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Otrada [13]
3 years ago
8

Need help please help please please please please

Mathematics
1 answer:
REY [17]3 years ago
7 0
<h3>1+3/4*1*3/5</h3><h3>7/4*8/5</h3><h3 /><h3>56/20</h3><h3>2+16/20</h3>
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(I have a deadline in 3 hours :c ) Ann's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs
max2010maxim [7]
A=3B
4.35A+5.40B=780.95
Substitute 3B into A for the second equation:
4.35(3B)+5.40B=780.90
13.05B+5.40B=780.90
18.45B=780.90
B=42.33
Plug into equation 1:
A=3B
A=3(42.33)
A=126.98
7 0
3 years ago
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Julie hung up 3 bird feeders in her yard each birdfeeder,has 12 perches how many perches are there altogether
Sedaia [141]
36 perches altogether.

I hope you like this answer, please Brainliest me, and have a great day! :D
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How many solutions does 2x + 3y = -6 and <br><br> 3x - 4y = -12 have
Ostrovityanka [42]

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8 0
3 years ago
The probability that a randomly selected 2 2​-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0
Mnenie [13.5K]

Answer:

a. Probability = 0.97735

b. Probability = 0.92294

c. P(At\ Least\ One) = 1

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

P(Live) = 0.98861

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)\ and\ P(Live)

Probability = 0.98861 * 0.98861

Probability = 0.9773497321

Probability = 0.97735 <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)^n

Where n = 7

Probability = 0.98861^7

Probability = 0.92294324145

Probability = 0.92294 <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

P(At\ Least\ One) = 1 - P(None).

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

P(Not\ Live) = 0.01139

The probability that none of the 7 lives up to 3 is:

P(None) = P(Not\ Live)^7

P(None) = 0.01139^7

Substitute this value for P(None) in

P(At\ Least\ One) = 1 - P(None).

P(At\ Least\ One) = 1 - 0.01139^7

P(At\ Least\ One) = 0.99999999999997513055642436060443621

P(At\ Least\ One) = 1 ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0
3 years ago
Use &gt;,&lt;,=.<br> 0.94___0.49
uranmaximum [27]
0.94>0.49 because 0.94 is the larger number.

5 0
3 years ago
Read 2 more answers
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