Can someone help i will give brainilest answer if correct !

Mathematics

2 answers:

Nikolay [14]3 years ago

7 0

The Answer is A i am pretty sure

ValentinkaMS [17]3 years ago

6 0

I think it would be C. Since if you are looking at a number line and you are going up on the line, you would be going into positive numbers. Not 100% but I'm pretty sure that's what it is. :)

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What is the next term of AP : √2, √8, √18, .......?

solmaris [256]

Answer:

$\sqrt {32}$

Step-by-step explanation:

Next term of the given AP can be obtained by adding common difference (d) in the last term $\sqrt {18}$

$d = \sqrt{18} - \sqrt{8} = 3 \sqrt{2} - 2 \sqrt{2} = \sqrt{2} \\ next \: term = \sqrt{18} + \sqrt{2} \\ = 3 \sqrt{2} + \sqrt{2} \\ = 4 \sqrt{2} \\ = \sqrt{ {4}^{2} \times 2 } \\ = \sqrt{16 \times 2} \\ \huge \red{ \boxed{next \: term = \sqrt{32} }}$

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3 years ago

I suck at math can someone help please??

Zielflug [23.3K]

6/7 × 4 would be the correct answer. This is because 6/7 × 4= 3 3/7 which is equal to 4/7 × 6= 3 3/7. No other expression would equal 4/7 × 6

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3 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

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