Listed below are the numbers of cricket chirps in 1 minute and the corresponding temperatures in F. Find the regression​ equation, letting chirps in 1 minute be the independent​ (x) variable. Find the best predicted temperature at a time when a cricket chirps times in 1​ minute, using the regression equation. What is wrong with this predicted​ temperature?
Chirps in 1 Min: 834, 1041, 904, 948, 1100, 1174, 1201, 1000
Temperature in Degree F: 70.5, 80.8, 76.1, 76.7, 85.6, 84.4, 89.9, 76.9

Answer: See explanation

Step-by-step explanation:

x=-(2-2*the square root of 6)/5, about 0.58

or

x=-(2+2*the square root of 6)/5, about -1.38

The values of the x from equation  \(5x^2+4x-4=0\) are  x = 0.5798 and -1.38.

Given that:

Equation:  \(5x^2+4x-4=0\)

To find the values of x that satisfy the equation \(5x^2+4x-4=0\), use the quadratic formula:

\(x = \dfrac{ -b \± \sqrt{b^2 - 4ac}}{ 2a}\)

Compare the equation with \(ax^2 + bx + c = 0\).

Here, a = 5, b = 4, and c = -4.

Plugging in the values to get,

\(x = \dfrac{-4 \± \sqrt{4^2 - 4 \times 5 \times (-4)}}{2 \times 5} \\x = \dfrac{-4 \± \sqrt{16 +80}}{10} \\x = \dfrac{-4 \± \sqrt{96}}{10}\\x = \dfrac{-4 \± {4\sqrt6}}{10}\)

So the solutions for x are calculates as:

Taking positive sign,

\(x = \dfrac{-4 + {4\sqrt6}}{10}\\x = \dfrac{-4 + {9.798}}{10}\\\)

x = 5.798/10

x = 0.5798

Taking negative sign,

\(x = \dfrac{-4 - {4\sqrt6}}{10}\\x = \dfrac{-4 - {9.798}}{10}\\\)

x = -13.798/10

x = -1.38

Hence, the exact solutions for the equation  \(5x^2 + 4x - 4 = 0\)  are x = 0.5798 and -1.38.

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The 90% confidence interval for the difference in the population means is -2.51 to 0.31

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

xˉ₁ = −17.1

s₁² = 8.4

n₁ = 22

​xˉ₂ = −16.0

s₂² = 8.7

n₂ = 24

Calculate the pooled variance using

P = (df₁ * s₁² + df₂ * s₂²)/df

Where

df₁ = 22 - 1 = 21

df₂ = 24 - 1 = 23

df = 22 + 24 - 2 = 44

So, we have

P = (21 * 8.4 + 23 * 8.7)/44

P = 8.56

Also, we have the standard error to be

SE = √(P/n₁ + P/n₂)

So, we have

SE = √(8.56/22 + 8.56/24)

SE = 0.86

The z score at 90% CI is 1.645, and the CI is calculated as

CI =  (x₁ - x₂) ± z * SE

So, we have

CI = (-17.1 + 16.0) ± 1.645 * 0.86

This gives

CI = -1.1 ± 1.41

Expand and evaluate

CI = (-2.51, 0.31)

Hence, the confidence interval is -2.51 to 0.31

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To find the value of x, we can use the fact that the sum of angles in a triangle is 180 degrees. Therefore, we can add up the given angles and subtract the sum from 180 degrees to find the value of x.

Sum of angles = t + a + c + s + r + m

180 degrees = 88 degrees + 145 degrees + 137 degrees + 130 degrees + 125 degrees + x degrees

Simplifying the equation:

180 degrees = 625 degrees + x degrees

To isolate x, we can subtract 625 degrees from both sides of the equation:

x degrees = 180 degrees - 625 degrees

x degrees = -445 degrees

Therefore, the value of x is -445 degrees.

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The area of one of the bases of the hexagonal prism is approximately 7.39 units^2.

To solve this problem, we can use the formula for the volume of a hexagonal prism:

V = (3√3/2) * a^2 * h

where V is the volume, a is the length of one side of the hexagonal base, and h is the height of the prism.

We are given that the height of the prism is 4 units and the volume is 98 units^3. Plugging these values into the formula, we can solve for the length of one side of the base:

98 = (3√3/2) * a^2 * 4
a^2 = 98 / (12√3)
a ≈ 2.95

Now that we know the length of one side of the base, we can find the area of the hexagon by using the formula:

A = (3√3/2) * a^2

where A is the area of one of the bases. Plugging in the value we found for a, we get:

A = (3√3/2) * (2.95)^2
A ≈ 7.39

Therefore, the area of one of the bases of the hexagonal prism is approximately 7.39 units^2.

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The maximum likelihood estimator for the variance, (ui|\(X_{2i}\), \(X_{3i}\), β₄\(X_{4i}\)), is unbiased.

To analyze the bias of the maximum likelihood estimator (MLE) for the variance, we need to consider the assumptions and properties of the regression model.

In the given regression model:

\(Y_i\) = β₁ + β₂\(X_{2i}\) + β₃\(X_{3i}\) + β₄\(X_{4i}\) + U\(_{i}\)

Here, \(Y_i\) represents the dependent variable, \(X_{2i}, X_{3i},\) and \(X_{4i}\) are the independent variables, β₁, β₂, β₃, and β₄ are the coefficients, U\(_{i}\) is the error term, and i represents the observation index.

The assumption of the classical linear regression model states that the error term, U\(_{i}\), follows a normal distribution with zero mean and constant variance (σ²).

Let's denote the variance as Var(U\(_{i}\)) = σ².

The maximum likelihood estimator (MLE) for the variance, σ², in a simple linear regression model is given by:

σ² = (1 / n) × Σ[( \(Y_i\) - β₁ - β₂\(X_{2i}\) - β₃\(X_{3i}\) - β₄\(X_{4i}\))²]

To determine the bias of this estimator, we need to compare its expected value (E[σ²]) to the true value of the variance (σ²). If E[σ²] ≠ σ², then the estimator is biased.

Taking the expectation (E) of the MLE for the variance:

E[σ²] = E[ (1 / n) × Σ[( \(Y_i\) - β₁ - β₂\(X_{2i}\) - β₃\(X_{3i}\) - β₄\(X_{4i}\))²]

Now, let's break down the expression inside the expectation:

[( \(Y_i\) - β₁ - β₂\(X_{2i}\) - β₃\(X_{3i}\) - β₄\(X_{4i}\))²]

= [ (β₁ - β₁) + (β₂\(X_{2i}\) - β₂\(X_{2i}\)) + (β₃\(X_{3i}\) - β₃\(X_{3i}\)) + (β₄\(X_{4i}\) - β₄\(X_{4i}\)) + \(U_{i}\)]²

= \(U_{i}\)²

Since the error term, \(U_{i}\), follows a normal distribution with zero mean and constant variance (σ²), the squared error term \(U_{i}\)² follows a chi-squared distribution with one degree of freedom (χ²(1)).

Therefore, we can rewrite the expectation as:

E[σ²] = E[ (1 / n) × Σ[\(U_{i}\)²] ]

= (1 / n) × Σ[ E[\(U_{i}\)²] ]

= (1 / n) × Σ[ Var( \(U_{i}\)) + E[\(U_{i}\)²] ]

= (1 / n) × Σ[ σ² + 0 ] (since E[ \(U_{i}\)] = 0)

Simplifying further:

E[σ²] = (1 / n) × n × σ²

= σ²

From the above derivation, we see that the expected value of the MLE for the variance, E[σ²], is equal to the true value of the variance, σ². Hence, the MLE for the variance in this regression model is unbiased.

Therefore, the maximum likelihood estimator for the variance, (ui|\(X_{2i}\), \(X_{3i}\), β₄\(X_{4i}\)), is unbiased.

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31) x = 25 degrees

Equation: (2x+15) + x + 90 = 180 degrees

--> 3x+105 = 180

-------> 3x = 75

-------------> x = 25 degrees

32) x = 17 degrees

Equation: (x+52) + (3x+21) + 39 = 180 degrees

--> 4x+112 = 180

-------> 4x = 68

-------------> x = 17 degrees

Answer:

20

Step-by-step explanation:

Given:

Lisa is checking out at the bookstore.

Each book (b) costs $1.50 and she has a $5 off coupon she can use.

She wants to spend $25 or less in the bookstore.

Solve:

Since there is a $5 off coupon..

$1.5b - $5.00<=$25.00

$1.5b<= $30.00

$1.5b/1.5 = $30.00/1.5

b=20

Lisa can buy 20 book.

~lenvy~

Answer:

1 1/8

Step-by-step explanation:

Answer:

1 1/8

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

Answer:

\( \sf \: j= \frac{−8}{7}s+ \frac{445}{7} , \sf \: s= \frac{−7}{8}j+ \frac{445}{8} \)

Step-by-step explanation:

\( \sf \: Let's solve for j. \\ \sf \: 7j+8s=445 \\ \sf \: Step 1: \: Add \: -8s \: to \: both \: s ides. \\ \sf \: 7j+8s+−8s=445+−8s \\ \sf \: 7j=−8s+445 \\ \sf \: Step \: 2: \: Divide \: both \: sides \: by \:7. \\ \sf \frac{7j}{7}= \frac{−8s+445}{7} \\ \sf \: j= \frac{−8}{7}s+ \frac{445}{7} \\\)

The probability that all three selected students are boys is approximately 0.3929 or 39.29%.

To calculate the probability that all three selected students are boys, we need to consider the total number of possible outcomes and the number of favorable outcomes.

In this case, there are 12 boys and 4 girls in the class, making a total of 16 students. We want to select three students, and we want all three of them to be boys.

The total number of ways to select three students from the class is given by the combination formula, which can be represented as:

Total Possible Outcomes = nCr(16, 3) = (16!)/((16-3)! * 3!) = 560

Now, let's consider the number of favorable outcomes where all three selected students are boys. Since there are 12 boys, we can choose three of them using the combination formula:

Favorable Outcomes = nCr(12, 3) = (12!)/((12-3)! * 3!) = 220

Therefore, the probability that all three selected students are boys is:

Probability = Favorable Outcomes / Total Possible Outcomes = 220 / 560 ≈ 0.3929, or approximately 39.29%.

Hence, the probability that all three selected students are boys is approximately 0.3929 or 39.29%.

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Answer:

1) -2

2) 8/7

Step-by-step explanation:

1) 10x+8=3(x-2)

  10x+8=3x-6

  10x-3x=-6-8

  7x=-14

  x=-2

2) 10x+8=3x

   10x-3x=8

   7x=8

   x=8/7

Hope it helps

Answer:

x = 46°

Step-by-step explanation:

Angles on a straight line sum to 180°.

Therefore, the interior angle of the triangle that forms a linear pair with the exterior angle marked 130° is:

⇒ 180° - 130° = 50°

The interior angle of the triangle that forms a linear pair with the exterior angle marked 96° is:

⇒ 180° - 96° = 84°

The interior angles of a triangle sum to 180°. Therefore:

⇒ 50° + 84° + x = 180°

⇒ 134° + x = 180°

⇒ 134° + x - 134° = 180° - 134°

⇒ x = 46°

Therefore, the size of angle x is 46°.

The value of the expression a^3 is 3 3/8

The given parameters can be represented as

If a = 3, then 5a = 15.

Also, we have the value of the variable a to be

a = 1 1/2

Express the value of the variable a as an improper fraction

So, we have

a = 3/2

When a = 1 1/2 or a = 3/2, the value of a³ is calculated using the following formula

a³ = (1 1/2)³ = (3/2)³

Rewrite the formula as

a³ = (3/2)³

Evaluate the exponent in the above equation

So, we have

a³ = (27/8)³

Evaluate the quotient in the above equation

So, we have

a³ = 3 3/8

Hence, the value of the expression a³ is 3 3/8

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Alex could place point H very close to one of the existing points (E, F, or G) such that the sum of lengths of EH, HF, and FG, should not exceed 22 - 15.21 = 6.79 units.

We have,

Given points:

E (3, 5)

F (6, 5)

G (9, 1)

We'll calculate the distances between these points to determine the perimeter of the fenced area:

\(Distance~ EF: \sqr((x_F - x_E)^2 + (y_F - y_E)^2)\\Distance ~FG: \sqrt((x_G - x_F)^2 + (y_G - y_F)^2)\\Distance ~GE: \sqrt((x_E - x_G)^2 + (y_E - y_G)^2)\)

Calculate the distances:

Distance EF: √((6 - 3)² + (5 - 5)²) = 3 units

Distance FG: √((9 - 6)² + (1 - 5)²) = √(9 + 16) = 5 units

Distance GE: √((3 - 9)² + (5 - 1)²) = √(36 + 16) = 2√13 units

Now, the total perimeter of the current fenced area is:

Perimeter = EF + FG + GE = 3 + 5 + 2√13 ≈ 8 + 7.21 ≈ 15.21 units.

Now, since you have 22 units of fencing available, the maximum additional length for the fourth side, which is the sum of lengths of EH, HF, and FG, should not exceed 22 - 15.21 = 6.79 units.

Thus,

Alex could place point H very close to one of the existing points (E, F, or G) such that the sum of lengths of EH, HF, and FG, should not exceed 22 - 15.21 = 6.79 units.

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Alex has used 8 units of fencing to enclose his plot from E to F to G, leaving him with 14 units to return from G to the final corner point H. Any point H must be within 14 units of G. H could be (9, -13) which is 14 units below G.

Alex can use the distance formula to calculate the distance between the points and see how much fencing he has already used. The distance between points E and F is 3 units (6 - 3 = 3). The distance between points F and G is 5 units (Square root of ((6 - 9)^2 + (5 - 1)^2)). This equates to 8 units of fencing already used. Therefore, Alex has 14 units of fencing left (22 - 8 = 14). To not require any additional fencing, any point H he chooses must be within 14 units of point G. Corner point H could be (9, -13), for example, which is exactly 14 units below G, so it would work perfectly. Alex cannot, however, return to point E (3, 5), because that's farther than 14 units.

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Answer:

5x – 30 = 176

Step-by-step explanation:

Answer:

(b) and (c)

Step-by-step explanation:

When the equation of a line is given in slope-intercept form (\(y=mx+b\)), m represents the slope of the line and b represents the y-intercept. As we can see, the two lines given have the same slope (-3), but different y-intercepts (the first line has a y-intercept of 4 and the second line has a y-intercept of -11).

Since the slopes of the two lines are the same, they will be parallel, and since their y-intercepts are different, they will not be the same line. Therefore, the two lines will not intersect (a solution to the system is any point where the two lines do intersect), and there is no solution to the system.

As a general rule:

Answer:

5.8

Step-by-step explanation:

2.5² + 5.2² = c² ≈ 5.8 units

Answer:

5.8

Step-by-step explanation:

√(2.5^2 + 5.2^2) = 5.77

5.77 to the nearest ones decimal point is 5.8

Answer:

6. 12cm

7. 11.2 cm

8. 92 degrees

9. 53 degrees

10. 37.5 cm

11. $ 693.12

Step-by-step explanation:

For questions 6-9 the triangle is congurent which means both their angles will be equal.

For the sides you can see that the second triangle is 1.5 times smaller than the first one as taking one of the sides 21/14= 1.5

So for Q. 6 we can multiply side PQ by 1.5.

For Q.7 divide CA by 1.5.

For Q. 10 we can do cross multiplication:

5/4 = x/30

= 150= 4x

= 37.5=x

For Q 11 we can see the size of the paper is 1.9 times bigger than the original one (38/20)

So the shorter side is 30.4 cm.

Then we have to find the area of the paper to multiply the two dimensions and we get 1155.2 square inches.

Then to figure out the money we multiply again and we get $693.12.

Answer:

1) 100 degrees F

2) 50 degrees F

Step-by-step explanation:

its just simple subtraction

Answer:

x - 9 = 12

Step-by-step explanation:

We can use x as "a number"

So the equation so far is just "a number", or x.

Decreased by 9 can also be said as subtracted by 9, so that makes the equation x - 9.

"is 12" can also be said as equals 12, which makes the equation x - 9 = 12

Answer:

_ - 9 = 12

Step-by-step explanation:

You can identify the starting number by adding 9 and 12 together to get 21. The completed equation would be 21 - 9 = 12.

Check the picture below.

\(4+10x=\cfrac{(9x+20)+10x}{2}\implies 8+20x=19x+20\implies x=12 \\\\[-0.35em] ~\dotfill\\\\ 4+10x\implies 4+10(12)\implies \stackrel{ \measuredangle DEC }{124^o}\)

The Laplace transform of h(t) is \(L{h(t)} = (6 + 8s)/(s^2 - 4) + 18/(s^2 + 9)\)

To use the table of Laplace transforms, we need to express the given function in terms of functions whose Laplace transforms are known. Recall that:

The Laplace transform of sinh(at) is \(a/(s^2 - a^2)\)

The Laplace transform of cosh(at) is \(s/(s^2 - a^2)\)

The Laplace transform of sin(bt) is \(b/(s^2 + b^2)\)

Using these formulas, we can write:

\(h(t) = 3 sinh(2t) + 8 cosh(2t) + 6 sin(3t)\\= 3(2/s^2 - 2^2) + 8(s/s^2 - 2^2) + 6(3/(s^2 + 3^2))\)

To find the Laplace transform of h(t), we need to take the Laplace transform of each term separately, using the table of Laplace transforms. We get:

\(L{h(t)} = 3 L{sinh(2t)} + 8 L{cosh(2t)} + 6 L{sin(3t)}\\= 3(2/(s^2 - 2^2)) + 8(s/(s^2 - 2^2)) + 6(3/(s^2 + 3^2))\\= 6/(s^2 - 4) + 8s/(s^2 - 4) + 18/(s^2 + 9)\\= (6 + 8s)/(s^2 - 4) + 18/(s^2 + 9)\)

Therefore, the Laplace transform of h(t) is:

\(L{h(t)} = (6 + 8s)/(s^2 - 4) + 18/(s^2 + 9)\)

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To find the Laplace transform of h(t) = 3 sinh(2t) 8 cosh(2t) 6 sin(3t), for t > 0, we can use the table of Laplace transforms. The Laplace transform of the given function h(t) is: L{h(t)} = (6/(s^2 - 4)) + (8s/(s^2 - 4)) + (18/(s^2 + 9))

First, we need to use the following formulas from the table:

- Laplace transform of sinh(at) = a/(s^2 - a^2)
- Laplace transform of cosh(at) = s/(s^2 - a^2)
- Laplace transform of sin(bt) = b/(s^2 + b^2)

Using these formulas, we can find the Laplace transform of each term in h(t):

- Laplace transform of 3 sinh(2t) = 3/(s^2 - 4)
- Laplace transform of 8 cosh(2t) = 8s/(s^2 - 4)
- Laplace transform of 6 sin(3t) = 6/(s^2 + 9)

To find the Laplace transform of h(t), we can add these three terms together:

L{h(t)} = L{3 sinh(2t)} + L{8 cosh(2t)} + L{6 sin(3t)}
= 3/(s^2 - 4) + 8s/(s^2 - 4) + 6/(s^2 + 9)
= (3 + 8s)/(s^2 - 4) + 6/(s^2 + 9)

Therefore, the Laplace transform of h(t) is (3 + 8s)/(s^2 - 4) + 6/(s^2 + 9).

To use a table of Laplace transforms to find the Laplace transform of the given function h(t) = 3 sinh(2t) + 8 cosh(2t) + 6 sin(3t) for t > 0, we'll break down the function into its components and use the standard Laplace transform formulas.

1. Laplace transform of 3 sinh(2t): L{3 sinh(2t)} = 3 * L{sinh(2t)} = 3 * (2/(s^2 - 4))
2. Laplace transform of 8 cosh(2t): L{8 cosh(2t)} = 8 * L{cosh(2t)} = 8 * (s/(s^2 - 4))
3. Laplace transform of 6 sin(3t): L{6 sin(3t)} = 6 * L{sin(3t)} = 6 * (3/(s^2 + 9))

Now, we can add the results of the individual Laplace transforms:

L{h(t)} = 3 * (2/(s^2 - 4)) + 8 * (s/(s^2 - 4)) + 6 * (3/(s^2 + 9))

So, the Laplace transform of the given function h(t) is:

L{h(t)} = (6/(s^2 - 4)) + (8s/(s^2 - 4)) + (18/(s^2 + 9))

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Answer:

1. -.005

2.-1/2

3. 0

4. 2

Step-by-step explanation:

if it is negative put positive and if it is positive put negative. for 0 it will always stay the same

The Linnaean class system provides a framework for understanding the diversity of life on Earth by grouping similar organisms together based on shared characteristics.

By comparing and contrasting the definitions of each class, we can see how different groups of animals are related to each other and how they differ in terms of their biological traits.

The Linnaean class system is a way of organizing living things based on shared characteristics. Let's compare and contrast the definitions of the Linnaean classes using a Venn diagram.

First, we have the class Mammalia, which includes all animals that have hair or fur, produce milk to feed their young, and have specialized teeth. This class overlaps with the class Aves, which includes all birds, because some birds have specialized beaks and feathers that are similar to mammalian hair and teeth. However, birds do not produce milk.

Next, we have the class Reptilia, which includes animals that are cold-blooded, lay eggs, and have scales or plates on their skin. This class overlaps with both Mammalia and Aves in terms of species that lay eggs, such as monotremes (platypus and echidnas) and some birds (ostriches and emus). However, reptiles lack specialized teeth and do not produce milk.

Finally, we have the class Amphibia, which includes animals that are cold-blooded, breathe through their skin, and undergo metamorphosis from a water-dwelling larval stage to a land-dwelling adult stage. This class overlaps with Reptilia in terms of some shared characteristics, but Amphibia also lacks specialized teeth and does not lay eggs with hard shells like reptiles.

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Answer:

10

Step-by-step explanation:

substitute the variables with their numbers

X2+Y

(3)2 +(4)

3x2=6

6+4

10

The Probability of winning based on the first two flips is:P(win) = P(X = 2) + P(X = 1) = 1/4 + 1/2 = 3/4The probability of winning is 3/4.

Jeremiah created a game where he flips a fair coin 3 times. Jeremiah wins the game if he flips heads at least 2 times. What are all the possible outcomes to win, based on the first two flips

The possible outcomes to win the game, based on the first two flips, are:

Head, Head (HH)Head, Tail (HT)Tail, Head (TH)The only possible outcome that does not lead to a win is Tail, Tail (TT).

What is the probability of winning, based on the first two flips

The probability of winning, based on the first two flips, can be calculated using the binomial distribution formula:P(X = k) = nCk * pk * (1-p)n-k

Where :n = 2 (since we are considering the first two flips)k = 2 or 1 (since we need at least 2 heads to win)P(head) = p = 1/2P(tail) = 1 - p = 1/2

Substituting the values in the formula:

P(X = 2) = 2C2 * (1/2)2 * (1/2)0 = 1 * 1/4 * 1 = 1/4P(X = 1) = 2C1 * (1/2)1 * (1/2)1 = 2 * 1/2 * 1/2 = 1/2

Therefore, the probability of winning based on the first two flips is:P(win) = P(X = 2) + P(X = 1) = 1/4 + 1/2 = 3/4The probability of winning is 3/4.

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