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1 year ago

A traveler averaged 52 miles per hour. How many hours were spent on a 182 mile trip?

Mathematics

1 answer:

m_a_m_a [10]1 year ago

3 0

The number of hours spent on the trip is 6 hours

<h3>How to calculate the number of hours spent during the trip ?</h3>

The traveller averaged 52 miles to be spent each hour

He plans on going for 182 miles

The number of hours he will spend during the 183 mile trip can be calculated as follows

= 182/52

= 6

Hence 6 hours will be spent on a 182 mile trip

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John is about to roll a six sided, fair number cube 60

sukhopar [10]

Answer:

No it is not a good prediction. Theoretical probability says that the answer would be 30 of the 60 (30/60) rolls would be even numbers.

Step-by-step explanation:

The probability of rolling an even number on a fair six sided die is 3/6 or 1/2 when simplified.

if john rolls the die 60 times, multiply the theoretical probability (1/2) by the number of rolls

(1 ÷ 2) x 60 = 30

∴ theoretical probability says that john would roll an even number 30 out of the 60 rolls

I hope this was helpful :-)

7 0

2 years ago

Read 2 more answers

The options are<br> 40<br> 80<br> 120<br> 140

Nataly [62]

According to given figure,

$\qquad \sf \dashrightarrow \: \angle ABC + 60° = 180°$

$\qquad \sf \dashrightarrow \: \angle ABC = 180 \degree - 60 \degree$

[ By linear pair ]

$\qquad \sf \dashrightarrow \: \angle ABC = 120 \degree$

now, we can see that :

$\qquad \sf \dashrightarrow \: \angle ABC + \angle BAC = \angle 1$

[ By Exterior angle property of Triangle ]

$\qquad \sf \dashrightarrow \: \angle 1 = 20 \degree + 120 \degree$

$\qquad \sf \dashrightarrow \: \angle 1 = 140 \degree$

4 0

2 years ago

The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100

gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

The mean is of 660, hence $\mu = 660$ .

The standard deviation is of 90, hence $\sigma = 90$ .

A sample of 100 is taken, hence $n = 100, s = \frac{90}{\sqrt{100}} = 9$ .

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

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

By the Central Limit Theorem

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