Answer:
P(z>1.3) = 0.9032
Step-by-step explanation:
We are given:
Mean = 5000
Standard deviation = 1000
x = 6300
P(x>6300)=?
z-score =?
z-score = x- mean/standard deviation
z-score = 6300 - 5000/1000
z- score = 1300/1000
z-score = 1.3
So, P(x>6300) = P(z>1.3)
Looking at the z-probability distribution table and finding value:
P(z>1.3) = 0.9032
So, P(z>1.3) = 0.9032
Answer:
C
Step-by-step explanation:
First, write 4 ½ in improper form. 2 × 4 + 1 = 9. So the fraction is 9/2.
Now divide:
9/2 ÷ 1/3
To divide by a fraction, multiply by the reciprocal:
9/2 × 3/1
27/2
13 1/2
Answer C.
Answer:
A reflection of D over line l would simply look like a backwards D. Reflecting it again over the line m would make it look like it did originally, a normal D.
Step-by-step explanation:
Wait ill come right back at u let me solve
49^2m-m : Not equivalent
7^2m-2m : Not equivalent
7^2m-m : Not equivalent
This is actually a trick question. All of the following are actually false statements. Want to know why? Let me show you.
For exponents, if you are dividing a number to some power (i.e 5^3) by the SAME number to a different power (i.e 5^2), then the expression is 5^3-2 or 5^1 = 5. This is true for any number a such that a^b ÷ a^c = a^b-c.
Since 7 and 49 are not the same number, this rule does not apply and thus cannot be simplified any further.
Let me prove why. 5^3 = 125, and 5^2 = 25, and 125 ÷ 25 = 5. This is also the same as 5^3-2 = 5^1 = 5. We just proved this as so.
But, what about 7 and 49, or 2 different numbers. Well it doesn't apply. 7^3 = 343, and 49^2 = 2401, and 343 ÷ 2401 ≈ 0.14. Thus, this is NOT equal to 7^3-2 which is 7. We just proved that a^y ÷ b^z ≠ a^y-z. Congratulations!
Hope this helped!
