3. -x²+-4x+-8
To solve:
−4x−5+−1(x²+3)
−4x+−5+−1x²+(−1)(3)
−4x+−5+−x²+−3
Combine like terms.
−4x+−5+−x²+−3
(−x²)+(−4x)+(−5+−3)
5.
It is not in standard form since this answer does NOT have x=, y=, #=. It only has the numbers. So, this is NOT in standard form.
Answer: SU = 4(1) + 1 = 5
Step-by-step explanation:
Since T is on segment SU we know the whole is eqaul to the sum of it’s parts.
ST + TU = SU substitute
3x - 1 + 3x = 4x + 1 simplify
6x - 1 = 4x + 1 solve for x
2x = 2
x = 1
ST = 3(1) -1 = 2
TU = 3(1) = 3
SU = 4(1) + 1 = 5
Answer: i eat fooood bro
Step-by-step explanation:
3n-5=-48-40n
Move -40n to the other side. Sign changes from -40n to +40n.
3n+40n-5=-48= -48-40n+40n
3n+40n-5=-48
Move -5 to the other side.
3n+40n-5+5=-48+5
3n+40n=-43
43n=-43
Divide by 43 for both sides
43n/43=-43/43
n=-1
Answer: n=-1
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: number of daily text messages a high school girl sends.
This variable has a population standard deviation of 20 text messages.
A sample of 50 high school girls is taken.
The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;δ²/n)
This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:
Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)
a.
P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836
b.
P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)
P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328
I hope you have a SUPER day!
