Find the sum or difference of (7-8i)+(-2-3i)

Given the function

Setler79 [48]

$lim \: f(x) = ( \infty + 4)( \infty - 2) {}^{2} \\ x - > \infty$

$lim \: f(x) = \infty \times \infty \\ x - > \infty$

$lim \: f(x)= \infty \\ x - > \infty$

$lim \: \frac{f(x)}{x} = \frac{x(1 - \frac{4}{x})(x - 2) {}^{2} }{x} \\ x - > \infty$

$lim \: \frac{f(x)}{x} = (1)( \infty - 2) {}^{2} \\ x - > \infty$

$lim \: \frac{f(x)}{x} = \infty \\ x - > \infty$

We can then say that the function f(x)=(x-4)(x-2)² admits an asymptotic direction parallel to the y-axis at +∞ and -∞ as well since we have to follow the same steps.

5 0

2 years ago

The first term of a geometric series is -3, the common ratio is 6, and the sum of the series is -4,665 Using a table of values,

vladimir1956 [14]

Answer:

There are 5 terms in the series.

Step-by-step explanation:

$S=a\frac{1-r^{n} }{1-r}\\a=-3\\r=6\\-4665=-3\frac{1-6^{n} }{1-6}\\1555=\frac{1-6^{n} }{-5}\\\-7775=1-6^{n}\\6^{n}=7776\\$

Take logs to get n = 5

8 0

3 years ago

Read 2 more answers

Sergio ate 3.5 cookies. Each cookie contained 5.7 grams of sugar. How many grams of sugar did Sergio eat?

FrozenT [24]

I’ll get it correct here :)

He ate 19.95 grams of sugar.

3.5 x 5.7 = 19.95

5 0

3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago

55÷ =5 what is answer

Andru [333]

Answer:

The answer is 11

Step-by-step explanation:

55÷5= 11

7 0

3 years ago