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MA_775_DIABLO [31]
3 years ago
8

Please helpppp i have a test tomorrow the problem is in the picture below

Mathematics
1 answer:
zimovet [89]3 years ago
8 0

Answer:

yes

Step-by-step explanation:

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Find the Slope <br><br> A. 2/5<br> B.5/2<br> C.-2/5<br> D.-5/2
Fiesta28 [93]
The answer is a bc it’s pointing upwards which makes it positive and the rise is 5 and the run making it 2/5
5 0
3 years ago
Read 2 more answers
In June 2005, a CBS News/NY Times poll asked a random sample of 1,111 U.S. adults the following question: "What do you think is
Shtirlitz [24]

Answer:

The appropriate null hypothesis is H_0: p = 0.25

The appropriate alternative hypothesis is H_1: p < 0.25

Step-by-step explanation:

Exactly a year prior to this poll, in June of 2004, it was estimated that roughly 1 out of every 4 U.S. adults believed (at that time) that the war in Iraq was the most important problem facing the country.

At the null hypothesis, we test if the proportion is still the same, that is, of \frac{1}{4} = 0.25. So

H_0: p = 0.25

We would like to test whether the 2005 poll provides significant evidence that the proportion of U.S. adults who believe that the war in Iraq is the most important problem facing the U.S. has decreased since the prior poll.

Decreased, so at the alternative hypothesis, it is tested if the proportion is less than 0.25, that is:

H_1: p < 0.25

3 0
3 years ago
The pet store sells bags of pet food. There are four bags of cat food. 1/6 of the bags of food are cat food. How many bags of pe
nikklg [1K]
If 1/6 of the bags of pet food are cat food, then the other 5/6 is other pet food.
Since there are 4 bags that make up 1/6 of the food, multiplying 4 by 6 should get the answer. 4 x 6 = 24
6 0
3 years ago
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1(2.5.3)(2.4.8)(6.2.9.4)
skelet666 [1.2K]
Explain further? Thanks....
6 0
3 years ago
Find maclaurin series
Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}

The first 3 terms of the series are

\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}

7 0
2 years ago
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