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2 years ago

Please help me with this

Mathematics

1 answer:

sergij07 [2.7K]2 years ago

3 0

Answer:

x=8

Step-by-step explanation:

The ratio of a side of PQRS to the corresponding side in TUVW. is 6:4=3:2.

The ratio of RS:VW=3:2. RS=12, and VW=x. 12:x=3:2. 3 multiplied by 4 is 12, so 2 multiplied by 4 is x. So, the makes x=8.

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1.64 written as fraction in simplest form

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2 years ago

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Answer:

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in the first third of the season the team played 18 games how many games will the team play during the whole season

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Carly placed 50 marbles in a bag. Fifteen of the marbles are pink, 8 marbles are black,

yuradex [85]

When you go into this problem, you want to figure out your marble ammount to 50 so in this case we will say C for color and 50 for the total ammount of marbles.

We know 15 are pink, 8 are black, 2 are green, 18 are clear, and 7 are striped

15P/50

8B/50

2G/50

18C/50

7S/50 for a total of 50 marbles

Now we use the chart to decide our awnsers

A. We know our propability of drawing a green and clear is 20/50 which if we simplify is a 2/5 ratio. If We put this in perspective 2/5 is rare and is unlikley to even.

B. We know a striped marble is 18/50 or 1.8/5 ratio which is mainly unlikely

C. We have 23/50 marbles that are black and pink, our propability is about 2.3/5 and gives us an even chance to get one of these

D. We know we have 33/50 marbles that are pink and clear and gives us a 3.3/5 chance of getting one of these and gives us an even to likely chance of getting one of these.

E. If we have a total of 17 marbles in these 3 colors, we have a 1.7/5 chance of getting one of these and is probably impossible to unlikey.

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1 year 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

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

$Z = \frac{670 - 660}{9}$

$Z = 1.11$

$Z = 1.11$ has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

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1 year ago