Help as quick and i will give the brainliest!!!

Answers

Answer 1

Answer:

I believe both answers are Shannon.

Answer 2

Answer: I think it’s sally and Shannon

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What is the least What is the least common multiple of four and 12

Answers

Answer: LCM(4,12) = 12

Step-by-step explanation:

This of this as singing all the multiples of 4

4, 8, 12.

4 + 4 + 4.

Answer:

Step-by-step explanation:

In tilapia, an important freshwater food fish from Africa, the males actively court females. They have more incentive to court a female who has already laid all of her eggs, but can they tell the difference? an experiment was done to measure the male tilapia's response to the smell of female fish. Water containing feces from females that were either pre-ovulatory (they still had eggs) or post-ovulatory (they had already laid their eggs) was washed over the gills of males hooked up to an electro-olfactogram machine which measured when the senses of the males were excited. The amplitude of the electro-olfactogram was used as a measure of the excitability of the males in the two different circumstances. Six males were exposed to the scent of pre-ovulatory females; their readings average 1.51 with a standard deviation of .25. Six different males were exposed to post-ovulatory females; their average readings of 0.87 with standard deviation is .31. Assume that the electro-olfactogram readings were approximately normally distributed within the groups.(A) test for a difference in the excitability of the males with exposure to these two types of females
(B) what is the estimated average difference in electro-olfactogram readings between the two groups? What is the 95% confidnece limit for the difference between population means?

Answers

Answer:

a) $t=\frac{1.51-0.87}{\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}}=3.936$

"=T.INV(1-0.025,10)", and we got $t_(critical)=\pm 2.28$

Statistical decision

Since our calculated value is higher than our critical value, $z_(calc)=3.936>2.28=t_(critical)$ , we have enough evidence to reject the null hypothesis at 5% of significance.

b) $(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}$

The degrees of freedom are given:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

$(1.51 -0.87) - 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 0.269$

$(1.51 -0.87) + 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 1.010$

Step-by-step explanation:

Part a

Data given and notation

$\bar X_(1)=1.51$ represent the mean for scent of pre ovulatory

$\bar X_(2)=0.87$ represent the mean for post ovolatory

$s_(1)=0.25$ represent the sample standard deviation for preovulatory

$s_(2)=0.31$ represent the sample standard deviation for postovulatory

$n_(1)=6$ sample size for the group preovulatory

$n_(2)=6$ sample size for the group postovulatory

z would represent the statistic (variable of interest)

$p_v$ represent the p value

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean's are different, the system of hypothesis would be:

H0: $\mu_(1) = \mu_(2)$

H1: $\mu_(1) \neq \mu_(2)$

If we analyze the size for the samples both are lower than 30, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We have all in order to replace in formula (1) like this:

$t=\frac{1.51-0.87}{\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}}=3.936$

Find the critical value

We find the degrees of freedom:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

In order to find the critical value we need to take in count that we are conducting a two tailed test, so we are looking for thwo values on the t distribution with df =10 that accumulates 0.025 of the area on each tail. We can us excel or a table to find it, for example the code in Excel is:

"=T.INV(1-0.025,10)", and we got $t_(critical)=\pm 2.28$

Statistical decision

Since our calculated value is higher than our critical value, $z_(calc)=3.936>2.28=t_(critical)$ , we have enough evidence to reject the null hypothesis at 5% of significance.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}$

The degrees of freedom are given:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

$(1.51 -0.87) - 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 0.269$

$(1.51 -0.87) + 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 1.010$

What is the value of y?

Answers

Answer:

80° because 50° 60° 70° and 80°

3. Draw a regular polygon with 8 vertices.

Answers

Answer:

A Regular Polygon with 8 vertices is Octogon

Step-by-step explanation:

If you count the sides of an octagon, It will be 8 sides

If you count the vertices of an octagon, it will be 8 vertices.

100 is 10 times much as

Answers

Answer:

?? If you answer is how does 10 become 100, you multiply 10 by 10?? Sorry

Step-by-step explanation:

Answer:

100 is 10 times as much as 10

Step-by-step explanation:

10 times 10 is 100.

Weights of men: 90% confidence; n = 14, x=161.5 lb, s =13.7 lb

Answers

Answer:

The 90% for the average weights of men is between 137.24 lb and 185.76 lb.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 14 - 1 = 14

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of $1 - (1 - 0.9)/(2) = 0.95$ . So we have T = 1.7709

The margin of error is:

M = T*s = 1.7709*13.7 = 24.26

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 161.5 - 24.26 = 137.24 lb

The upper end of the interval is the sample mean added to M. So it is 161.5 + 24.26 = 185.76 lb

The 90% for the average weights of men is between 137.24 lb and 185.76 lb.