#1Write each of the following as a rational number.a. 11b. 3student work?C.-4

Answer:

254 ft

Step-by-step explanation:

first find the legs of the triangle.

you already know 65, and you would get another leg by 83-50, which equals 33.

Next use pythagorean theorem to find the other leg

a^2+b^2=c^2

rearrange so you find a leg

a^2=c^2-b^2

a^2=65^2-33^2

a^2=4225-1089

a^2=3136

a=56

Now you know that the missing leg is 56. Since that leg is the same as the short side of the rectangle, you 83+50+65+56=254

If circle with center N with m∠MNP=124 ∘ and MN=13,  the area of sector MNP is approximately equal to 194.86 square units.

To find the area of a sector of a circle, we need to use the formula:

Area of sector = (central angle/360°) x πr²

Where r is the radius of the circle.

In this problem, we know that the central angle m∠MNP is 124° and MN, which is also the radius of the circle, is 13. So we can substitute these values into the formula:

Area of sector = (124/360) x π(13)²

Area of sector ≈ 194.86

Therefore, the area of sector MNP is approximately equal to 194.86 square units.

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Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions are solved using the z-score formula.

In a set with mean \(\mu\) and standard deviation \(\sigma\), the zscore of a measure X is given by:

\(Z = \frac{X - \mu}{\sigma}\)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\(\mu = 51200, \sigma = 8200\)

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

\(Z = \frac{X - \mu}{\sigma}\)

\(Z = \frac{65000 - 51200}{8200}\)

\(Z = 1.68\)

\(Z = 1.68\) has a pvalue of 0.954

X = 55000

\(Z = \frac{X - \mu}{\sigma}\)

\(Z = \frac{55000 - 51200}{8200}\)

\(Z = 0.46\)

\(Z = 0.46\) has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

\(Z = \frac{X - \mu}{\sigma}\)

\(Z = \frac{48000 - 51200}{8200}\)

\(Z = -0.39\)

\(Z = -0.39\) has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

\(Z = \frac{X - \mu}{\sigma}\)

\(Z = \frac{41000 - 51200}{8200}\)

\(Z = -1.24\)

\(Z = -1.24\) has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

\(Z = \frac{X - \mu}{\sigma}\)

\(Z = \frac{61200 - 51200}{8200}\)

\(Z = 1.22\)

\(Z = 1.22\) has a pvalue of 0.889

X = 41200

\(Z = \frac{X - \mu}{\sigma}\)

\(Z = \frac{41200 - 51200}{8200}\)

\(Z = -1.22\)

\(Z = -1.22\) has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

As a result, option two (3.50) is correct.

Step-by-step explanation:                                                    
                                   
The code chunk that allows you to filter the data frame for chocolate bars with at least 70% cocoa and a rating of at least 3.5 pounds is as follows

∴ filter(Cocoa, percent>='70%'&Rating>=3.5)


Therefore, the code you write is facet wrap(Cocoa. Percent). Facet wrap() is a function in this code chunk that allows you to wrap a variable's facets. All of the ingredients are derived from cocoa beans. Cacao solids (the "brown" part, which contains health benefits and the unmistakable chocolatey flavor) and cacao butter (the "white" part, which contains the chocolate's fatty component) are split in proportion.


As a result, 3.5 appears in row 1 of your Tibble.

Other options I (iii), and (iv) are incorrect because the question states that any rating greater than or equal to 3.5 points can be considered a
high rating, and the rating here should be 3.5, not 3.75, 4.25, or 4

So, option (ii) is correct.  


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Answer:

17/25

Step-by-step explanation:

The equation for the probability of two events that are not mutually exclusive is:

p(A ∨ B) = p(A) + p(B) - p(A ∧ B)

A = the number is prime

B = the number is prime

The numbers are:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25

Here are the 8 prime numbers that satisfy event A:

3, 5, 7, 11, 13, 17, 19, 23

p(A) = 8/25

Here are the 13 numbers that are greater than 12 that satisfy event B:

13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25

p(B) = 13/25

Here are the 4 numbers that satisfy both event A and event B:

13, 17, 19, 23

p(A ∧ B) = 4/25

p(A ∨ B) = p(A) + p(B) - p(A ∧ B)

p(A ∨ B) = 8/25 + 13/25 - 4/25

p(A ∨ B) = 17/25

The probability that the number drawn is prime or greater than 12 = \(\frac{18}{25}\)

"Probability is a branch of mathematics which deals with finding out the likelihood of the occurrence of an event."

P(A) = n(A)/n(S)

where,  n(A) is the number of favorable outcomes, n(S) is the total number of events in the sample space.

For given question,

In a lottery game, a single ball is drawn at random from a container that contains 25 identical balls numbered from 1 through 25.

n(S) = 25

Let event A: the number drawn is prime

The prime numbers from 1 to 25 are:

2, 3, 5, 7, 11, 13, 17, 19, 23

So, n(A) = 9

The  probability that the number drawn is prime,

\(P(A)=\frac{n(A)}{n(S)}\\\\ P(A)=\frac{9}{25}\)

Let event B: the number drawn is greater than 12

So, n(B) = 13

The probability that the number drawn is greater than 12,

\(P(B)=\frac{n(B)}{n(S)}\\\\ P(B)=\frac{13}{25}\)

The number drawn is prime as well as greater than 12.

Such numbers are : 13, 17, 19, 23

n(A ∩ B) = 4

So, the probability that the number drawn is prime as well as greater than 12,

\(P(A\cap B)=\frac{n(A\cap B)}{n(s)}\\\\ P(A\cap B)=\frac{4}{25}\)

Using the equation P(AUB) = P(A) + P(B) - P(A ∩ B) to find the probability that the number drawn is prime or greater than 12,

\(\Rightarrow P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\\Rightarrow P(A\cup B)=\frac{9}{25}+ \frac{13}{25} -\frac{4}{25} \\\\\Rightarrow P(A\cup B)=\frac{9+13-4}{25}\\\\ \Rightarrow P(A\cup B)=\frac{18}{25}\)

Therefore, the probability that the number drawn is prime or greater than 12 = \(\frac{18}{25}\)

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Answer:D

Step-by-step explanation:

Answer:

196

Step-by-step explanation:

Answer:

196

Step-by-step explanation:

Answer: Changing the radius of an object by a given amount has a greater effect on the volume than changing the height by the same amount. The volume of a cylinder is given by the formula V = πr²h, where V is the volume, r is the radius, and h is the height. If we change the radius by a given amount, say x, the new radius would be r+x. Hence, the new volume would be V' = π(r+x)²h = π(r²+2rx+x²)h = V + 2πrxh + πx²h. We can see that the volume change equals 2πrxh + πx²h. The first term is proportional to both the radius and the height, whereas the second term is proportional to the square of the radius and the height. Assuming that the height change is also x, the new volume would be V'' = πr²(h+x) = V + πr²x. We can see that the volume change is proportional to the radius squared and the change in height. Therefore, changing the radius by a given amount has a greater effect on the volume than changing the height by the same amount.

Answer:

e-5=2f

take '-5' to the other side where '2f' is

e=2f+5

Step-by-step explanation:

\(y = 6 \tan( \frac{x}{2} ) - 3\)

Set y=0

\(0 = 6 \tan( \frac{x}{2} ) - 3\)

\(3 = 6 \tan( \frac{x}{2} ) \)

\( \frac{1}{2} = \tan( \frac{x}{2} ) \)

\( \tan {}^{ - 1} ( \frac{1}{2} ) + \pi = \frac{x}{2} \)

Multiply both sides by 2

\(2 \tan {}^{ - 1} ( \frac{1}{2} ) + 2\pi = x\)

So the x intercepts of the function is

\(2 \tan {}^{ - 1} ( \frac{1}{2} ) + 2\pi\)n

Let’s solve the equation x^2 - 4x = 5 by factoring:

First, we’ll move all the terms to one side of the equation:

x^2 - 4x - 5 = 0

Now, we’ll factor the left side of the equation. We’re looking for two numbers that multiply to -5 and add to -4. Those numbers are -5 and 1. So we can write:

(x - 5) (x + 1) = 0

Now we’ll use the zero-product property to solve for x. This property states that if the product of two numbers is zero, then at least one of the numbers must be zero. So we have:

x - 5 = 0 or x + 1 = 0

Solving each equation separately, we find that x = 5 or x = -1.

So, the solutions to the equation x^2 - 4x = 5 are x = 5 and x = -1.

Answer: C

Step-by-step explanation:

1. divide 3 by 2 to find out how many pounds of blueberries she picked each day. 3/2 is 1.5

2. multiply 1.5 by 16 to convert to ounces. 1.5x16 is 24.

3. she picked 24 ounces of blueberries each day.

The quadratic equation so formed will be \(x^{2}\) - 2x - 15 = 0.

What is a quadratic equation?

Any algebraic equation that can be expressed in standard form as where x represents an unknown value and where a, b, and c represent known values, where a 0 is a quadratic equation.

We know that the standard form of a quadratic equation is represented as: a\(x^{2}\) + bx + c = 0.

Here, we are given a = 4, b = -8 and c = -60.

On substituting these values, we get

⇒ a\(x^{2}\) + bx + c = 0

⇒ 4\(x^{2}\) - 8x - 60 = 0

On dividing both sides by 4, we get

⇒ \(x^{2}\) - 2x - 15 = 0

Hence, the quadratic equation so formed will be \(x^{2}\) - 2x - 15 = 0.

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Step-by-step explanation:

Given expressions:

1. x² + x

2. 5x² - 25x

3. x² - 16

4. x² + 4x + 4

5. x² - 5x + 6

   Problem: Factor each of the quadratic expression

Solution:

A quadratic expression is an expression whose highest power is two.

1. x² + x

     For this expression:

       x² + x

    Factorizing x since  it is a common term;

       x(x + 1)          

2. 5x² - 25x

    For this expression;

   Factorize 5x since it common to both of them;

      5x(x - 5)

3. x² - 16

    For this expression;

  Apply the differences of two squares;

     x² - 16

     x² - 4²;

    Note;  x² - y²  = (x-y)(x+y)

So, x² - 4² = (x-4)(x + 4)

4. x² + 4x + 4

   To solve this,

     Find two numbers whose product will be 4 and sum also 4;

  x² + 4x + 4 = x² + 2x + 2x + 4

                    = x(x + 2) + 2(x + 2)

                     = (x + 2)(x + 2)

5. x² - 5x + 6

    To factorize this expression, simply find two numbers whose sum is -5 and product is 6;

            x² - 5x + 6 = x² - 3x - 2x + 6

                              = x (x - 3) -2(x - 3)  

                              = (x -2)(x-3)

The correct answer is that the data set {3, 7, 4} is represented in the given histogram.(option-a)

The given histogram represents the number of children in each age group who need to travel to school. Since the histogram has only three bars, we can conclude that there are only three age groups.

The first bar represents children aged 5-10, of which there are 3. The second bar represents children aged 11-13, of which there are 7. The third bar represents children aged 14-18, of which there are 4.

Therefore, the data set that is represented in the histogram is:

{3, 7, 4}

None of the other data sets given match the values in the histogram. The first data set has duplicate values and is not sorted by age group. The second data set includes ages that are not represented in the histogram. The third data set has values for ages 6, 11, 12, 13, 14, 15, 17, and 18, but the histogram does not have bars for all those ages. (option-a)

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The percentage of buyers is approximately 68.26% of buyers of new houses paid between \($149,000\) and \($151,000\) .

We are given that the prices of the new homes are normally distributed with a mean of \($150,000\) and a standard deviation of $1000.

Using the 68-95-99.7 rule, we know that: approximately 68% of the data falls within one standard deviation of the mean approximately 95% of the data falls within two standard deviations of the mean, approximately 99.7% of the data falls within three standard deviations of the mean.

In order to determine the proportion of customers who spent between $149,000 and , we must first determine the z-scores for these values:

z1 = (149,000 - 150,000) / 1000 = -1 z2 = (151,000 - 150,000) / 1000 = 1

Now, we can determine the proportion of data that falls between z1 and z2 using the z-table or a calculator. The region to the left of z1 is 0.1587, and the area to the left of z2 is 0.8413, according to the z-table. Thus, the region bounded by z1 and z2 is:

0.8413 - 0.1587 = 0.6826

We can get the percentage of consumers who spent between by multiplying this by 100% is  \($149,000\)  and \($151,000\):

0.6826 x 100% = 68.26%

Therefore, the standard deviation of customers who paid between is \($149,000\) and \($151,000\)  for this model of new homes.

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Answer:

(b), (d) and (e)

Step-by-step explanation:

Given

\(p(t) = 1 * e^t\)

See attachment for \(y = p(t)\)

Required

Select true statements from the given options

(a) \(\ln(30)\) = days the bacteria reaches 30000

We have:

\(p(t) = 1 * e^t\)

In this case:

\(t = \ln(30)\) and \(p(t) = 30000\)

So, we have:

\(30000 = 1 * e^{\ln(30)}\)

\(30000 = e^{\ln(30)}\)

Using a calculator, we have:

\(e^{\ln(30)} = 30\)

So:

\(30000 = 30\)

The above equation is false.

(a) is not true

(b) The graph shows that \(\ln(20) \approx 3\)

We have:

\(p(t) = 1 * e^t\)

Let t = 3

So;

\(p(3) = 1 * e^3\)

From the graph, \(p(3) = 20\)

So:

\(20 = 1 * e^3\)

\(20 = e^3\)

Take natural logarithm of both sides

\(\ln(20) = \ln(e^3)\)

This gives:

\(\ln(20) = 3\)

(b) is true

(c) \(\ln(t) = y\) is the logarithm form of \(y = e^t\)

We have:

\(y = e^t\)

Take natural logarithm of both sides

\(\ln(y) = \ln(e^t)\)

This gives:

\(\ln(y) = t\)

\(\ln(y) = t\)  \(\ne\) \(\ln(t) = y\)

(c) is false

(d) \(e^4 > 50\) and  \(\ln(50) < 4\)

From the graph, we have:

\(e^4 = 54\) --- rough readings

This implies that:

\(e^4 > 50\) is true

Because \(54 > 50\)

Take natural logarithm of both sides

\(\ln(54) > \ln(50)\)

Rewrite as:

\(\ln(50) < \ln(54)\)

We have:

\(e^4 = 54\)

Take natural logarithm of both sides

\(\ln(e^4) = \ln(54)\)

\(4 = \ln(54)\)

\(\ln(54) = 4\)

Substitute \(\ln(54) = 4\) in \(\ln(50) < \ln(54)\)

\(\ln(50) < 4\)

(d) is true

(e) The graph shows that \(10 \approx \ln(2.3)\)

We have:

\(p(t) = 1 * e^t\)

Let t = 2.3

So;

\(p(2.3) = 1 * e^{2.3}\)

From the graph,

\(p(2.3) = 10\) ---- rough readings

So:

\(10 = 1 * e^{2.3}\)

\(10 = e^{2.3}\)

Take natural logarithm of both sides

\(\ln(10) = \ln(e^{2.3})\)

This gives:

\(\ln(10) = 2.3\)

(e) is true

The value of this constant, by assuming that the pressure at some reference height y₀ is p₀, is p - p₀ = p.g(y₀ - y).

dp = -pg dy

At height y = y₀ and p = p₀

∫dp = ∫-pg dy

⇒p = -pgy + C        Eqn(1)

Where c is the constant of integration

We know that y = y₀ and p = p₀

⇒p₀ = -pgy₀ + C

⇒C = p₀ + pgy₀

Put the value of C in Eqn(1)

⇒p = -pgy + p₀ + pgy₀

⇒p - p₀ = -pgy + pgy₀

⇒p - p₀ = pg(y₀ - y)

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Answer:

d = 7h +5

Step-by-step explanation

If you go to K-12 then this is the right answer I just took the quiz.

Answer:

d = 7h +5

Answer:

The answer is 32

Step-by-step explanation:

The formula for finding the area of a trapezoid is:

         ((base 1 + base 2)/ 2 )* h

Now all you have to do is substitute the numbers in.

Note: bases will always be the ones like 12 and 4 in this case. We have just named then 1 and 2.

Answer:

32 square units

Step-by-step explanation:

\(\displaystyle A=\frac{1}{2}(b_1+b_2)h=\frac{1}{2}(12+4)(4)=\frac{1}{2}(16)(4)=\frac{1}{2}(64)=32\)

Note that \(b_1\) and \(b_2\) are the lengths of each base of the trapezoid, so it doesn't matter which is which.

Let's make a system of equations:

144/x=y

168/(x+5)=y

So,

144/x=168/(x+5)

Solving using cross multiplication,

x=30

So, her average speed for driving 144 miles was 30 miles per hour.

-Hunter

Answer:

Your math problem isnt showing up

Answer:  55 degrees  (choice B)

============================================================

Explanation:

Let x be the measure of each congruent base angle.

The three interior angles of any triangle always add to 180 degrees.

x+x+70 = 180

2x+70 = 180

2x = 180-70

2x = 110

x = 110/2

x = 55 degrees

Answer:

55 degrees

Step-by-step explanation:

Since the triangle is isosceles there are two congruent base angles and 1 vertex angle. The three angles add up to 180 degrees.

180 minus the vertex angle of 70 degrees leaves 110 degrees for the remaining two angles. Divide 110 by 2 to get the measure of each base angle.

180 - 70 = 110/2 = 55 degrees

Answer:

For which questions dud

Answer:

if you can provide the figure it would be quite easy for the people giving answers

Answer:

  1/60 gallon

Step-by-step explanation:

You want to know what is left of 4/5 gallon of paint if 3/4 gallon was used to paint a doghouse, and 2/3 of the remaining amount was used to paint a door.

To find the amount remaining we subtract the amount used for the doghouse from the original amount:

  4/5 -3/4 = (16 -15)/20 = 1/20 . . . . . gallon

2/3 of that was used, so 1/3 of it remains:

  (1/3)(1/20 gallon) = 1/60 gallon

There is 1/60 of a gallon of paint left in the can.

__

Additional comment

In a standard size gallon paint can, that's about 0.11 inches of paint in the bottom.

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Approximately 138.3 grams of the compound will remain after 8 hours. The decay of a radioactive compound is an exponential process, which can be modeled by the equation:

N(t) = N₀ * e^(-λt)

where N(t) is the amount of the compound remaining at time t, N₀ is the initial amount of the compound, λ is the decay constant, and e is the mathematical constant known as Euler's number.

To find the amount of the compound remaining after 8 hours, we need to plug in the given values into the equation above. We know that the initial mass of the compound is 260 grams, and the decay rate is 3.8% per hour, which means that the decay constant λ is equal to 0.038. Therefore, the equation becomes:

N(8) = 260 * e^(-0.038 * 8)

Simplifying this equation gives:

N(8) = 260 * e^(-0.304)

N(8) ≈ 138.3 grams

Therefore, approximately 138.3 grams of the compound will remain after 8 hours.

It's important to note that this calculation assumes that the decay rate remains constant over time, which may not always be the case in reality. Additionally, the actual amount of the compound remaining may vary due to experimental error or other factors.

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Answer:

They must mix 70 kilograms of the 10% chocolate and 30 kilograms of the 60% chocolate.

Step-by-step explanation:

Given that a candy distributor needs to mix a 10% fat-content chocolate with a 60% fat-content chocolate to create 100 kilograms of a 25% fat-content chocolate, to determine how many kilograms of each kind of chocolate must they use, the following calculations must be performed:

10 x 0.5 + 60 x 0.5 = 35

10 x 0.6 + 60 x 0.4 = 30

10 x 0.7 + 60 x 0.3 = 25

100 x 0.7 = 70

Thus, they must mix 70 kilograms of the 10% chocolate and 30 kilograms of the 60% chocolate.

11, - 6, -1, 4, 9, 14, 19, ... are mapped onto 4. ... A;-l = (-ltQn (mod N),. (A5.34) ... 131 2, 14, 34, 38, 42, 78, 90, 178, 778, 974(1000).

Step-by-step explanation:

the nearest .1/2 incp, we say the 'unit 131/2 incl. 1.4 a measUrement is-stated tp- be 546 inch, this UM= the