All categories

3 years ago

The explicit rule for a sequence and one of the specific terms is given. Find the position of the given

Mathematics

1 answer:

AlekseyPX3 years ago

7 0

Answer:

14th term

Step-by-step explanation:

You might be interested in

Employer-Based Health Insurance A report from a Gallup poll29 in 2011 started by saying, "Forty-five percent of American adults

eduard

Hdhdhdhdgdgdgggggggggg

8 0

3 years ago

Rationalization<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B6%7D%7B%20%5Csqrt%7B3%7D%20%7D%20" id="TexFormula1" title=" \f

rosijanka [135]

Answer:

2√3

Step-by-step explanation:

To rationalize, we multiply the denominator and numerator by the surd at the base

We have it that;

6/√3 * 1

= 6/√3 * √3/√3

= 6 √3/3 = 2√(3

7 0

3 years ago

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago

Find<br> f (2) if f(x)=-5x2if x <-6

viva [34]

Answer: -180 for -6 and -20 for 2

Please give me Brainliest

3 0

3 years ago

Help Me, I only have 2 minutes

ehidna [41]

Answer:

$6 \frac{2}{13}$