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3 years ago

Distance between (-5,2) and (-5,-4)

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

7 0

Answer:

The distance between these two points is 6

Step-by-step explanation:

distance formula

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5 2/5+a=71/2 what is the answer to this problem

Gre4nikov [31]

Answer:
30 1/10

Step by step explanation:
71/2-5 2/5= 30 1/10
30 1/10 divided by a (also know as 1) = 30 1/10

3 0

3 years ago

Please help fast!!!!!!!!!!!!!!

denis-greek [22]

Answer:

The third option, -1/2(30) = -15

Step-by-step explanation:

If it decreases by 1/2, we know it has to be -1/2, and the days we are given are 30. We need to multiply -1/2 by 30 in order to find the water level after 30 days, which would be -15 (inches?) I hope this helped.

5 0

3 years ago

In Pennsylvania the average IQ score is 101.5. The variable is normally distributed, and the population standard deviation is 15

blsea [12.9K]

Answer:

We conclude that the Pennsylvania school district have an IQ higher than the average of 101.5

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 101.5

Sample mean, $\bar{x}$ = 106.4

Sample size, n = 30

Alpha, α = 0.05

Population standard deviation, σ = 15

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 101.5\\H_A: \mu > 101.5$

We use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{106.4 - 101.5}{\frac{15}{\sqrt{30}} } = 1.789$

Now, $z_{critical} \text{ at 0.05 level of significance } = 1.96$

Since,

$z_{stat} > z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that the Pennsylvania school district have an IQ higher than the average of 101.5

5 0

3 years ago

What value of r is a solution to this equation? 387 divide by r = 9

Vlad1618 [11]

Answer:

A. 43

Step-by-step explanation:

8 0

3 years ago

Let f(x) = [infinity] xn n2 n = 1. find the intervals of convergence for f. (enter your answers using interval notation. ) find

inna [77]

Best guess for the function is

$\displaystyle f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}$

By the ratio test, the series converges for

$\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n}\right| = |x| \lim_{n\to\infty} \frac{n^2}{(n+1)^2} = |x| < 1$

When $x=1$ , $f(x)$ is a convergent $p$ -series.

When $x=-1$ , $f(x)$ is a convergent alternating series.

So, the interval of convergence for $f(x)$ is the closed interval $\boxed{-1 \le x \le 1}$ .

The derivative of $f$ is the series

$\displaystyle f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n^2} = \frac1x \sum_{n=1}^\infty \frac{x^n}n$

which also converges for $|x|$ by the ratio test:

$\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac n{x^n}\right| = |x| \lim_{n\to\infty} \frac{n}{n+1} = |x| < 1$

When $x=1$ , $f'(x)$ becomes the divergent harmonic series.

When $x=-1$ , $f'(x)$ is a convergent alternating series.

The interval of convergence for $f'(x)$ is then the closed-open interval $\boxed{-1 \le x < 1}$ .

Differentiating $f$ once more gives the series

$\displaystyle f''(x) = \sum_{n=1}^\infty \frac{n(n-1)x^{n-2}}{n^2} = \frac1{x^2} \sum_{n=1}^\infty \frac{(n-1)x^n}{n} = \frac1{x^2} \left(\sum_{n=1}^\infty x^n - \sum_{n=1}^\infty \frac{x^n}n\right)$

The first series is geometric and converges for $|x|$ , endpoints not included.

The second series is $f'(x)$ , which we know converges for $-1\le x$ .

Putting these intervals together, we see that $f''(x)$ converges only on the open interval $\boxed{-1 < x < 1}$ .

6 0

2 years ago