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n200080 [17]
3 years ago
12

Which description matches the data in the table?

Mathematics
1 answer:
Naddik [55]3 years ago
3 0

Answer:

A student startes out with five cards and adds four cards each week

Step-by-step explanation:

pls mark brainliest

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Find the midpoint of the segment with the given endpoints (5,-9) and (-2,-2)
QveST [7]
Another way to solve this is to use the Midpoint Formula.  The midpoint of a segment joining points (x_1,y_1) and (x_2,y_2) is

\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} \right)

So the midpoint of your segment is

\left(\frac{5+(-2)}{2},\frac{-9+(-2)}{2}\right) = \left(\frac{3}{2},-\frac{11}{2} \right)

Perhaps it helps to see that the x-coordinate of the midpoint is just the average of the x-coordinates of the points.  Ditto for the y-coordinate of the midpoint; just average the y's.
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
A linen shop sells sheets and pillow cases .Linen from easiest company offer a choice in each set of sheets and pillow cassia pl
mariarad [96]

Answer:

polkadot's strips

plain stripes

polkadot's plain

Step-by-step explanation:

3

8 0
3 years ago
Solving a Multi-Step Equation
VLD [36.1K]

Answer:

x = 15

Step-by-step explanation:

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3 years ago
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What’s the answer <br> A. 2<br> B.4<br> C.6<br> D. 12
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The Answer is C) 6 I believe
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3 years ago
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