Mr. Johnson instructed his students to use the Pythagorean Theorem to draw a line that is precisely 13−−√ units long from the star to another grid point. The dots are one unit apart. Using right triangles to determine the distance, how many different lines could Mr. Johnson’s students create without going off the paper?

Answer:

1 mile = 0.5 cm

Step-by-step explanation:

I hope this helps!

Hey again!

The answer to your question is 2 cm = 1 mile

We can solve this by doing 9/4.5, and we get 2. This means that every 2 centimeters in the scale drawing is equal to 1 mile in the actual.

Hope this helps!

The average rate of change, in simplest form, is -2/5.

What is average rate ?

Divide the change in y-values by the change in x-values to determine the average rate of change. Identifying changes in quantifiable parameters like average speed or average velocity calls for the knowledge of the average rate of change.

The rate of change of the function is its gradient or slope.

The formula for calculating the gradient of a function is expressed as:

\(m=\frac{d y}{d x}=\frac{y_2-y_1}{x_2-x_1}$$\)

Using the coordinate points from the table (0,41) and (15,35)

Substitute the coordinate into the expression:

\($$\begin{aligned}& m=\frac{35-41}{15-0} \\& m=\frac{-6}{15} \\& m=\frac{-2}{5}\end{aligned}$$\)

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There is a significant relationship between the number of cats she owns and the number of times an old widow was married (r = +0.91, p < 0.05, r² = 0.82).

Given, the Pearson correlation coefficient of +0.91,

There appears to be a strong +ve correlation between the number of cats she owns and the number of times an old widow was married.

It suggests that the more times a widow was married,the more cats she tends to own.

Approximately 82% of the variance in the number of cats owned can be explained by the number of times a widow was married is indicated by the coefficient of determination (r²).

Hence, we can say that there is a significant relationship between the number of cats she owns and the number of times an old widow was married (r = +0.91, p < 0.05, r² = 0.82).

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We need to determine whether each given value of w is a solution to the equation 2(w - 5) = -20.

To check if a given value of w is a solution to the equation, we substitute the value of w into the equation and see if both sides are equal.

Let's evaluate the equation for each given value of w:

1. For w = 0:

  2(0 - 5) = -20

  -10 = -20

Since -10 is not equal to -20, w = 0 is not a solution to the equation.

2. For w = 5:

  2(5 - 5) = -20

  2(0) = -20

  0 = -20

Since 0 is not equal to -20, w = 5 is not a solution to the equation.

3. For w = 10:

  2(10 - 5) = -20

  2(5) = -20

  10 = -20

Since 10 is not equal to -20, w = 10 is not a solution to the equation.

Therefore, none of the given values of w (0, 5, 10) are solutions to the equation 2(w - 5) = -20.

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Answer: A

Step-by-step explanation:

Ok, in the first expression we have:

f - (g - h)

this can be written as:

f + (-1)*(g - h)

Now we can distribute that -1, because this is a scalar.

Now we have:

f + (-1)*g - (-1)*h

This is equal to:

f - g + h = (f - g) + h.

So the relation:   ƒ −(g − h) = (ƒ − g) − h is wrong, and this is because the subtraction operation dos not have the distributive property. (when we "break" the parentheses, the sign of h should change)

The mean crack size is 1.25 mm.

(a) To sketch the PDF and CDF, we can plot the given probability density function (PDF) on a graph paper.

The PDF f_X(x) is defined as follows:

f_X(x) = {

x/8 for 0 < x ≤ 2,

1/4 for 2 < x ≤ 5,

0 elsewhere

}

First, let's plot the PDF on the graph paper:

        |       .     .

   1/4  |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

   0.2  |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

        |       .     .

   0.1  |   .   .   .   .

        | . . . . . . . .

        +----------------

          0   2   4   6

The height of the PDF corresponds to the probability density at a given value of x.

Next, let's calculate the cumulative distribution function (CDF) to sketch it on the graph paper.

The CDF is obtained by integrating the PDF from negative infinity to x:

F_X(x) = ∫[0,x] f_X(t) dt

For 0 ≤ x ≤ 2:

F_X(x) = ∫[0,x] (t/8) dt = (1/8) * ∫[0,x] t dt = (1/8) * (t^2/2)|[0,x] = (1/8) * (x^2/2) = x^2/16

For 2 < x ≤ 5:The height of the PDF corresponds to the probability density at a given value of x.

Next, let's calculate the cumulative distribution function (CDF) to sketch it on the graph paper.

The CDF is obtained by integrating the PDF from negative infinity to x:

F_X(x) = ∫[0,x] f_X(t) dt

For 0 ≤ x ≤ 2:

F_X(x) = ∫[0,x] (t/8) dt = (1/8) * ∫[0,x] t dt = (1/8) * (t^2/2)|[0,x] = (1/8) * (x^2/2) = x^2/16

For 2 < x ≤ 5:The height of the PDF corresponds to the probability density at a given value of x.

Next, let's calculate the cumulative distribution function (CDF) to sketch it on the graph paper.

The CDF is obtained by integrating the PDF from negative infinity to x:

F_X(x) = ∫[0,x] f_X(t) dt

For 0 ≤ x ≤ 2:

F_X(x) = ∫[0,x] (t/8) dt = (1/8) * ∫[0,x] t dt = (1/8) * (t^2/2)|[0,x] = (1/8) * (x^2/2) = x^2/16

For 2 < x ≤ 5:F_X(x) = ∫[0,2] (t/8) dt + ∫[2,x] (1/4) dt = (1/8) * ∫[0,2] t dt + (1/4) * ∫[2,x] dt = (1/8) * (t^2/2)|[0,2] + (1/4) * (t)|[2,x] = (1/8) * 2 + (1/4) * (x-2) = 1/4 + (1/4) * (x-2) = 1/4 + (x-2)/4 = (x+1)/4

For x > 5:

F_X(x) = 1

Now, let's plot the CDF on the same graph paper:

        | . . . . . . . .

   1    | . . . . . . . .

        | . . . . . . . .

        | . . . . . . . .

        | . . . . . . . .

   0.8  | . . . . . . . .

        | . . . . . . . .

        | . . . . . . . .

        | . . . . . . . .

   0.6  | . . . . . . . .

        | . . . . . . . .

        | . . . . . . . .

        | . . . . . . . .

   0.4  | . . . . . . . .

        |

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The ratio for the Illustration a+b/a-b=5/3 is 5:4.

Ratio demonstrates how many times one number can fit into another number. Ratios contrast two numbers by ordinarily dividing them. A/B will be the formula if one is comparing one data point (A) to another data point (B).

From the information given, the equation is written thus:

a + b / a - b = 5 / 3

Cross multiply

5(a - b) = 3 (a + b)

5a - 5b = 3a + 3b

Collect the like terms

5a - 3a = 3b + 5b

2a = 8b

Divide

a = 8b / 2

a = 4b

Therefore (a + b) / a will be:

= (4b + b) / 4b

= 5b / 4b

= 5/4

= 5 : 4

The ratio is 5:4.

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Solve for x.

41. (x + a)/b + b = a

1. Subtract b.

(x + a)/b = a - b

2. Multiply by b.

x + a = ab - b^2

3. Subtract a.

x = ab - b^2 -a

42. bx + a = dx + c

1. Subtract a.

bx = dx + c - a

2. Subtract dx.

bx - dx = c - a

3. Factor x out.

x(b -d) = c - a

4. Divide by (b-d).

x = (c-a)/(b-d)

43. cx - b = ax + d

1. Add b.

cx = ax + d + b

2. Subtract ax.

cx - ax = d + b

3. Factor x out.

x(c -a) = d + b

4. Divide by (c-a).

x = (d +b)/c-a).

44. a(x -3) + 8 = b(x - 1)

1. Distribute everything.

ax - 3x + 8 = bx - b

2. Subtract 8.

ax - 3a = bx - b -8

3. Subtract bx.

ax - bx - 3a = -b - 8

4. Factor x out.

x ( a - b) - 3a = -b - 8

5. Add 3a.

x ( a - b) = -b - 8 + 3a

6. Divide by (a-b).

x =  (-b - 8 + 3a) / (a - b)

45. c ( x + 2) - 5 = b ( x -3)

1. Distribute everything.

cx + 2c -5 = bx - 3b

2. Add 5.

cx + 2c = bx - 3b + 5

3. Subtract bx.

cx - bx + 2c = -3b + 5

4. Factor x out.

x ( c - b ) + 2c = -3b + 5

5. Subtract 2c.

x ( c - b) = -3b + 5 - 2c

6. Divide by ( c - b).

x = (-3b + 5 - 2c)/(c-b)

46. a (3tx - 2b) = c (dx -2)

1. Distribute everything

3txa - 2ba = cdx - 2c

2. Subtract cdx.

3txa - cdx - 2ba = -2c

3. Add 2ba.

3txa - cdx = 2ba - 2c

4. Factor x out.

x ( 3ta - cd) = 2ba - 2c

5. Divide by (3ta - cd).

x = (2ba - 2c)/(3ta - cd)

47. b ( 5px - 3c) = a (qx -4)

1. Distribute everything.

5bpx - 3bc = apx - 4a

2. Subtract apx.

5bpx - apx - 3bc = -4a

3. Add 3bc.

5bpx - apx = -4a + 3bc

4. Factor x out.

x ( 5bp - ap ) = -4a + 3bc

5. Divide by (5bp - ap).

x = (-4a + 3bc)/(5bp - ap)

48. [a/b(2x - 12)]= c/d

1. Multiply b first.

a ( 2x - 12) = c/d

2. Multiply by d.

ad(2x - 12) = cb

3. Distribute ad.

2adx - 12ad = cb

4. Add 12ad.

2adx = cb + 12ad

5. Divide by 2 ad.

x = (cb + 12ad)/2ad

Answer:

V = 90π in³ or 282.7 in³

Step-by-step explanation:

The formula for the volume of a cylinder is V = hπr²

Since we are given height(h) and the radius(r), we simply plug these in and solve for the volume(V).

V = (10)π(3²)

V = (10)π(9)

V = 90π in³ or 282.7 in³

Answer: The X intercept is 3 and the Y intercept is -6

Answer:

-6 + 12q.

Step-by-step explanation:

Let's start by simplifying the expression 18-20+2q6q-4q.

First, we can combine the numerical terms 18 and -20 to get -2.

Next, we can combine the q terms by factoring out a common factor of q:

2q6q - 4q = 2q(6q - 2)

Now we can substitute this expression back into our original expression:

18-20+2q(6q - 2)

And finally, we can simplify further by using the distributive property:

18 - 20 + 12q - 4 = -6 + 12q

The simplified expression is -6 + 12q.

Under the null hypothesis. Using a standard normal table or a calculator, we find that the P-value is between 0.05 and 0.10.

The answer is c) between 0.05 and 0.10.

In statistics, the null hypothesis (H0) is a statement that assumes that there is no significant difference between two or more groups, samples, or populations.

To test for the equality of the proportions of streams that fail to meet minimum requirements in the two areas, we can use a two-sample test of proportions.

Let p1 be the true proportion of streams that fail to meet minimum requirements in group 1 (Eastern Virginia), and p2 be the true proportion of streams that fail to meet minimum requirements in group 2 (Western Virginia). The null hypothesis is that the two proportions are equal, i.e., H0: p1 = p2, and the alternative hypothesis is that they are not equal, i.e., Ha: p1 ≠ p2.

We can use the pooled estimate of the proportion, p, to test the null hypothesis. The formula for the pooled estimate of the proportion is:

p = (x1 + x2) / (n1 + n2)

where x1 and x2 are the numbers of streams that fail to meet minimum requirements in groups 1 and 2, respectively, and n1 and n2 are the sample sizes.

The test statistic is:

z = (p1 - p2) / √(p * (1 - p) * (1/n1 + 1/n2))

Under the null hypothesis, the test statistic follows a standard normal distribution.

The observed values are x1 = 17, n1 = 85, x2 = 24, n2 = 80.

The pooled estimate of the proportion is:

p = (17 + 24) / (85 + 80) = 0.202

The test statistic is:

z = (17/85 - 24/80) / √(0.202 * (1 - 0.202) * (1/85 + 1/80)) = -1.78

The P-value for a two-sided test is the probability of observing a test statistic as extreme as -1.78 or more extreme, under the null hypothesis. Using a standard normal table or a calculator, we find that the P-value is between 0.05 and 0.10.

Therefore, the answer is c) between 0.05 and 0.10.

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Answer: The answer is option C: $17,925.00.

Step-by-step explanation: Using the simple interest formula:

I = Prt

where I is the interest, P is the principal, r is the interest rate, and t is the time in years.

We can find the interest earned over the 6 years:

I = P * r * t = $15,000 * 0.0325 * 6 = $2,925

Adding the interest to the principal, we get the balance at the end of the 6 years:

Balance = Principal + Interest = $15,000 + $2,925 = $17,925

Therefore, the answer is option C: $17,925.00.

Step-by-step explanation:

We can solve the system by substitution. Since the second equation is already solved for y, we can substitute 3x + 12 for y in the first equation and get:

3x + (3x + 12) = 12

Simplifying this equation, we get:

6x + 12 = 12

Subtracting 12 from both sides, we get:

6x = 0

Dividing both sides by 6, we get:

x = 0

Now that we know x, we can substitute it into either equation to find y. Using the second equation, we get:

y = 3(0) + 12 = 12

Therefore, the solution to the system is (x,y) = (0,12).

Since there is only one solution, the system has one solution.

Step-by-step explanation:

Principal = 250,000

Rate = 4%

Simple interest = 850,000

Time = ?

\(t = \frac{100 \times interest}{principal \times rate} \\ t = \frac{100 \times 850000}{250000 \times 4} \\ t = \frac{100 \times 85}{25 \times 4} \\ t = \frac{8500}{100} \\ t = 85years\)

Answer: +24

Step-by-step explanation: To multiply (-8)(-3), remember that a negative times a negative is always a positive. So (-8)(-3) is +24.

Answer:

+24

Step-by-step explanation:

took quiz

The fetus experiences tactile stimulation in the womb as a result of: several factors including movement, pressure, and the mother's digestive and respiratory systems.

Tactile stimulation is the sense of touch. The fetus can experience a sense of touch even while still in the womb. The sense of touch can be evoked by several factors including movement, pressure, and the mother's digestive and respiratory systems.In the womb, the fetus is in a dark, warm, and quiet environment.

Therefore, they can feel when their mother touches her stomach or when someone touches her from outside the belly. The tactile stimulation also occurs when the fetus moves around or kicks and stretches. The fetus' tactile sensitivity has been shown to be well-developed by the end of the first trimester.

The fetus is also sensitive to pressure changes. This is because the amniotic fluid in which they are suspended is influenced by changes in pressure. For instance, if the mother is sitting, standing, or lying down, this causes changes in the pressure of the amniotic fluid.

These changes cause the fetus to move or shift their position. This movement, in turn, stimulates the fetus' tactile senses.

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\(a {y}^{2} + by + c = 0 \)

\(3 {y}^{2} + 8y + 2 = 0 \)

\(a = 3\)

\(b = 8\)

\(c = 2\)

_________________________________

\(∆ = {b}^{2} - 4ac \)

\(∆ = ({8})^{2} - 4 \times (3) \times (2) \)

\(∆ = 64 - 24\)

\(∆ = 40\)

_________________________________

\(y = \frac{ -b ± \sqrt{∆} }{2a} \\ \)

##############################

\(y(1) = \frac{ - 8 + \sqrt{40} }{6} \\ \)

\(y(1) = \frac{ - 8 + \sqrt{4 \times 10} }{6} \\ \)

\(y(1) = \frac{ - 8 + 2 \sqrt{10} }{6} \\ \)

\(y(1) = \frac{2( - 4 + \sqrt{10}) }{2 \times 3} \\ \)

\(y(1) = \frac{ - 4 + \sqrt{10} }{3} \\ \)

+++++++++++++++++++++++++++++++++++++++

\(y(2) = \frac{ - 8 - \sqrt{40} }{6} \\ \)

\(y(2) = \frac{ - 8 - \sqrt{4 \times 10} }{6} \\ \)

\(y(2) = \frac{ - 8 - 2 \sqrt{10} }{6} \\ \)

\(y(2) = \frac{2( - 4 - \sqrt{10}) }{2 \times 3} \\ \)

\(y(2) = \frac{ - 4 - \sqrt{10} }{3} \\ \)

##############################

_________________________________

And we're done.....♥️♥️♥️♥️♥️

Answer:

y= mx+ c is the linear equation for y= 2 x+1

Answer:

15

Step-by-step explanation:

"There is a two-digit number such that the sum of its digits is 6"

The digits can be

0, 6   possible numbers: 06, 60

1, 5    possible numbers: 15, 51

2, 3    possible numbers: 23, 32

3, 3     possible numbers: 33

"the product of the digits is 1/3 of the original number"

That means we need a number that is 3 times the product of the digits.

0, 6   product of the digits: 0     3 times product: 0

1, 5    product of the digits: 5     3 times product: 15

2, 3    product of the digits: 6     3 times product: 18

3, 3     product of the digits: 9     3 times product: 27

The only number that appears with both conditions is 15.

Answer: 15

Answer:

Step-by-step explanation:

j) (2g+9h-5)-(6g-4h+2); g=-2 ; h=5

(2*(-2)+9*5-5)-(6*(-2)-4*5+2)

(-4+45-5)-(-12-20+2)

36-(-32+2)

36-(-30)

36+30

66

Answer:

C. 54cm²

Step-by-step explanation:

Split the figure into 2. A=LW. the top shape is easy, 8x5 = 40

As for the bottom one, we need to figure out the height. The whole left side is 12cm, and the part in the top shape is 5cm since it is across from the labeled side, and is a rectangle. 12-5= 7, the height of the smaller shape.

From there, we use LW to figure that out. 7x2 = 14

Now we know the area of both shapes, so we must add them together. Think of it as finding the area of each shape seperately, which is what we did. 40 + 14 = 54 and don't forget the label!

Answer:

1/4

Step-by-step explanation:

In probabilty,

"AND" means "multiplication"

"OR" means "addition"

Here, we want probability of "rolling number less than 4" "AND" "rolling number greater than 3". Thus, we find the inidvidual probabilities and then "MULTIPLY" (since "AND").

In a 6-sided die, the numbers are 1,2,3,4,5, and 6.

Thus,

Probability of rolling a number less than 4 = 3/6 = 1/2 (the numbers are 1,2, and 3)

and

Probability of rolling a number greater than 3 = 3/6 = 1/2 (the numbers are 4,5, and 6)

Thus:

P(less than 4) and P(greater than 3) = 1/2 * 1/2 = 1/4

The probability is 1/4

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The length of time it would take for the tithe to double in value, given the rate it is invested is 9.16 years.

The annual interest rate required for the amount to accumulate would be 4.82%.

The principal or present value that would return the amount in the Guaranteed Investment Certificate (GIC) is $29,409.92.

To find how long it will take for the tithe to double in value, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Since we want the tithe to double, we have:

2P = P(1 + r/n)^(nt)

2 = (1 + r/n)^(nt)

2 = (1 + 0.0775/12)^(12t)

(1 + 0.0775/12)^(12t) = 2

12t x log(1 + 0.0775/12) = log(2)

t = log(2) / (12 x log(1 + 0.0775/12))

t = 9.16 years

To find the annual interest rate (APR) required for $22,500 to accumulate to $50,000 in 14 years with interest compounded semi-annually, we can use the same compound interest formula:

A = P(1 + r/n)^(nt)

50,000 = 22,500(1 + r/2)^(2*14)

(1 + r/2)^(28) = 50,000/22,500

28 * log(1 + r/2) = log(50,000/22,500)

r/2 = 10^(log(50,000/22,500) / 28) - 1

r = 0.0482

r = 4.82%

To find the present value (principal) that will return $40,000 in 9 years with a 4% APR compounded quarterly, use the compound interest formula:

A = P(1 + r/n)^(nt)

We want to find the principal (P) when the future value (A) is $40,000, the annual interest rate (r) is 4%, or 0.04 in decimal form, the interest is compounded quarterly (n = 4), and the investment period (t) is 9 years.

40,000 = P(1 + 0.04/4)^(4 x 9)

40000 = P(1 + 0.01)^(36)

40000 = P(1.01)^36

P = 40000 / (1.01)^36

P = 29,409.92

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Answer:

-193 is correct.

The length and width of the postcard are 5.5±0.5  and \(\frac{31}{8} \pm \frac{3}{8}\)

Given that, the machine will accept postcards whose length is between 5 inches and 6 inches. Let the length of the postcard be represented as x±y, therefore

x - y =5    equation 1

x+y=6      equation 2

Add equations (1) and (2), and we get

2x=11

x=5.5

Putting the value of x=5.5 into equation (1), we get y=0.5

Therefore, the length of the postcard can be represented by 5.5±0.5 inches

Given, that the machine will accept a postcard whose width is between \(\frac{7}{2}\) and \(\frac{17}{4}\) inches. Let the width of the postcard be represented as x±y, therefore

​x -y = \(\frac{7}{2}\)

x + y = \(\frac{17}{4}\)

adding both the equation

2x = \(\frac{31}{4}\)

x = \(\frac{31}{8}\)

Putting the value of x=\(\dfrac{31}{8}\)  into equation (1), we get \(y=\dfrac{3}{8}\) . Therefore the width of the postcard can be represented as \(\dfrac{31}{8} \pm \dfrac{3}{8}\) inches.

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Answer:

∡3 ≅ ∡5 because they are alternate interior angles

∡3 ≅ ∡6 because they are alternate interior angles

∡3 ≅ ∡7 because they are corresponding angles

Answer:

what grade is this

Step-by-step explanation:

I'm very confused is this algebra?

In the case of a suspected myocardial infarction (MI) in a client presenting in the Emergency Department, certain medications should be administered as soon as possible to improve outcomes.

When a client presents with a suspected MI, time is of the essence in initiating appropriate treatment to minimize damage to the heart muscle and improve outcomes. The administration of aspirin is a priority as it helps prevent further blood clot formation. Aspirin inhibits platelet aggregation, reducing the risk of thrombus formation in the coronary arteries.

Nitroglycerin is another medication that should be given promptly. Nitroglycerin helps relieve chest pain by dilating coronary arteries and improving blood flow to the heart. It also helps reduce preload and afterload, thereby decreasing the workload on the heart.

In cases where the client experiences severe pain despite aspirin and nitroglycerin administration, morphine can be administered to alleviate pain and reduce anxiety. Morphine also helps dilate blood vessels, which can further improve blood flow to the heart.

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"A client presents in the Emergency Department with a suspected myocardial infarction (MI). Which medications should be given as soon as possible to improve outcomes?"