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3 years ago

What is the amswer to -4k-2(k-3)=-6k+8

Mathematics

1 answer:

SashulF [63]3 years ago

4 0

Answer:

k = 1/2

Step-by-step explanation:

-4k-2(k)-2(-3)=-6k+8

-4k-2k+6=-6k+8

-2k+6=-6k+8

4k+6=8

4k=2

4k/4= 2/4

k=1/2

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Which equation represents exponential decay?

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3 years ago

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

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