All categories

3 years ago

When both parties discuss the terms,of a contract they are ___.

Mathematics

2 answers:

QveST [7]3 years ago

6 0

I believe the answer is C.

horsena [70]3 years ago

6 0

D or a

try those that might just work

You might be interested in

16-2t=3/2t+9 solve for t

In-s [12.5K]

answer: t=2

steps:

16-2t = 3/2t+9

7 = 3/2t + 2t

2 is 4/2

7 = 3/2t + 4/2t

7 = 7/2t

7/2t = 7

(2/7)7/2t=7(2/7)

t=7(2/7)

t=2

7 0

1 year ago

6) Supplementary Exercise 5.51

tresset_1 [31]

Answer:

$P(X \le 4) = 0.7373$

$P(x \le 15) = 0.0173$

$P(x > 20) = 0.4207$

$P(20\ge x \le 24)= 0.6129$

$P(x = 24) = 0.0236$

$P(x = 15) = 1.18\%$

Step-by-step explanation:

Given

$p = 80\% = 0.8$

The question illustrates binomial distribution and will be solved using:

$P(X = x) = ^nC_xp^x(1 - p)^{n-x}$

Solving (a):

Given

$n =5$

Required

$P(X\ge 4)$

This is calculated using

$P(X \le 4) = P(x = 4) +P(x=5)$

This gives:

$P(X \le 4) = ^5C_4 * (0.8)^4*(1 - 0.8)^{5-4} + ^5C_5*0.8^5*(1 - 0.8)^{5-5}$

$P(X \le 4) = 5 * (0.8)^4*(0.2)^1 + 1*0.8^5*(0.2)^0$

$P(X \le 4) = 0.4096 + 0.32768$

$P(X \le 4) = 0.73728$

$P(X \le 4) = 0.7373$ --- approximated

Solving (b):

Given

$n =25$

Required

$P(X\le 15)$

This is calculated as:

$P(X\le 15) = 1 - P(x>15)$ --- Complement rule

$P(x>15) = P(x=16) + P(x=17) + P(x =18) + P(x = 19) + P(x = 20) + P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)$

$P(x > 15) = {25}^C_{16} * p^{16}*(1-p)^{25-16} +{25}^C_{17} * p^{17}*(1-p)^{25-17} +{25}^C_{18} * p^{18}*(1-p)^{25-18} +{25}^C_{19} * p^{19}*(1-p)^{25-19} +{25}^C_{20} * p^{20}*(1-p)^{25-20} +{25}^C_{21} * p^{21}*(1-p)^{25-21} +{25}^C_{22} * p^{22}*(1-p)^{25-22} +{25}^C_{23} * p^{23}*(1-p)^{25-23} +{25}^C_{24} * p^{24}*(1-p)^{25-24} +{25}^C_{25} * p^{25}*(1-p)^{25-25}$

$P(x > 15) = 2042975 * 0.8^{16}*0.2^9 +1081575* 0.8^{17}*0.2^8 +480700 * 0.8^{18}*0.2^7 +177100 * 0.8^{19}*0.2^6 +53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0$