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Lapatulllka [165]
10 months ago
12

is

C%3A%20the%20%5C%3A%20power%20%5C%3A%20of%20%5C%3A%2099%7D%20" id="TexFormula1" title=" \frac{3to \: the \: power \: of \: 100}{3to \: the \: power \: of \: 99} " alt=" \frac{3to \: the \: power \: of \: 100}{3to \: the \: power \: of \: 99} " align="absmiddle" class="latex-formula">greater than,less than, or equal to 3
Mathematics
1 answer:
Norma-Jean [14]10 months ago
7 0

Remember the laws of exponents:

\frac{a^n}{a^m}=a^{n-m}

Then:

\begin{gathered} \frac{3^{100}}{3^{99}}=3^{100-99} \\ =3^1 \\ =3 \end{gathered}

Therefore:

\frac{3^{100}}{3^{99}}=3

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The graph does not support his claim. The vertical scale does not start At 0 which distorts the percent of change.
4 0
3 years ago
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The vertices of PQRS have coordinates P(–1,5) , Q(3,4) , R(2,−4) , and S(−3,−2) .
aleksandrvk [35]

Answer:

P(5,1), Q(4,-3), R(-4,-2), S(-2,3)

Let me know if it’s correct !!

8 0
2 years ago
Tamara is planting bushes around her house. So far today she has planted 5 bushes and she wants to plant at least 12 bushes toda
bixtya [17]

Answer:

b+5\geq 12

Step-by-step explanation:

Given:

Number of bushes planted = 5

Minimum number of bushes to be planted = 12

Let the number of bushes planted after planting 5 bushes be b.

Since, Tamara has already planted 5 bushes, total number of bushes planted is given as:

b+5

Now, as per question, total number of bushes should be 12 or greater than 12.

Therefore, b+5\geq 12.

6 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
In a class of students, the following data table summarizes
Ksenya-84 [330]

Answer:

Step-by-step explanation:

you have t as a total amount of the whole class.

f= did not pass and you place the total number next to it like this;  f/t represents that group. You continue this in your workings.

f/t = probability failed.

p/t = probability passed

c/t = probability completed

dc/t = probability did not complete

and draw a tree.

The random would be either of these, we know there s a possibility of 1/4 but the odds stay central. We look for a new way after this data is converted as a  fraction and show these fractions as your answer.

4 0
3 years ago
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