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weqwewe [10]
1 year ago
9

Question 15

Mathematics
1 answer:
larisa86 [58]1 year ago
4 0

Answer:

The correct option is (C).

Step-by-step explanation:

According to the Central Limit Theorem if we have n unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 3.5

<em>σ</em> = 1.7078

<em>n</em> = 400 = number of times the experiment is repeated.

As the sample size is quite large, i.e. <em>n</em> = 400 > 30 the central limit theorem can be used to approximate the sampling distribution of the sample mean.

The mean of the distribution of sample means is:

\mu_{\bar x}=\mu=3.5

The standard deviation of the distribution of sample mean is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{1.7078}{\sqrt{400}}=0.0854

The distribution of the sample mean is:

\bar X\sim N(\mu = 3.5,\ \sigma=0.0854).

Thus, the correct option is (C).

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Approximately 68% of a normal distribution lies within one standard deviation of the mean, so this corresponds to students with scores between (57.5 - 6.5, 57.5 + 6.5) = (51, 64)
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Jenna dances for 3 hours on sunday, 2 hours on Monday and Tuesday, 1 hour on THursday,1.5 hours oon friday, and 2 hours on Satur
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Answer: The average number of hours she danced per day is 1.9 hours (rounded to the nearest tenth)

Step-by-step explanation: We start by calculating how many hours she danced all together which can be derived as follows;

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The number of days she danced which is the observed data is 6 days (she did  not dance at all on Wednesday).

The average (or mean) hours she danced each day can be calculated as

Average = ∑x ÷ x

Where ∑x is the summation of all data and x is number of observed data

Average = (3+2+2+1+1.5+2) ÷ 6

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Approximately, average hours danced is 1.9 hours (to the nearest tenth)

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1 year ago
The probability of flowering plants bearing flowers in January is given the table. If a plant has flowered in January, what is t
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Answer:

Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

Step-by-step explanation:

We know that:

Probability =Total favourable outcomes/ Total possible outcomes

Probability of flowering plants bearing flowers in January is given in the table. It means that out of 100 peony  70 peony flowers in January.

There are five kinds of flowers.

We have to select peony and find it's Probability.

As all flowers are independent, their flowering phenomenon does not depend on each other.

So,Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

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Answer:

a)one lens: 111,212.66

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b)y=111,200+12.66x

slope: 12.66

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d) 20.07

e) 19.2

f)decreases

Step-by-step explanation:

a) Cost of producing one lens

fixed expenses+variable expenses=111,200+12.66=111,212.66

Cost of producing 15,000 lenses

fixed expenses+variable expenses=111,200+12.66(15,000)=301,100

b)Expense function:

let y be the expense cost

let x be the number of lenses produced

y=111,200+12.66x

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The function is  a line equation: y=mx+b where m is the slope.

In this case b=111,200 and m=12.66. So the slope is 12.66

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d) The cost of producing 15000 lenses is:

111,200+12.66(15,000)=301,100

If DiMonte produces 15000 lenses, the average cost per lens is:

301,100/15000=20.073≈20.07

e)The cost of producing 17000 lenses is:

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If DiMonte produces 17000 lenses, the average cost per lens is:

326,420/17,000=19.2

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Step-by-step explanation:

<u>As we have to use area model :</u>

So, we

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