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zubka84 [21]
2 years ago
7

Only the evaluate question. please answer quickly ​

Mathematics
1 answer:
dem82 [27]2 years ago
4 0

Answer:

2m²-6m

Step-by-step explanation:

First simplify the numerator.

Then cancel the common factor of m+3 and use distributive property.

Multiply m by m and lastly multiply -3 by 2.

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Integrate x<br><img src="https://tex.z-dn.net/?f=x4" id="TexFormula1" title="x4" alt="x4" align="absmiddle" class="latex-formula
Dmitrij [34]

Answer:

The result of the integration is \frac{x^5}{5}

Step-by-step explanation:

Integration of a power of x:

The integration of a power of x is:

\int x^{n} dx = \frac{x^{n+1}}{n+1}

In this question:

n = 4. So

\int x^{4} dx = \frac{x^{4+1}}{4+1} = \frac{x^5}{5}

The result of the integration is \frac{x^5}{5}

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 <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
What is 3/4 -7/10 -3/4 and 8/10 In order from least to greatest
zaharov [31]
-3/4, -7/10, 3/4, 8/10 is least to greatest
3 0
3 years ago
Read 2 more answers
Find the product 1/1/4OOOOJOOOO...!!! Casi todas las cuentas de facebook están siendo duplicadas. Gracias por no aceptar una nue
gizmo_the_mogwai [7]
Okay, tu no tienes un Facebook...tienes una pregunta?
4 0
3 years ago
Draw an isosceles Triangle with exactly one 40 degrees angle. Is this the only possibility or can you draw another triangle that
vaieri [72.5K]

Answer:

The correct answer is yes, we can draw an isosceles triangle with only one 40° angle. NO this is not the only possibility.

Step-by-step explanation:

plz mark brainliest

7 0
3 years ago
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