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Brut [27]
4 years ago
15

Explain how to write the number that is one million more than 458,610,008

Mathematics
2 answers:
ch4aika [34]4 years ago
8 0
458,610,008+100,000,000
Kay [80]4 years ago
5 0
1,000,000,000+458,610,008= 1,458,610,008
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Mandarinka [93]

Answer:  Attached.

Step-by-step explanation:

Either solve the equations directly, or graph them and look for the point of intersection.  

8 0
3 years ago
What is the value 8x=5x+21
cupoosta [38]
X = 7

8(7) = 5(7) + 21

56 = 35 + 21

56 = 56
8 0
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Each month, a phone company charges $30 for 500 minutes of talk time plus $0.45 for each additional minute. Write an expression
damaskus [11]

Answer:

  30 +0.45(m -500) . . . . for m > 500

Step-by-step explanation:

If talk time is "m" minutes, then the number of "additional minutes" is ...

   m -500

The cost of those is $0.45 each, so (in dollars), the charge for additional minutes is ...

   0.45(m -500)

The charge of $30 applies regardless of the number of minutes, so is added to the minutes charge. The total cost of a bill for talking more than 500 minutes is ...

   30 +0.45(m -500)

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The cost for talking 500 minutes or less is simply 30.

8 0
4 years ago
Y=5x−2 into polar form
Alinara [238K]

Answer:

r·sin Ф = 5(r·cos Ф) - 2

Step-by-step explanation:

y = r·sin Ф and x = r·cos Ф, and so

y = 5x - 2 becomes

r·sin Ф = 5(r·cos Ф) - 2

8 0
3 years ago
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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
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