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Rzqust [24]
3 years ago
15

Find the vertical asymptote(s) of f(x) = 5x2 + 3x + 6/ x2 - 100

Mathematics
1 answer:
Maslowich3 years ago
4 0

Answer:

The answer is x=-10,10

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baherus [9]

Answer:

3) 8

4) 5.25 cups

Step-by-step explanation:

3) cross multiply:

36 × 10 = 360

45 × x = 45x

equate them both:

360 = 45x

divide both sides by 45

360 ÷ 45 = 8

x = 8

4) 8 people : 3 cups of flour

14 people : how many cups?

14 ÷ 8 = 1.75

to get from 8 people to 14 people we multiply by 1.75, so do the same thing to the 3 cups:

3 × 1.75 = 5.25 cups

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Answer:

-7x^3+8x^2+x-5

Step-by-step explanation:

We are simply adding the two functions:

f(x) + g(x) = (2x^2-5x^3+x-7) + (6x^2-2x^3+2) = 8x^2-7x^3+x-5 = -7x^3+8x^2+x-5

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2 years ago
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zalisa [80]

Answer:

D. 240

Step-by-step explanation:

We have 20 stamps in 1 book.

We are give 12 books.

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Does a statistics course improve a student's mathematics skills,as measured by a national test? Suppose a random sample of 13 st
viva [34]

Answer:

There is not enough evidence to support the claim that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

P-value=0.042.

Step-by-step explanation:

The question is incomplete:

The data of the scores for each student is:

Before    After

430        465

485        475

520        535

360        410

440        425

500        505

425        450

470        480

515        520

430        430

450        460

495        500

540        530

We will generate a sample for the difference of scores (before - after) and test that sample.

The sample of the difference is [35 -10 15 50 -15 5 25 10 5 0 10 5 -10]

This sample, of size n=13, has a mean of 9.615 and a standard deviation of 18.423.

The claim is that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

Then, the null and alternative hypothesis are:

H_0: \mu=0\\\\H_a:\mu> 0

The significance level is 0.01.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{18.423}{\sqrt{13}}=5.11

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{9.615-0}{5.11}=\dfrac{9.615}{5.11}=1.882

The degrees of freedom for this sample size are:

df=n-1=13-1=12

This test is a right-tailed test, with 12 degrees of freedom and t=1.882, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>1.882)=0.042

As the P-value (0.042) is bigger than the significance level (0.01), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

4 0
2 years ago
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