80% of x = 16
"of" translates to multiplication and 80% becomes 80/100.

80/100 × x = 16

0.8x = 16

0.8x/0.8 = 16/0.8

x = 16/0.8

x = 20

Your final answer is 80% of 20 is 16.

4 0

3 years ago

Abit is planning a birthday party for his grandfather. He bought a cake for \$18$18. He also wants to buy some balloons, which a

Temka [501]

Answer:

Step-by-step explanation: if he try to buy as many balloons as passible with the money he has it would be 3$

6 0

2 years ago

What is the 5-digit code?

kvv77 [185]

9514 1404 393

Answer:

4 2 6 1 7 or 1 2 4 3 7

Step-by-step explanation:

4 are correct of 1,3,4,6,7

3 are correct of 0,2,3,6,7

1 is correct of 0,3,6,8,7 -- so not both 3 and 6, meaning 1, 4, 7 are correct

Correct digits so far are 1, 4, 7 and one of {3, 6}. Line 2 tells another is 0 or 2; line 3 tells it is not 0. This adds 2 to the list of correct digits.

This makes the correct digits 1, 2, {3, 6}, 4, 7. We don't know whether the last digit is 3 or 6.

Looking at digits correctly placed, line 2 tells us the code is either ...

_ 2 6 _ 7 or _ 2 _ 3 7

Then line 1 fills in those blanks as ...

4 2 6 1 7 or 1 2 4 3 7

Either of these codes agrees with all of the clues.

3 0

2 years ago

44. Express each of these system specifications using predicates, quantifiers, and logical connectives. a) Every user has access

DENIUS [597]

Answer:

a. ∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

b. FileSystemLocked → ∀x Access(x, SystemMailbox)

c. ∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))

d. ∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

Step-by-step explanation:

Let the domain be users and mailboxes. Let User(x) be “x is a user”, let Mailbox(y) be “y is a mailbox”, and let Access(x, y) be “x has access to y”.

∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

(b)

Let the domain be people in the group. Let Access(x, y) be “x has access to y”. Let FileSystemLocked be the proposition “the file system is locked.” Let System Mailbox be the constant that is the system mailbox.

FileSystemLocked → ∀x Access(x, SystemMailbox)

(c)

Let the domain be all applications. Let Firewall(x) be “x is the firewall”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”.

∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))

(d)

Let the domain be all applications and routers. Let Router(x) be “x is a router”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”. Let ThroughputNormal be “the throughput is between 100kbps and 500 kbps”. Let Functioning(y) be “y is functioning normally”.

∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

4 0

3 years ago