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kykrilka [37]
3 years ago
5

Please help me! x-1/12 = 7/2 a: 2 b: 6 c: 41 d: 43

Mathematics
1 answer:
Shalnov [3]3 years ago
3 0
X=43/12 If you need work shown please tell me in the comments. Also please mark me as brainliest.
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PLEASE ANSWER ASAP DETAILS ARE BELOW
amid [387]
Is to long to explain so I'm doing quite

find the triangle side so 2.1 square minus 1.4 square it is equal to 2.45

now you just multiple 2.45 by 2 because if you notice 2.45 is the triangle side that is missing and this is only the half so multiple by 2 for get the X value, I hope this helps you.
8 0
3 years ago
Five out of six residents of Mayville have library cards. Which tool will best allow Darren to simulate this scenario and predic
Doss [256]

Answer:

a 5-sector spinner

Step-by-step explanation:

There are 5 people so I think that would be best

4 0
3 years ago
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fair coin is tossed three times. What is the probability that all three tosses land on heads,given that:(a) the first toss lands
maxonik [38]

Answer:

(a) The probability that all three tosses land on heads given that the 1st toss lands on Heads is \frac{1}{4}.

(b) The probability that all three tosses land on heads given that at least one toss lands on heads is \frac{1}{7}.

Step-by-step explanation:

The sample space of tossing 3 coins is:

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} = 8 outcomes.

(a)

The sample space such that the 1st toss lands on Heads is:

S₁ = {HHH, HHT, HTH, HTT} = 4 outcomes.

The probability that all three tosses land on heads given that the 1st toss lands on Heads is:

P (3 Heads | 1st toss is Heads) = \frac{P(3\ Heads\ \cap \ 1st\ toss\ is\ Heads)}{P(1st\ toss\ is\ Heads)}

                                                  =\frac{\frac{1}{4} }{\frac{4}{8} } \\=\frac{1}{8}\times \frac{8}{4}\\  =\frac{1}{4}

Thus, the probability that all three tosses land on heads given that the 1st toss lands on Heads is \frac{1}{4}.

(b)

The sample space such that at least one toss lands on heads is:

S₂ = {HHH, HHT, HTH, THH, HTT, THT, TTH} = 7 outcomes

The probability that all three tosses land on heads given that at least one toss lands on heads is,

P (3 Heads | At least 1 Heads) =\frac{P(3\ Heads\ \cap\  At\ least\ 1\ Heads)}{P( At\ least\ 1\ Heads)}

                                                  =\frac{\frac{1}{7} }{\frac{7}{8} } \\=\frac{1}{8}\times \frac{8}{7}\\  =\frac{1}{7}

Thus, the probability that all three tosses land on heads given that at least one toss lands on heads is \frac{1}{7}.

4 0
3 years ago
What type of function can approach zero as x decreases without end?
kherson [118]

Answer: c. Exponential

3 0
3 years ago
Simplify 8 + 2(10 – r)
mixer [17]
Distribute
a(b+c)=ab+ac
2(10-r)=20-2r

now we have
8+20-2r
28-2r
6 0
3 years ago
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