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3 years ago

PLZ HELP URGENT GET THE BRAINLIEST AND MORE POINTS

Mathematics

1 answer:

lord [1]3 years ago

7 0

Answer:

The 27th term is 22

Step-by-step explanation:

Let the first term is a and common difference is d.

The nth term is:

aₙ = a + (n - 1)d

We have:

a₆ = 15
a₉ = 16

The difference of these terms is:

(a + 8d) - (a + 5d) = 16 - 15
3d = 1
d = 1/3

Then the first term is:

a + 5*1/3 = 15
a = 15 - 5/3 = 13 1/3

The nth term equation is:

aₙ = 13 1/3 + 1/3(n - 1) = 1/3n + 13

If the nth term is 22, find n:

1/3n + 13 = 22
1/3n = 22 - 13
1/3n = 9
n = 9*3
n = 27

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Five companies (A, B, C, D, and E) that make elec- trical relays compete each year to be the sole sup- plier of relays to a majo

NNADVOKAT [17]

Answer:

$P(a | e') = 0.22$

$P(b | e') = 0.28$

$P(c | e') = 0.33$

$P(a | e' , d' , b') = 0.57$

Step-by-step explanation:

From the question we are told that

The probabilities are

Supplier chosen A B C

Probability P(a) = 0.20 P(b) = 0.25 P(c) = 0.15

D E

P(d) = 0.30 P(e) = 0.10

Generally the new probability of companies A being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(a | e') = \frac{P (a \ and \ e')}{P(e')}$

$P(a | e') = \frac{P (a)}{P(e')}$

$P(a | e') = \frac{P (a)}{1- P(e)}$

=> $P(a | e') = \frac{ 0.20}{1- 0.10}$

=> $P(a | e') = 0.22$

Generally the new probability of companies B being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(b | e') = \frac{P (b \ and \ e')}{P(e')}$

$P(b | e') = \frac{P (b)}{P(e')}$

$P(b | e') = \frac{P (b)}{1- P(e)}$

=> $P(b | e') = \frac{ 0.25}{1- 0.10}$

=> $P(b | e') = 0.28$

Generally the new probability of companies C being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(c | e') = \frac{P (c \ and \ e')}{P(e')}$

$P(c | e') = \frac{P (c)}{P(e')}$

$P(c | e') = \frac{P (c)}{1- P(e)}$

=> $P(c | e') = \frac{ 0.15}{1- 0.10}$

=> $P(c | e') = 0.17$

Generally the new probability of companies D being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(d | e') = \frac{P (d \ and \ e')}{P(e')}$

$P(d | e') = \frac{P (d)}{P(e')}$

$P(d | e') = \frac{P (d)}{1- P(e)}$

=> $P(d | e') = \frac{ 0.30}{1- 0.10}$

=> $P(c | e') = 0.33$

Generally the probability that B, D , E are not chosen this year is mathematically represented as

$P(N) = 1 - [P(e) +P(b) + P(d) ]$

=> $P(N) = 1 - [0.10 +0.25 +0.30 ]$

=> $P(N) = 0.35$

Generally the probability that A is chosen given that E , D , B are rejected this year is mathematically represented as

$P(a | e' , d' , b') = \frac{P(a)}{P(N)}$

=> $P(a | e' , d' , b') = \frac{0.20 }{0.35 }$

=> $P(a | e' , d' , b') = 0.57$

5 0

3 years ago

A new truck that sells for 29,000 depreciates 12% each year. What is the decay factor b?

Katena32 [7]

$\bf y=a(b)^t\qquad or\qquad A=I(1-r)^t\qquad \textit{let's check}\\\\
-------------------------------\\\\
\qquad \textit{Amount for Exponential Decay}\\\\
A=I(1 - r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &29000\\
r=rate\to 12\%\to \frac{12}{100}\to &0.12\\
t=\textit{elapsed time}\\
\end{cases}
\\\\\\
A=29000(1-0.12)^t\implies 
\begin{array}{llll}
A=&29000(0.88)^t\\
&\uparrow \qquad\quad \uparrow \\
&a\qquad\quad b
\end{array}$