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TiliK225 [7]
3 years ago
14

Match the expression with its name. 6x4 – 10x + 9 A) not a polynomial B) Cubic binomial C) Fourth -degree trinomial D) quadratic

monomial
Mathematics
1 answer:
Alla [95]3 years ago
6 0

Answer:

Fourth - Degree trinomial

Step-by-step explanation:

6x^4 - 10x + 9

It is a polynomial because there is no negative exponent or fractional exponent

We have highest exponent 4. so it is not cubic. For cubic polynomial highest exponent should be 3

Given expression have highest exponent 4 so its fourth degree. We have three terms 6x^4 , -10x  and 9. so we say its trinomial.

Hence we name it as Fourth - Degree trinomial

We have 3 terms so this is not a monomial


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Items are inspected for flaws by two quality inspectors. Both inspectors inspect every item and the probability that an item has
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a)0.976

b)0.00926

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d)0.35

Step-by-step explanation:

Let X_i be an item passed by inspector i

Let Y be the event that there is a fault in an item

The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1

If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e.P(\bar{X_1}|Y)=0.92

So, P(X_1|Y)=1-0.92=0.08

If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e.P(\bar{X_2}|Y)=0.7

So,P(X_2|Y)=1-0.7=0.3

If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e. P(X_1|\bar{Y}) = 0.95

So, P(\bar{X_1}|\bar{Y}) = 1-0.95=0.05

If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e. P(X_2|\bar{Y}) = 0.8

So, P(\bar{X_2}|\bar{Y}) = 1-0.8=0.2

a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )

P(found by atleast one inspector | It has flaw )=1-P(X_1|Y) P(X_2|Y)

P(found by atleast one inspector | It has flaw )=1-0.08 \times 0.3

P(found by atleast one inspector | It has flaw )=0.976

Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976

b)P(Y|X_1)=\frac{P(X_1|Y) P(Y)}{P(X_1|Y) P(Y)+P(X_1|\bar{Y}) P(\bar{Y})}

P(Y|X_1)=\frac{0.08 \times 0.1}{0.08 \times 0.1+0.95 \times 0.9}=0.00926

C)P( two inspectors draw different conclusions on the same item)=P(X_1 \cap \bar{X_2} \cap Y)+P(\bar{X_1} \cap X_2 \cap Y)+P(X_1 \cap \bar{X_2} \cap \bar{Y})+P(\bar{X_1} \cap X_2 \cap \bar{Y})

P( two inspectors draw different conclusions on the same item)=0.2402

D)

P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2)}\\P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2 \cap Y)+P(X_1 \cap X_2 \cap \bar{Y})}\\P(Y|(X_1 \cap X_2))=0.35

3 0
3 years ago
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