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4 years ago

13,18=31 7,25=31 12,30=42 whats 26,13=?

Mathematics

1 answer:

Aleksandr-060686 [28]4 years ago

7 0

The answer is 39. You have to add the two numbers together.

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Step 1 -3(3x+5)-6<6

Mnenie [13.5K]

Error is between steps 4 and 5. It should say:

Step 4: -9x < 27
Now divide by -9
Step 5: x > -3
Note that the inequality changes direction.
When you mean multiply or divide an inequality by a negative number, the inequality changes direction.

For example:
4 < 5
Multiply by -1:
-4 > -5

Another example:
-5 < 0
5 > 0

6 0

3 years ago

Pls help this is due today

Viktor [21]

Answer:

Option C sounds correct to me.

Step-by-step explanation:

Have a nice day!

and can I maybe have brainliest?

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3 years ago

Gwen wants to create a congruent shape to the one she made. Her regular pentagon has a perimeter of 28.6 cm. What is going to be

mario62 [17]

Length of the sides in the shape

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4 years ago

Read 2 more answers

(3x ^ 2 - 4 + x ^ 3) / (x - 1) Synthetic Division

nadya68 [22]

Answer:

i love algebra anyways

the ans is in the picture with the steps

(hope it helps can p lz have brainlist it will make my day

Step-by-step explanation:

4 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:

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

$Probability = 0.98861 * 0.98861$

$Probability = 0.9773497321$

$Probability = 0.97735$ --- Approximated

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 $n = 7$

$Probability = 0.98861^7$

$Probability = 0.92294324145$

$Probability = 0.92294$ --- Approximated

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