The answer is Tuesday there will be 15 teachers for 179 students and 12 times 15 is 180
Answer:
f(n) = -n^2 -3n +5
Step-by-step explanation:
Suppose the formula is ...
f(n) = an^2 +bn +c
Then we have ...
f(1) = 1 = a(1^2) +b(1) +c
f(2) = -5 = a(2^2) +b(2) +c
f(3) = -13 = a(3^2) +b(3) +c
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Here's a way to solve these equations.
Subtract the first equation from the second:
-6 = 3a +b . . . . . 4th equation
Subtract the second equation from the third:
-8 = 5a +b . . . . . 5th equation
Subtract the fourth equation from the fifth:
-2 = 2a
a = -1
Then substituting into the 4th equation to find b, we have ...
-6 = 3(-1) +b
-3 = b
and ...
1 = -1 +(-3) +c . . . . . substituting "a" and "b" into the first equation
5 = c
The formula is ...
f(n) = -n^2 -3n +5
We can't use P = 2L + 2W because the sides have different lengths.
P = a + b + c + d
a = x
b = 2x
c = 2x - 2
d = x + 5
P = x + 2x + (2x - 2) + (x + 5)
P = <u>x + 2x + 2x</u> - 2 <u>+ x</u> + 5
P = 6x + 3
if x = 4:
P = 6 · 4 + 3
P = 24 + 3
P = 27
Answer:x= 8/√3, y= 16/ √3
Step-by-step explanation:
See the picture below
Answer:
(a)3
(b)4
(c)6
(d)5
Step-by-step explanation:
(a)Josie rolls a six-sided die 18 times.
P(she rolls a two)=1/6
Therefore, the estimated number of times she rolls a two in 18 trials
=1/6 X 18
=3
(b)Slips of paper are numbered 1 through 10.
P(the number 10 appear)=1/10
If one slip is drawn and replaced 40 times, expected number of 10
=1/10 X 40
=4
(c)A spinner consists of 10 equal- sized spaces: 2 red, 3 black, and 5 white.
P(red)=2/10
If the spinner is spun 30 times, expected number of red space
=2/10 X 30
=6
(d)A card is picked from a standard deck
P(drawing an ace)=4/52
If the card is picked 65 times and replaced each time.
Expected Number of Aces =4/52 X 65 =5
