Choose the solutions of the following systems of equations x^2+y^2=6 and x^2-y=6

Answer:

89

Step-by-step explanation:

Given that

2 O, 2 T and 1 M

Now based on this, the following arrangements exist  

1

Three arrangements i.e. {M,T,O}

2

XX or XY

XX in 2C1 = two arrangements i.e. {OO or TT}

XY in 3C2 × 2! = six arrangements

3

XXY or XYZ

XXY in 2C1 × 2C1 × 3! ÷ 2! = twelve arrangements

XYZ in 3C3 × 3! = six arrangements

4

XXYY or XXYZ

XXYY = 4! ÷ (2! × 2!) = six arrangements

XXYZ in 2C1 ×  4! ÷ 2! = twenty four arrangements

5

= 5! ÷ (2! ×2!)

= 120 ÷ 4

= 30

Thus, the overall total is

= 3 + (2 +6)+ (12 +6) + (6 +24) + 30

= 89

1) 4.55

2) Short hit

Step-by-step explanation:

1)

The score and corresponding probabilities are shown in the following table:

Score 3 4 5 6 7

Probability 0.15 0.40 0.25 0.15 0.05

Let us define

X = Miguel's score on the Water Hole

The expected value of the variable X can be represented as:

where

represents the different possible outcomes of X

denotes the associated probabilities

Accordingly, the expected value of Miguel's score is calculated as follows:

2)

Here, we will again consider:

X = Miguel's score on the Water Hole

In this case:

- For a successful long hit, the expected value of X is

- Conversely, if the long hit isn't successful, the expected value of X becomes

We also know that the likelihood of succeeding with a long hit is

Consequently, the chance of failure in a long hit is

Thus, the expected value of X when opting for the long hit strategy is:

In the first part of this question, we calculated the expected value using the short hit strategy, yielding:

Given that the expected value for X is smaller (which is better) when utilizing the short hit option, we can conclude that this approach is superior.

Answer:

The mean and center will be nearly identical, while the variability in simulation B will exceed that of simulation A.

Step-by-step explanation:

Solution

In general, the size of the sample significantly influences the distribution.

As a guideline, by quadrupling the sample size, the sampling error typically reduces by half.

Here, simulation B has a sample size that is twice that of simulation A.

Thus, the sampling error related to the mean is affected, and it stands that A does not have a greater value.

However, its sample size is considerable for this context, making them nearly equal.

Additionally, increased trials lead to decreased variability in simulation. A displays lesser variability.

As trial numbers increase, variability in simulation declines, hence A has lesser variability.

225 goes to the bakery
300 goes to the grocery store

When the stretch is proportional to the weight, it can be expressed as:
weight/(49 cm) = (135 N)/(21 cm)
weight = (49/21)×135 N
weight = 315 N

The fish's weight would be 315 N.