This is a system of equations. You can solve it by substitution or elimination. I'm going to use substitution x - 2y = -4.5; add 2y to each side x - 2y + 2y = -4.5 + 2y; simplify x = 2y - 4.5 2(2y - 4.5) + 3y = 12 4y - 9 + 3y = 12 7y - 9 = 12 7y - 9 + 9 = 12 + 9 7y = 21 y = 3 <--------first answer x - 2y = -4.5 x - 2(3) = -4.5 x - 6 = -4.5 x -6 + 6 = -4.5 + 6 x = 1.5 <--------second answer Check: 2(1.5) + 3(3) = 12? 3 + 9 = 12? 12 = 12; It checks 1.5 - 2(3) = -4.5? 1.5 - 6 = -4.5? -4.5 = -4.5; It checks x = 1.5 and y = 3 <-----------Final Answer
4x^2-3x+4y^2+4z^2=0
here we shall proceed as follows:
x=ρcosθsinφ
y=ρsinθsinφ
z=ρcosφ
thus
4x^2-3x+4y^2+4z^2=
4(ρcosθsinφ)^2-3(ρcosθsinφ)+4(ρsinθsinφ)^2+4(ρcosφ)
but
ρ=1/4cosθsinφ
hence we shall have:
4x^2-3x+4y^2+4z^2
=1/4cosθsinθ(cosθ(4-3sinφ))+4sin^2(φ)
Answer:
So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
Step-by-step explanation:
In this data we have
Mean= u = 18
X= 38
Standard deviation = s= 6
1) We formulate the null and alternate hypothesis as
H0: u = 18 against Ha : u > 18 One tailed test .
2) The significance level alpha = ∝= 0.05 and Z alpha has a value ± 1.645 for one tailed test.
3)The test statistics used is
Z= X- u / s
z= 38-18/6= 3.333
4) The calculated value of z = 3.33 is greater than the z∝ = 1.645
5) So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
First we set the criteria for determining the true of value of the variable. That whether the rats learn in less or more than 18 trials.
Then we find the value of z for the given significance value given and the test about to be checked.
Then the test statistic is determined and calculated.
Then both value of z and z alpha re compared. If the test statistics falls in the rejection region reject the null hypothesis and conclude alternate hypothesis is true.
The figure shows that the calulated z value lies outside the given z values
Answer:
Step-by-step explanation:
The probability of getting one even and one odd is the sum of probability of getting even and of getting odd and even. Since there are 3 even or odd on each die out of six numbers you have
P(EO)(3/6)(3/6)=1/4 and P(OE)=(3/6)(3/6)=1/4 so 1/4+1/4= 1/2
or you can find the probability of getting two odds or two evens and subtract that from 1
P(EE)=(3/6)(3/6)=1/4, P(OO)=(3/6)(3/6)=1/4
P(EO)=1-1/4-1/4=1/2
Isolate the variable by dividing each side by factors that don't contain the variable.
x > 6
