MATHEMATICS COLLEGE

What is the y-intercept of the line your equation X-1/2Y = -6A.-6

B.-3

C.0

D.12

Answers

Answer 1
Answer:

9514 1404 393

Answer:

  D.  12

Step-by-step explanation:

Set x=0 and solve for y.

  0 -1/2y = -6

  y = (-2)(-6) = 12 . . . . . . multiply by -2 to clear the fraction

The y-intercept is 12.

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Answers

Answer:

So i beiieve the answer is B

Step-by-step explanation:

Because if you thinl about the ride for A is $2 plus 25 cents so its 2 plus 25m thats per mile and b 3 dollars and 15 so u keep adding up until u get the same from both equations if u need further explanation let me know

Please helpppp!!! it’s timed!!!! thank u for helping!!!!!

Answers

Answer:

ADMC

Step-by-step explanation:

The secant of a circle is a line that cuts the cuts two points of the circumference with one end point external to the circle;

Fro the given diagram, the two lines that cuts the circumference of the circle at two points are AEB and ADMC

Hence the one that contains diameter with point M is line ADMC

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 a simplified form of the expression 2(–x + 2) + 3x?

Answers

In the simplified form the expression is x+4

Answer:

x+4

Step-by-step explanation:

The ratio of Sunita's age to Mark's age is currently 3 to 4, and in 12 years, it will be 5 to 6. What is Mark's current age?A) 18

B)24

C)30

D)36

Answers

Answer:

24 years.

B is correct option.

Step-by-step explanation:

Let Sunita's current age is x and that of Mark's is y.

Hence, we have

(x)/(y)=(3)/(4)\n\nx=(3)/(4)y.......(1)

Now, in 12 years the ratio will be 5 to 6. Thus, we have

(x+12)/(y+12)=(5)/(6)

Cross multiplying, we get

6x+72=5y+60

Plunging, the value of x from equation (1)

6((3)/(4)y)+72=5y+60\n\n(9y)/(2)+72=5y+60\n\n(y)/(2)=12\n\ny=24

Therefore, Mark's current age is 24 years.

Mark's current age is 24  

The soccer team is making pizzas for a fundraiser. They put1/4 of a pound of cheese on each pizza. If they have 12 one-pound packages of cheese, how many pizzas can they make?

Answers

4 pizza equal 1 pound of cheese multiply 4×12=48
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