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vitfil [10]
3 years ago
8

On Saturday Marianna practice piano for 3/4 an hour,watched tv for 2 1/2 hours, and cooked dinner for 1/4 hour m.hiw long did Ma

rianna do all above activities in all

Mathematics
2 answers:
Kobotan [32]3 years ago
6 0
I know the answer my math teacher said it would be 3 1/2 hours
katovenus [111]3 years ago
5 0
2 1/2= 2 2/4
3/4 + 1/4 =4/4 or 1
convert 2 2/4 into an improper fraction: 10/4
10/4+4/4=14/4
14/4 simplifies into 3 2/4 hours
3 and 2/4 simplifies into 3 and 1/2
Mariana spent 3 and 1/2 hours doing all the activities
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Evaluate each numerical expression. 3^4 - (-4)^2
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<u>65.</u>

Step-by-step explanation:

1. Write the expression.

3^{4} -(-4)^{2}

2. Expand the expression.

3x3x3x3 -[(-4)x(-4)]

3. Simplify.

81 -[16]\\\\65.

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Which statement is true about this argument?
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In a certain population of mussels (Mytilus edulis), 80% of the individuals are infected with an intestinal parasite. A marine b
kirza4 [7]

Answer:

The probability that 85% or more of the sampled mussels will be infected is 0.1057.

Step-by-step explanation:

Let <em>X</em> = number of mussels infected with an intestinal parasite.

The probability that a random selected mussel is infected with an intestinal parasite is, <em>p</em> = 0.80.

A random sample of <em>n</em> = 100 mussels from the population are examined by a marine biologist.

The random variable <em>X</em> follows a Binomial distribution with parameters n = 100 and p = 0.80.

But the sample selected is too large, i.e. <em>n</em> = 100 > 30.

So a Normal approximation to binomial can be applied to approximate the distribution of \hat p<em>, </em>the sample proportion of mussels infected with an intestinal parasite, if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

n × p = 100 × 0.80 = 80 > 10

n × (1 - p) = 100 × (1 - 0.80) = 20 > 10

Thus, a Normal approximation to binomial can be applied.

So,  the distribution of \hat p is:

\hat p\sim N(p, \frac{p(1-p)}{n}  )

Compute the probability that 85% or more of the sampled mussels will be infected as follows:

Apply continuity correction:

P(\hat p\geq 0.85)=P(\hat p>0.85+0.50)

                   =P(\hat p>0.90)\\

                   =P(\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}>\frac{0.85-0.80}{\sqrt{\frac{0.80(1-0.80)}{100}}})

                   =P(Z>1.25)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that 85% or more of the sampled mussels will be infected is 0.1057.

5 0
3 years ago
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