation:
Let p be the proportion of people who leave one space after a period.
Given: Sample size : n= 525
Number of people responded that they leave one space. =440
i.e. sample proportion: ![\hat{p}=\dfrac{440}{525}\approx0.838](https://tex.z-dn.net/?f=%5Chat%7Bp%7D%3D%5Cdfrac%7B440%7D%7B525%7D%5Capprox0.838)
z-score for 90% confidence level : 1.645
Formula to find the confidence interval :
![\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}](https://tex.z-dn.net/?f=%5Chat%7Bp%7D%5Cpm%20z%5E%2A%5Csqrt%7B%5Cdfrac%7B%5Chat%7Bp%7D%281-%5Chat%7Bp%7D%29%7D%7Bn%7D%7D)
![0.838\pm (1.645)\sqrt{\dfrac{0.838(1-0.838)}{525}}\\\\=0.838\pm (1.645)\sqrt{0.00025858285}\\\\=0.838\pm (1.645)(0.01608)\\\\= 0.838\pm0.0265\\\\=(0.838-0.0265,\ 0.838+0.0265)\\\\=(0.8115,\ 0.8645)](https://tex.z-dn.net/?f=0.838%5Cpm%20%281.645%29%5Csqrt%7B%5Cdfrac%7B0.838%281-0.838%29%7D%7B525%7D%7D%5C%5C%5C%5C%3D0.838%5Cpm%20%281.645%29%5Csqrt%7B0.00025858285%7D%5C%5C%5C%5C%3D0.838%5Cpm%20%281.645%29%280.01608%29%5C%5C%5C%5C%3D%200.838%5Cpm0.0265%5C%5C%5C%5C%3D%280.838-0.0265%2C%5C%200.838%2B0.0265%29%5C%5C%5C%5C%3D%280.8115%2C%5C%200.8645%29)
Hence, a 90% confidence interval for the proportion of people who leave one space after a period: (0.8115, 0.8645)