Pearson's product-moment correlation coefficient is represented by the following letter.
Group of answer choices
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1. See attachment for the labelled tree diagram

2. The outcomes are {HR, HY, HB, TR, TY, TB}

3. The sample space is {HR, HY, HB, TR, TY, TB}

4. The outcomes in the event "Toss a tails, spin yellow" is 1

5. The probability of tossing tails and spinning yellow is 1/6

Given that

Coin = Head or Tail

Spinner = Red, Yellow, Blue

Next, we complete the columns using the above

See attachment

Using the following key

From the completed tree diagram, the outcomes are

Outcomes = {HR, HY, HB, TR, TY, TB}

This means that the total number of outcomes is 6

And each outcome has a probability of 1/6

This is the same as the outcomes written inside the rectangles

So, we have

Sample space = {HR, HY, HB, TR, TY, TB}

Here, we have

"Toss a tails, spin yellow"

This is represented as TY

So, the outcomes in the event "Toss a tails, spin yellow" is 1

In (b), we have

Each outcome has a probability of 1/6

This means that the probability of tossing tails and spinning yellow is 1/6

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Answer:

Step-by-step explanation:

True. But there is a very high chance of not happening

The present value (PV) is $3000.45 when it is rounded to two decimal places.

We can use the following formula for the present value of an annuity:

PV = PMT * [(1 - (1 + i)^(-n)) / i]

Here, n = 20, i = 0.027, and PMT = $207. Now, plug in the values:

PV = 207 * [(1 - (1 + 0.027)^(-20)) / 0.027]

First, calculate the values inside the parentheses:

(1 + 0.027)^(-20) = 0.60829 (rounded to 5 decimal places)

Next, subtract this value from 1:

1 - 0.60829 = 0.39171 (rounded to 5 decimal places)

Now, divide the result by the interest rate:

0.39171 / 0.027 = 14.50704 (rounded to 5 decimal places)

Finally, multiply this value by the payment amount:

207 * 14.50704 = 3000.45

So, the present value (PV) is $3000.45 when rounded to two decimal places.

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By using  piecewise functions the values of  [tex]\lim_{\theta \to \pi^+} h(\theta)[/tex] is 1 and  [tex]\lim_{\theta \to \pi/2^-} h(\theta)[/tex] is -1

The given functions are h(θ)=cos2θ, θ<π/2

h(θ)=tanθ/2,   π/2<θ≤π

h(θ)=sinθ/2, θ ≥π

Now let us find the value of [tex]\lim_{\theta \to \pi^+} h(\theta)[/tex]

[tex]\lim_{\theta \to \pi^+} \frac{ sin(\theta)}{2}[/tex]

This is a right hand limit which we take the values greater than π.

Apply the limit theta as pi.

sinπ/2

We know that sin90 degrees is 1.

[tex]\lim_{\theta \to \pi^+} h(\theta)[/tex]=1

Now [tex]\lim_{\theta \to \pi/2^-} h(\theta)[/tex]

This is a left hand limit which we take the values lesser than π/2.

[tex]\lim_{\theta \to \pi/2^-} cos(2\theta)[/tex]

Now apply the limit theta as π/2.

cos2(π)/2

[tex]\lim_{\theta \to \pi/2^-} h(\theta)[/tex] = -1

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The value of x in the logarithm equation 3.2 = log(x/1) is 1584.89

From the question, we have the following parameters that can be used in our computation:

3.2 = log(x/1)

Evaluate the quotient of x and 1

So, we have

3.2 = log(x)

Take the exponent of both sides

[tex]x = 10^{3.2[/tex]

Evaluate the exponent

x = 1584.89

Hence, the value of x in the equation 3.2 = log(x/1) is 1584.89

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i) The counting numbers which are multiples of 5 and less than 50 can be represented using the roster method as:

{5, 10, 15, 20, 25, 30, 35, 40, 45}

ii) The set of all-natural numbers x for which x+6 is greater than 10 can be represented as:

{5, 6, 7, 8, 9, 10, 11, 12, 13, ...}

iii) The set of all integers x for which 30/x is a natural number can be represented as:

{-30, -15, -10, -6, -5, -3, -2, -1, 1, 2, 3, 5, 6, 10, 15, 30}

Note that in the third set, we include both positive and negative integers that satisfy the condition.

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(a) Proved If [tex]a = a^-1[/tex], then det(a) = ±1.

(b) The determinant must be +1 or -1

When a square matrix a is equal to its inverse [tex]a^-1[/tex], the determinant of a is either +1 or -1. This can be explained as follows:

(a) In the case where [tex]a = a^-1[/tex], we can multiply both sides of the equation by a to obtain [tex]a^2 = I[/tex], where I is the identity matrix.

Taking the determinant of both sides, we have[tex]det(a^2) = det(I)[/tex], and since [tex]det(I) = 1,[/tex] we get [tex](det(a))^2 = 1.[/tex]

This implies that det(a) is either +1 or -1.

(b) The determinant of a matrix represents the scaling factor of the transformation it represents.

If [tex]a = a^-1[/tex], it means that applying the transformation twice results in the identity transformation, which preserves the shape and orientation of vectors.

Therefore, the determinant must be +1 or -1 to maintain this property.

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We can see that we are looking for the probability of getting a 5 or a factor of 48 from a 6-sided dice. Thus, the probability is 1.

Probability is a way to gauge or quantify how likely something is to happen. It reflects the likelihood or potential for an event to occur, with values ranging from 0 (impossible) to 1. (certain).

We can see here that the probability of getting a 5 or a factor of 48 is:

P(5) = 1/6

Factors of 48 are:  1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

The factors of 48 found in the dice are: 1, 2, 3, 4, 6

Thus, P(factor of 48) = 5/6

Thus, P(5 or factor of 48) = P(5) +  P(factor of 48) = 1/6 + 5/6 = 6/6 = 1

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The probability that it will rain on exactly one of the five days is approximately 0.395 .

To find the probability that it will rain on exactly one of the five days, we can use the binomial probability formula.

The probability of rain on any given day is 1/4, and we want to find the probability of rain on exactly one day out of the five.

Using the binomial probability formula, the probability of rain on exactly one day can be calculated as follows:

P(X = 1) = C(5, 1) * (1/4)^1 * (3/4)^4

where C(5, 1) represents the number of ways to choose one day out of the five.

C(5, 1) = 5! / (1! * (5-1)!) = 5

Plugging in the values, we have:

P(X = 1) = 5 * (1/4)^1 * (3/4)^4

Calculating this expression gives us approximately 0.395 .

Therefore, the probability that it will rain on exactly one of the five days they are there is approximately 0.395 , rounded to the nearest thousandth.

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The value of the line integral ∫c f · dr is 1.885.

To evaluate the line integral ∫c f · dr, where c is given by the vector function r(t) and f(x,y) = xyi + 6y²j, we need to first parameterize the curve c using the given vector function r(t).

r(t) = 15t⁵i + t⁵j

The curve c starts at the point (0,0) when t=0 and ends at the point (15,1) when t=1.

Next, we need to calculate the differential of r(t) with respect to t:

dr/dt = 75t⁴i + 5t⁴j

We can now substitute the parameterization of c and the differential dr/dt into the formula for line integrals to get:

∫c f · dr = ∫[0,1] f(r(t)) · (dr/dt) dt

= ∫[0,1] (15t⁶)(t⁵)i + 6(t⁵)²(j) · (75t⁴i + 5t⁴j) dt

= ∫[0,1] (15t¹¹) dt + ∫[0,1] (6t¹⁰) dt

=[tex](15/12)t^{12} |_0^1 + (6/11)t^{11} |_0^1[/tex]

= (15/12) + (6/11)

= 1.885

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