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Semmy [17]
3 years ago
11

(a)A train completed a journey of 850 kilometres with an average speed of 80 km/hr.

Mathematics
1 answer:
Yanka [14]3 years ago
6 0

Answer:

a. 850/85 = 10.625 hr or 10 hr 37 min 30 sec

b. 6:08 AM

C. Average time is (10.625 + 10.8)/2 = 10.7125

or 10 hour 42 min 45 sec

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Multiply (2-5i)(2+5i)
Ksenya-84 [330]

Answer:

29

Step-by-step explanation:

4 - 25i^2

4 - 25x(-1)

4 + 25 = 29

the ^2 means exponent 2

7 0
3 years ago
SOMEBODY PLEASE HELP ME it’s due at 11:59
lapo4ka [179]

Answer:

c

Step-by-step explanation:

given: miles traveled and cost

to find: which one is correct

solution:

miles traveled =25

And cost =80

80/25 = 3.2

similarity all $3.20 cost increase

7 0
3 years ago
Please help me complete the blanks!!
DaniilM [7]

Step-by-step explanation:

If half of the tuna sandwiches were on white bread then 21 of them would be white, meaning the other half would be on brown bread.

Then if 25% of ham sandwiches were on brown bread then 25% of 32 would be 8, so the final 75% would be on white bread, which would be 24.

Finally, with 20 more white bread sandwiches sold than brown bread it couldn’t be 80 and 20, it would be 40 brown bread in total and 60 white bread in total adding up to 100, as it says white sells 20 more than brown not only 20.

To finish this question you would add up the tuna and ham for white (21+24=45) then minus that from 60 (60-45=15) so there were 15 cheese sandwiches on white bread. And then for brown (21+8=29) then minus that from 40 (40-29=11) so therefore there are 11 cheese sandwiches on brown bread.

Answer:

Brown:

Tuna-21

Cheese-11

Ham-8

Total-40

White:

Tuna-21

Cheese-15

Ham-24

Total-60

Hope that helps!

7 0
4 years ago
Majesty Video Production Inc. wants the mean length of its advertisements to be 26 seconds. Assume the distribution of ad length
Paladinen [302]

Answer:

a) By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.

b) s = 0.44

c) 0.84% of the sample means will be greater than 27.05 seconds

d) 98.46% of the sample means will be greater than 25.05 seconds

e) 97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds

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 normally distributed samples are 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(also called standard error) 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.

In this problem, we have that:

\mu = 26, \sigma = 2, n = 21, s = \frac{2}{\sqrt{21}} = 0.44

a. What can we say about the shape of the distribution of the sample mean time?

By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.

b. What is the standard error of the mean time? (Round your answer to 2 decimal places)

s = \frac{2}{\sqrt{21}} = 0.44

c. What percent of the sample means will be greater than 27.05 seconds?

This is 1 subtracted by the pvalue of Z when X = 27.05. So

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

By the Central Limit Theorem

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

Z = \frac{27.05 - 26}{0.44}

Z = 2.39

Z = 2.39 has a pvalue of 0.9916

1 - 0.9916 = 0.0084

0.84% of the sample means will be greater than 27.05 seconds

d. What percent of the sample means will be greater than 25.05 seconds?

This is 1 subtracted by the pvalue of Z when X = 25.05. So

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

Z = \frac{25.05 - 26}{0.44}

Z = -2.16

Z = -2.16 has a pvalue of 0.0154

1 - 0.0154 = 0.9846

98.46% of the sample means will be greater than 25.05 seconds

e. What percent of the sample means will be greater than 25.05 but less than 27.05 seconds?"

This is the pvalue of Z when X = 27.05 subtracted by the pvalue of Z when X = 25.05.

X = 27.05

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

Z = \frac{27.05 - 26}{0.44}

Z = 2.39

Z = 2.39 has a pvalue of 0.9916

X = 25.05

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

Z = \frac{25.05 - 26}{0.44}

Z = -2.16

Z = -2.16 has a pvalue of 0.0154

0.9916 - 0.0154 = 0.9762

97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds

8 0
3 years ago
Help me with the problem plz 6p6
satela [25.4K]

i need more information to answer the question

5 0
3 years ago
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