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yaroslaw [1]
3 years ago
10

Approximating values mean to

Mathematics
1 answer:
tino4ka555 [31]3 years ago
6 0

Answer:

come close to the value.

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What is the product in lowest terms?
Alisiya [41]
-5/12 × 8/13

First, we need to start out by applying the multiplication rule towards fractions. If you haven't learned or don't remember the rules, you can always look up fraction rules. The rule is: a/b × c/d = ac/bd. Basically, we are combining the numerators and both of the denominators.
- \frac{5 \times 8}{12 \times 13}

Second, let's now multiply what we have in the fraction. (5 × 8 = 40) and (12 × 13 = 156). Doing so will create a new fraction for us to use.
-\frac{40}{156}

Third, now obviously, we can simplify the fraction we just got into lower terms. To do that, we have to collect the greatest common factor (GCF) of both the numerator and denominator and list their factors to find the common factor that is the greatest.

Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40
Factors of 156: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156

Out of the listed factors, which of those are the common ones? 1, 2, and 4 are the common factors. Since we are looking for the greatest common factor, it would be 4 since that is the higher number out of the commons. The GCF is 4.

Fourth, we can now divide our numerator (40) and our denominator (156) by the GCF we just found out, which was 4. 
40 \div 4 = 10  \\ 156 \div 4 = 39

Fifth, our last step is to create our new simplified fraction. All we have to do is take our new numerator and denominator to see the fraction. 

Answer in fraction form: \fbox {-10/39}
Answer in decimal form: \fbox {-0.2564}
4 0
3 years ago
A class of 30 swim team members took a lifesaving test. Eighteen of the members had an average score of 92 points. The remaining
inysia [295]

Answer:

88.8%

Step-by-step explanation:

1.) 30-18= 12 which is the number of member that had an average of 84

2.)Then multiply  12 times 84 = 1008

3.) You multiply 18 by 92 which equals 1656

4.)Add 1008 and 1656 together  and get 2664.

5.)Divide 2664 / 30 and you get the average of 88.8

Hope this helps :)

plz brainly & like

8 0
3 years ago
Read 2 more answers
Solve the equation. Please show your math work. <br> 2/3A  24 <br> Answer:
Korvikt [17]
What is the ribbon thing for?
4 0
3 years ago
Lock 1: Write your response in Alphabetical Order and in ALL CAPITAL LETTERS (Do not put any spaces or commas.) You must choose
Dennis_Churaev [7]

Answer:

I believe A C and E.

7 0
3 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:

a)  

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
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