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Zigmanuir [339]
3 years ago
5

2.8-5.3n - 4n + 5 =plss help again Thank you!​

Mathematics
2 answers:
Mademuasel [1]3 years ago
5 0

Answer:

In simplest form: 7.8 - 9.3n

Step-by-step explanation:

Equation: 2.8 - 5.3n - 4n + 5

Combine like terms: 2.8 + 5 = 7.8

-5.3n + -4n = -9.3n

Put it all together: 7.8 - 9.3n

Hope this helps! :)

ra1l [238]3 years ago
4 0

Answer:

7.8-9.3n

Step-by-step explanation:

2.8-5.3n - 4n + 5

Combine like terms

2.8 +5 -5.3n -4n

7.8-9.3n

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Apple products have become a household name in America with 51% of all households owning at least one Apple product (CNN, March
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Answer:

(a) The events of "household with kids" and "household without kids" are mutually exclusive and exhaustive.

(b) The probability that a household is without kids is 0.317.

(c) The probability that a household is with kids and owns an Apple product is 0.417.

(d) The probability that a household is without kids and does not own an Apple product is 0.152.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = number of households with kids

<em>A</em> = number of households that own an Apple product.

It is provided that of the 1,200 households in a representative community, 820 are with kids.

The information provided is:

P(A|X)=0.61\\P(A|X^{c})=0.48\\P(X)=\frac{820}{1200}=0.683\\

(a)

Mutually exclusive events are those events that cannot occur together.

The event of "household with kids" cannot take place along with the event "household without kids".

Thus, the events of "household with kids" and "household without kids" are mutually exclusive.

Exhaustive events are those events of which at least one will definitely occur.

A household will either have kids or not have kids.

Thus, the events of "household with kids" and "household without kids" are exhaustive.

(b)

Compute the probability that a household is without kids as follows:

P(X^{c})=1-P(X)=1-0.683=0.317

Thus, the probability that a household is without kids is 0.317.

(c)

Compute the probability that a household is with kids and owns an Apple product as follows:

P(X\cap A)=P(A|X)P(X)=0.61\times0.683=0.41663\approx0.417

Thus, the probability that a household is with kids and owns an Apple product is 0.417.

(d)

Compute the probability that a household is without kids and does not own an Apple product as follows:

P(X^{c}\cap A)=P(A|X^{c})P(X^{c})=0.48\times0.317=0.15216\approx0.152

Thus, the probability that a household is without kids and does not own an Apple product is 0.152.

6 0
3 years ago
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