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1 year ago

I forgot how to do this. Can anyone help me please

Mathematics

2 answers:

Svetach [21]1 year ago

7 0

Answer:

Step-by-step explanation:

180 since it is supplementary

then minus 100 which is 80 then minus 12 which is 68 then half that which is 34

Nezavi [6.7K]1 year ago

3 0

Answer:

X°+X°+12°+100°=180°{Being strangle angle}

2x°+112°=180°

2x=180°-112°

2x= 68°

X=68°/2

X=34°

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7.<br>(-3x³)(-2x³y⁴z)(-3z²)÷(4x³z)(-3yz) (-3xyz)

serg [7]

Answer:

$-\frac{x^2y^2}{2}$

Step-by-step explanation:

We have an extremely large equation and are asked to divide it, so let's solve it step-by-step :

Remove the parenthesis to make it easier to read :

$\frac{-3x^3 *2x^3y^4z *3z^2}{4x^3z*3yz*3xyz}$

Multiply the numerators :

$\frac{-18x^6y^4z^3}{4*3*3x^3yyzzz}$

Multiply the denominators :

$\frac{-18x^6y^4z^3}{36x^4y^2z^3}$

Apply the negative rule :

$-\frac{18x^6y^4z^3}{36x^4y^2z^3}$

Cancel the common factor which is 18 :

$-\frac{x^6y^4z^3}{2x^4y^2z^3}$

Apply the addition exponent rule :

$\frac{y^4z^3x^{6-4} }{2y^2z^3}$

Subtract :

$\frac{x^2y^4z^3 }{2y^2z^3}$

Apply the rule for y :

$\frac{x^2y^{4-2} z^3 }{2z^3}$

Subtract :

$\frac{x^2y^2z^3 }{2z^3}$

Cancel the common factor of z^3 :

$-\frac{x^2y^2}{2}$

5 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

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