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Lilit [14]
3 years ago
5

Help me ..................

Mathematics
1 answer:
Law Incorporation [45]3 years ago
6 0
What is the question for the thing s
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The temperature in a city was recorded over a ten-hour period. The graph below shows the relationship between the temperature an
harkovskaia [24]

Answer:

A

Step-by-step explanation:

From the graph, we can see that the <em>temperature increased initially</em>, then <em>started to decrease</em>. <u>All of these NOT in a constant rate, though</u>.

Roughly, the temperature was increasing from start till around 5.5 hours. Then on the temperature decreased.

So we can eliminate B, C, and D. That leaves us with correct answer as A.

5 0
3 years ago
Concerns about the climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g. from r
natta225 [31]

Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

(a) If the true mean is 0.9370 with a standard deviation of 0.0090 within what interval will 99% of the sample means fail?

Samples of 34 means that n = 34

We have that \mu = 0.937, \sigma = 0.009

By the Central Limit Theorem, s = \frac{0.009}{\sqrt{34}} = 0.0015

Within what interval will 99% of the sample means fail?

Between the (100-99)/2 = 0.5th percentile and the (100+99)/2 = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

-2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = -2.575*0.0015

X = 0.933

99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

Z = \frac{X - \mu}{s}

2.575 = \frac{X - 0.937}{0.0015}

X - 0.937 = 2.575*0.0015

X = 0.941

99% of the sample means will fall between 0.933 and 0.941.

(b) If the true mean 0.9370 with a standard deviation of 0.0090, what is the sampling distribution of ¯X?

By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

6 0
3 years ago
Find the value of the expression 12 ÷ v for v = 2
Zarrin [17]

Answer:

Step-by-step explanation:

12 / v....when v = 2

12 / 2 = 6 <==

3 0
3 years ago
Read 2 more answers
Which of the following options have the same value as 2% of 90? Choose 2 answers: А 0.2.90 B 0.02.90 200.90 2.90 E 2 . 90 100​
Wewaii [24]

Answer:

0.02.90 and 2.90

Step-by-step explanation:

2% of 90 equal to both 0.02.90 because are worth same.

7 0
2 years ago
6
8090 [49]

Answer:

James bought 11 good tickets and 5 bad tickets.

Step-by-step explanation:

Given that:

Cost of each good ticket = $8

Cost of each bad ticket = $5

Total amount spent = $113

Total tickets bought = 16

Let,

x be the number of good tickets bought

y be the number of bad tickets bought

x+y=16     Eqn 1

8x+5y=113    Eqn 2

Multiplying Eqn 1 by 5

5(x+y=16)

5x+5y=80    Eqn 3

Subtracting Eqn 3 from Eqn 2

(8x+5y)-(5x+5y)=113-80

8x+5y-5x-5y=33

3x=33

Dividing both sides by 3

\frac{3x}{3}=\frac{33}{3}\\x=11

Putting x=11 in Eqn 1

11+y=16

y=16-11

y=5

Hence,

James bought 11 good tickets and 5 bad tickets.

3 0
2 years ago
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