Max height is the vertex
convert to vertex form (y=a(x-h)^2+k) by completeing the square

h(t)=-16(t²-4t)+80
h(t)=-16(t²-4t+4-4)+80
h(t)=-16((t-2)²-4)+80
h(t)=-16(t-2)²+64+80
h(t)=-16(t-2)²+144

vertex is (2,144)
at t=2, the height is 144

max height is 144ft

4 0

3 years ago

Hamburgers come packed 12 in a package. Hamburger buns come packed 10 in a package. If we want one hamburger for each bun for a

BlackZzzverrR [31]

Answer:

5 hamburger packages and 6 bun packages

Step-by-step explanation:

It is asking for the highest common multiple of 12 and 10. the first time a multiple of 12 is divisible by 10 is at 12x5, which is 60. now divide 60 by 12 and 10, 5 and 6.

4 0

2 years ago

A bicycle lock has a four-digit code. The possible digits, 0 through 9, cannot be repeated. What is the probability that the loc

kozerog [31]

First let's find the number of the elements in the sample space, that is the total number of codes that can be produced.

The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one.
and so on...

so in total there are 10*9*8*7 = 5040 codes.

a. What is the probability that the lock code will begin with 5?

Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.

Thus, there are 1*9*8*7=504 codes which start with 5.

so

P(first digit is five)= $\frac{n(first -digit- is- 5)}{n(all-codes)}= \frac{1*9*8*7 }{10*9*8*7 }= \frac{1}{10}=0.1$

b. <span>What is the probability that the lock code will not contain the number 0?

from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}

from which we can form in total 9*8*7*6 codes which do not contain 0.

P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6

Answer:

0.1 ; 0.6</span>

4 0

3 years ago

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