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Marta_Voda [28]
3 years ago
9

Solve this equation. Enter your answer in the box. 1/3 (y+7)=3(y−1)

Mathematics
2 answers:
Mashcka [7]3 years ago
8 0

Answer:

y=2

Step-by-step explanation:

y+7=9(y-1)

y+7=9y-9

16=8y

2=y

pshichka [43]3 years ago
3 0

Answer:

y=2

Step-by-step explanation:

https://www.mathpapa.com/algebra-calculator.html?q=1%2F3(y%2B7)%3D3(y%E2%88%921)

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Apples come in bags of 4. Oranges come in bags of 6.
muminat

Answer: Apples come in bags of 4. Oranges come in bags of 6.

so it going to be Multiple 4x6=24 I explain it, is this how you want it

Step-by-step explanation: hope this help

3 0
3 years ago
Read 2 more answers
A particular telephone number is used to receive both voice calls and fax messages. suppose that 20% of the incoming calls invol
Phoenix [80]

Answer:

0.1091 or 10.91%

Step-by-step explanation:    

We have been given that a particular telephone number is used to receive both voice calls and fax messages. suppose that 20% of the incoming calls involve fax messages and consider a sample of 20 calls. We are asked to find the probability that exactly 6 of the calls involve a fax message.  

We will use Bernoulli's trials to solve our given problem.

P(X=x)=^nC_x\cdot P^x(1-P)^{n-x}

P(X=6)=^{20}C_6\cdot (0.20)^6(1-0.20)^{20-6}

P(X=6)=\frac{20!}{6!(20-6)!}\cdot (0.20)^6(0.80)^{14}

P(X=6)=\frac{20!}{6!(14)!}\cdot (0.000064)(0.04398046511104)

P(X=6)=38760\cdot (0.000064)(0.04398046511104)

P(X=6)=0.1090997009730

P(X=6)\approx 0.1091\\

Therefore, the probability that exactly 6 of the calls involve a fax message would be approximately 0.1091 or 10.91%.

4 0
3 years ago
What is a number that when you multiply it by 0.9 and
lora16 [44]
0.9x12=10.8,, 10.8-6.3=4.5. The number is 12
8 0
3 years ago
A lock has a three number code made up of 17 numbers if none of the numbers are allowed to repeat how many different ways can yo
Marrrta [24]

Answer:

680

Step-by-step explanation:

Use the binomial coefficient where you choose k=3 numbers out of n=17 possible numbers and find the total amount of combinations since order does not matter:

\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}\\ \\\binom{17}{3}=\frac{17!}{3!(17-3)!}\\\\\binom{17}{3}=\frac{17!}{3!(14)!}\\\\\binom{17}{3}=\frac{17*16*15}{3*2*1}\\\\\binom{17}{3}=680

Thus, you can make 680 three-non-repeating-number codes

8 0
2 years ago
Using the extreme value theorem, which graph has a maximum of 3 and minimum of –9?
klasskru [66]

Answer:

the 1st one

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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