All categories

2 years ago

(9x^2+5x^2+11x)+(-2x^3+9x-8x^2)

Mathematics

1 answer:

lidiya [134]2 years ago

6 0

Answer:

First step is to remove those unnecessary brackets :

$\longrightarrow\tt \: 9 {x}^{3} + 5 {x}^{2} + 11x - 2 {x}^{3} + 9x - 8 {x}^{2}$

Combine the like terms :

$\longrightarrow \tt \: 9 {x}^{3} - 2 {x}^{3} + 5 {x}^{2} - 8 {x}^{2} + 11x + 9x$

$\longrightarrow \boxed{\tt \: 7 {x}^{3} - 3 {x}^{2} + 20x}$

Our final answer is 7x^3 - 3x^2 + 20x .

--------- HappY LearninG <3 --------

You might be interested in

A centroid of a triangle is always inside the triangle because the

Alik [6]

The answer is B. Hope this helps!

4 0

3 years ago

Cornel buys a video game from a store. The video game's original Price is $60 All the video games are on sale for 25% of the ori

Mama L [17]

Answer:

Cornel pays $44.75 for his video game.

Step-by-step explanation:

First, you have to find the price after the 25% discount by calculating 25% of the initial price and subtracting that result from the original price:

60*25%=15

60-15=45

Then, you have to find the price after the 12% discount by calculating 12% of the price after the 25% discount and subtracting that result from the price after the 25% discount:

45*12%=5.4

45-5.4=39.6

Finally, you have to find the price including the tax by calculating 13% of the price after the discounts and add the result to that price:

39.6*13%=5.15

39.6+5.15=44.75

According to this, Cornel pays $44.75 for his video game.

3 0

4 years ago

PLS HELP! ITS DUE SOON ILL GIVE BRAINLIEST

lutik1710 [3]

Answer:

I think it’s a but if not a then d

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

5.4*7/4 I need help plz help me thanks

Lady bird [3.3K]

Decimal: 9.75
Fraction: 189/20
Mixed Number: 9 9/20

7 0

3 years ago

Membership in Mensa requires an IQ score above 131.5. Nine candidates take an IQ test, and they have a mean IQ score of 133. IQ

rodikova [14]

Answer:

a) $P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)$

And we can find this probability using the complement rule:

$P(Z\geq 2.2)=1- P(z$

b) $P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)$

And we can find this probability using the complement rule:

$P(Z\geq 6.6)=1- P(z$

c) No. The mean can be lower than 131.5 if we find the probability:

$P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)$

$P(Z\leq 6.3) \approx 1$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

A. If 1 person is randomly selected from thegeneral population, find the probability of getting someone with anIQ score of at least 133.

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(100,15)$

Where $\mu=100$ and $\sigma=15$

We are interested on this probability

$P(X\geq 133)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)$

And we can find this probability using the complement rule:

$P(Z\geq 2.2)=1- P(z$

And the probability is calculated from the normal standard table or with excel.

B. If 9 people are randomly selected,find the probability that their mean IQ score is at least 133.

For this case since the distribution for the random variable X is normal then the distribution for the sample mean is also normal and given by:

$\bar X = \sim N(\mu= 100 ,\sigma_{\bar x}= \frac{15}{\sqrt{9}}=5)$

The new z score is defined as

$z=\frac{x-\mu}{\sigma_{\bar x}}$

If we apply this formula to our probability we got this:

$P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)$

And we can find this probability using the complement rule:

$P(Z\geq 6.6)=1- P(z$

And the probability is calculated from the normal standard table or with excel.

C. Although the results are available,the individual IQ test scores have been lost. Can it be concluded that all 9 candidates have IQ scores above 131.5 so that they allare eligible for Mensa membership?

No. The mean can be lower than 131.5 if we find the probability:

$P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)$

$P(Z\leq 6.3) \approx 1$

6 0

3 years ago

Read 2 more answers