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3 years ago

Two different functions are represented.

Mathematics

2 answers:

Crank3 years ago

7 0

Answer:

Neither function A nor function B has an x-intercept.

Step-by-step explanation:

Function A has a y-intercept at (0, 5).

Function B is defined for negative numbers as well as positive numbers, but not for x=0.

These observations eliminate all answer choices but the first one.

ziro4ka [17]3 years ago

5 0

Answer:

The answer is A.

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Which fraction has a terminating decimal as it's the small expansion?

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B. 1/5

When turned into a decimal, 1/5 is 0.2 which is a decimal that terminates.

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4 years ago

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Which decimal is equivalent to (5 × 10) + (2 × 1/100)? a 50.002 b 5.02 c 50.20 d 50.02

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Answer:

d 50.02

Step-by-step explanation:

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What is the product in lowest terms?

brilliants [131]

C
because you would take 3/4 and multiply it by 3/3 to get the denominator to 12
and you would multiply 1/3 by 4/4 again to get the denominator to 12.
You are getting the denominator to 12 because 12 is 3 and 4s Most Common multiple!
Now that you have multiplied them you should have 9/12 and 4/12
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4 years ago

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The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

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3 years ago

Which of the following is a perfect square?

o-na [289]

Greeting's!

<span>B. 279
_____
</span>

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3 years ago