is 2 the only even prime number

After simplifying the number, the number 4.38 × 10^{–3} = 0.00438 is the smallest number. So the number b is the smallest number.

We simplify the numbers to find the smallest number.

The numbers are

a. 4.38 × 10^3

We can write 10^3 as 1000

Now the number is

= 4.38 × 1000

Multiply

= 4380

b. 4.38 × 10^{–3}

The number of times to divide the base number is indicated by negative power. For each power of 10, the decimal points are moved to the left when multiplying the integer by the negative power of 10.

So we can write that number as

= 4.38 × 1/1000

Simplify

= 0.00438

c. 4.38 × 10^{–2}

We can write that number as

= 4.38 × 1/100

Simplify

= 0.0438

d. 438

e. 4.38 × 10^2

We can write that number as

= 4.38 × 100

Multiply

= 438

We can see that the number 4.38 × 10^{–3} = 0.00438 is the smallest number.

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The complete question is:

Which number is the smallest?

a. 4.38 × 10^3

b. 4.38 × 10^{–3}

c. 4.38 × 10^{–2}

d. 438

e. 4.38 × 10^2

Answer:

[tex]\frac{1}{x^2}[/tex]

Step-by-step explanation:

first, simplify [tex](x^{10})^\frac{1}{5}[/tex]  [tex]\frac{10}{1} =\frac{1}{5}[/tex] , 5 and 10 cancel each other, so [tex]\frac{2}{1}[/tex] or 2.  [tex]x^2[/tex]

secondly, [tex]\sqrt{x^8}[/tex], we need to only do [tex]\sqrt{8}[/tex] which is 4, so answer is [tex]x^4[/tex].

then divide [tex]\frac{x^2}{x^4}[/tex] , minus 2 - 4= -2, [tex]x^-2[/tex]

but if there was any negative exponent in math it needs to be down so answer is [tex]\frac{1}{x^2}[/tex].

Answer:

1/x² = x^(-2)

Step-by-step explanation:

((x¹⁰)^(1/5))/sqrt(x⁸)

4 things are important to remember here :

[tex] {x}^{a \div b} = \sqrt[b]{ {x}^{a} } [/tex]

so, for example, a square root is the same as 1/2 as exponent.

now we can simplify.

let's start with the numerator :

(x¹⁰)^(1/5) = x^(10×1/5) = x²

then the denominator :

sqrt(x⁸) = x^(8×1/2) = x⁴

the whole fraction looks like

x²/x⁴

remember about the subtraction of exponents when dividing :

x²/x⁴ = x^(2-4) = x^-2 = 1/x²

FYI

when you remember, what an exponent actually is : the number how often the base had to be multiplied by itself, ...

x×x / x×x×x×x

the top 2 x can be reduced to 1 by division by 2 of the bottom x.

what is left is 1/x²

D. N = 24, I% = 3.6, PV =, PMT = -415, FV = 0, P/Y = 12, C/Y = 12, and PMT:END are the values in the group that would provide the same result as the given equation.

The present value of a sum of money is its current value. For instance, the present value is what $110 is currently worth if you are promised $110 in a year.

In the TVM solver, we have;

I = The annual percentage rate

N = n × t

t = The number of years

PV = Present value

PMT = Payment

P/Y = Payments made annually = n

C/Y equals the number of compounding periods in a year.

The following is how the monthly payment formula is stated.

M = P(r/n)(1+r/n)ⁿˣⁿ/(1+r/n)ⁿˣⁿ-1

 which gives,

P = M [(1+r/n)ⁿˣⁿ-1] / (r/n)(1+r/n)ⁿˣⁿ

Therefore, we get:

where:

M = PMT = -415

P = PV

r = I

P/Y = n = 12

Therefore,

0.003 = I/12

I = 0.003 × 12 = 0.036 = 3.6%

N = n×t = 24

P = (415)[(1+I/12)²⁴- 1] /(I/12)(1+I/12)²⁴

  = ( 415) [(1+0.003)²⁴ - 1] / (0.003)(1+0.003)²⁴ = ?

The present value, PV, is the equation's value.

We have FV = 0 when payments are paid based on the PV.

The set of values with the same value as ($415)(1+0.003)24 - 1 / (0.003)(1+0,003)

24 has the following values when entered into a calculator's TVM solver: D. N = 24, I% = 3.6, PV =, PMT = -415, FV = 0 and PMT:END.

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Answer:

Step-by-step explanatio

23.2

The volume of B will be 1350 cm³.

What is volume?

In three-dimensional space, volume is the area occupied by an object within its borders. Another name for it is an object's capacity.

If two solids are similar, the ratio of their volumes is equal to the cube of the ratio of their corresponding sides.

we have given that,

Height of A = 10 cm.

Volume of A = 400cm³.

Height of B = 15 cm.

Both A and B solids are similar .

from above told concept we get,

→ V(A) / V(B) = (Height of A)³ / (Height of B)³

Putting all given values we get,

→ 400/ V(B) = (10)³/(15)³

→ 400/V(B) = (1000/3375)

→ V(B) = (400 * 3375) / 1000

→ V(B) = (4 * 3375) / 10

→ V(B) = 1350 cm³. (Ans.)

Hence, Volume of B will be 1350 cm³.

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Each DVD cost is $17.80

This is a simple equation

From the question we have the following information :

1. Total spent for 2 dvd ( before tax ) = $17.80

2. Sales Tax for those spending = $2.85

Let's call the cost of each DVD "x".

The total cost of both DVDs would be 2x.

We know that the total cost of both DVDs and the sales tax was $38.45, so we can set up an equation:

2x + $2.85 = $38.45

Solving for x:

2x = $38.45 - $2.85 = $35.60

x = $35.60 / 2 = $17.80

Hence from explanation above, each DVD cost $17.80.

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Given the lengths of the two sides and the angle between them, only one triangle can be created under the given circumstances.

1. Given that angle B is 32.5°, side AB is 37 cm, side AC is 26 cm, etc.

2. Calculate side BC using the Law of Cosines:

BC = (2(AB)(AC)cosB) + (AB)(AC)2

3. Input the values that are known: BC = (37 2 + 26 2 - 2(37)(26)cos32.5°)

4. Condense: BC = (1369 plus 676 minus 1848 cos 32.5 °)

5. Determine BC =. (2095 - 1539.07)

6. Condense: BC = 556.93

7. Determine BC as 23.701 cm.

8. Since the lengths of the two sides and the angle between them are specified, only one triangle can be formed under the current circumstances.

By applying the Law of Cosines, we can determine the length of the third side, BC, given that side AB is 37 cm, side AC is 26 cm, and angle B is 32.5°. In order to perform this, we must first determine the cosine of angle B, which comes out to be 32.5°. Then, we enter this value, together with the lengths of AB and AC, into the Law of Cosines equation to obtain BC.BC = (AB2 + AC2 - 2(AB)(AC)cosB) is the equation. BC is then calculated by plugging in the known variables to obtain (37 + 26 - 2(37)(26)cos32.5°). By condensing this formula, we arrive at BC = (1369 + 676 - 1848cos32.5°). Then, we calculate BC as BC = (2095 - 1539.07), and finally, we simplify to obtain BC = 556.93. Finally, we determine that BC is 23.701 cm. Given the lengths of the two sides and the angle between them.

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Answer:

9

Step-by-step explanation:

Let's suppose "my number" is x

If you add 21 to my number: 21 + x

Multiply the result by 3: 3(21+x)

You will get 84 more than two thirds of my number

[tex]3(21 + x) = 84 + \frac{2}{3} x \\ 63 + 3x = 84 + \frac{2}{3} x \\ \frac{7}{3} x = 21 \\ 7x = 63 \\ x = 9[/tex]

Therefore my number is 9

If you need any help with the algebra part feel free to let me know!

The linear equation of f(n) can be written as mn + b, where m is the slope of the line and b is the y-intercept.

A linear equation is one that has the formula Ax + B = 0, where A and B are constants and x is a variable. When graphed, these equations produce a straight line, which is why they are termed linear equations. Linear equations are used to describe the connection between two variables known as the independent variable (x) and the dependent variable (y) (y). When the value of the independent variable is known, the equation may be used to discover the value of the dependent variable.

The linear equation of f(n) can be written as: f(n) = mn + b, where m is the slope of the line and b is the y-intercept.

To find the equation, we need to calculate the slope m and the y-intercept b. We can do this by using the given values of n and f(n).

First, let's calculate the slope m. We can use the formula m = (y2 - y1)/(x2 - x1), where (x1, y1) and (x2, y2) are two points on the line. In this case, we can use the points (6, -1) and (4, 2):

m = (2 - (-1))/(4 - 6) = 3/-2 = -1.5

Next, let's calculate the y-intercept b. We can use the formula b = y - mx, where (x,y) is a point on the line and m is the slope. In this case, we can use the point (2, 5):

b = 5 - (-1.5)(2) = 5 + 3 = 8

Therefore, the linear equation of f(n) is f(n) = -1.5n + 8.

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There are 3,265,920 arrangements of the letters in mathematics.

Consider T+H together which is easy to solve, and also consider the “TH” as one letter. There are two Ts in maThemaTics, and they can be together as TH or HT.

so there are actually 2*2=4 ways to build the TH itself. Now just plug a short hold on the “not followed by the h” and let us know how many ways to build it with just the TH restriction.

So with TH together, we can construct 10 letters in the word, two As, two Ms, and 1 as each of the remaining 6 letters. So it means there are

10!2!2!(1!)6=10!2!2!=36288004=907200

Now there are 4 ways to construct the TH, so 907200∗4=3,628,800.

To then add the “not followed by h” restriction, we should first determine all of the combinations where you have “THE” or “HTE” in the word. There are 4 ways to construct this letter grouping.

Then we have the remaining 9 letters in the word, two As, two Ms, and 1 of each of the remaining 5 letters.

So there are 4∗9!2!2!(1!)5=4∗9!4=3628802=362,880 ways to construct the word where you force THE/HTE together.

Finally, subtract the arrangements we have THE/HTE in the word from the arrangements where you just have TH/HT in order to get the arrangements that have TH/HT that is not immediately followed by h. 3,628,800–362,880=3,265,920.

Therefore, there are 3,265,920 arrangements.

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.

Yes, a triangle can be formed with side lengths of 3 inches, 3 inches, and 5 inches.

A triangle can be formed if and only if the sum of the lengths of any two sides is greater than the length of the third side. In this case, the sum of the lengths of sides 3 inches and 3 inches (6 inches) is greater than the length of the third side (5 inches), so a triangle can be formed.

In summary, a triangle can be formed with side lengths 3 inches, 3 inches, and 5 inches because the sum of the lengths of any two sides is greater than the length of the third side.

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Each person get nine and two sixths gummy worms

Is a number that expresses the portion of some number over a total. The number that expresses the portion is known as the numerator and the number that expresses the total is known as the denominator.

Information about the problem:

Calculating how many gummy worms each person get:

Gummy worms per person = 56/6

Gummy worms per person = 9.33

Rounding down = 9 gummy worms per person

Total gummies gave = 9 * 6

Total gummies gave = 54

Gummies left = 56 - 54

Gummies left = 2

Gummy worms per person = 9 2/6

Nine and two sixths gummy worms

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Answer:

When x=4, ln x-ln y=y-4, so ln 4-ln 4=4-4, which is true. Therefore, when x=4, y=4, and dy/dx=0.

Step-by-step explanation:

So, when x = 4, the value of the differentiable function dy/dx is 0.

A differentiable function of one real variable is one that has a derivative at each point in its domain. In other words, a differentiable function's graph has a non-vertical tangent line at each interior point in its domain.

Here,

Given the equation ln x - ln y = y - 4, we can rearrange it to get ln(x/y) = y - 4. Taking the derivative of both sides with respect to x using the chain rule:

d/dx (ln(x/y)) = d/dx (y - 4)

(1/x) (dx/dx) - (1/y) (dy/dx) = dy/dx

dy/dx = (1/y) (dx/dx) + (1/x) (dy/dx) = (1/y) + (1/x) (dy/dx)

Rearranging and solving for dy/dx:

(x/y) (dy/dx) = (1/y) - (1/x)

dy/dx = (x/y^2) (1/x - 1/y) = (x/y^2) (1/x - 1/y)

We can substitute x = 4 and y = 4 into the expression to find the value of dy/dx when x = 4:

dy/dx = (4/16) (1/4 - 1/4) = 0

So, the value of differentiable function dy/dx when x = 4 is 0.

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The truth value of the given statements if the universe of discourse of each variable is the set of real numbers are 1. false, 2. false, 3. true, 4. false, 5. false, 6. false, 7. true, 8. true, 9. false, 10. true, 11. true, and 12. true.

1. False. There is no real number x such that x^2 = -1.

2. False. For any x in the real numbers, there exists a y such that xy ≠ 1. For example, if x ≠ 0, then y = 1/x will satisfy xy ≠ 1.

3. True. For any real number x, the real number y = x^2 satisfies the equation x^2 = y.

4. False. For any x and y in the real numbers, x + y = y + x.

5. False. For any x in the real numbers, there exists a y such that xy ≠ 0. For example, if x ≠ 0, then y = 1/x will satisfy xy ≠ 0.

6. False. There is no real number y such that y^2 = -1.

7. True. For any x and y in the real numbers, the real number z = x + y^2 satisfies the equation z = x + y^2.

8. True. For any x ≠ 0 in the real numbers, the real number y = 1/x satisfies the equation xy = 1.

9. False. There is no real number x such that x^2 = 2.

10. True. For any x in the real numbers, the real number y = 1 - x satisfies the equation x + y = 1.

11. True. The solution to the system of equations x + 2y = 2 and 2x + 4y = 5 is x = -1 and y = 3.

12. True. For any x in the real numbers, the real number y = (1 - x)/2 satisfies the equation (x + y = 2) and (2x - y = 1).

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The result after the endpoint of a line segment is reflected is (-5. -10)

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

Point = (-5, 10)

The reflection rule is given as

Reflection across the x-axis

Mathematically, this is represented as

(x, y) = (x, -y)

So, we have

Image = (-5, -10)

Hence, the image is (-5, -10)

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After finding the z score, total 49.865% you could expect to score between 60 and 120.

Math exams mean(μ) = 120

Math exams standard deviation(σ) = 20

We have to determine the percent you could expect to score between 60 and 120.

The area total included to the left of any score or value is indicated in the Z-score table (x). The top row and the first column of the Z-table represent the Z-values, while all of the numbers in the middle represent the areas.

P(60 < x < 120) = P((60 - 120)/20 < (x - μ)/σ < (120 - 120)/20)

Simplifying

P(60 < x < 120) = P((-60)/20 < (x - μ)/σ < 0/20)

P(60 < x < 120) = P(-3 < Z < 0)

We can write this as

P(60 < x < 120) = P(Z > -3) - P(Z < 0)

P(60 < x < 120) = 0.99865 - 0.5

P(60 < x < 120) = 0.49865

P(60 < x < 120) = 49.865%

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The probability the ball drawn from the second box is red, is 17/56.

In a box are 7 red balls and 8 blue balls.

From this box are drawn 4 balls and placed in a second box.

We have to determine the probability the ball drawn from the second box is red.

The total number of balls = Red Balls + Blue Balls

The total number of balls = 7 + 8

The total number of balls = 15

The number of balls placed in second balls = 4

The number of combination of balls in box 2.

BBBB, BBBR, BBRR and BRRR

Then, one ball is drawn from the second box.

The probability the ball drawn from the second box is red.

P(Ball is red in the second box) = P(BBBB) + P(BBBR) + P(BBRR) + P(BRRR)

P(Ball is red in the second box) = [tex]\frac{^{5}C_{4}}{^{8}C_{0}}(0)+\left(\frac{^{5}C_{3}\times ^{3}C_{1}}{^{8}C_{4}}\right)\times\frac{^{1}C_{1}}{^{4}C_{1}}+\left(\frac{^{5}C_{2}\times ^{3}C_{2}}{^{8}C_{4}}\right)\times\frac{^{2}C_{1}}{^{4}C_{1}}+\left(\frac{^{5}C_{1}\times ^{3}C_{3}}{^{8}C_{4}}\right)\times\frac{^{3}C_{1}}{^{4}C_{1}}[/tex]

P(Ball is red in the second box) = [tex]0+\left(\frac{30}{70}\times\frac{1}{4}\right)+\left(\frac{30}{80}\times\frac{1}{2}\right)+\left(\frac{5}{70}\times\frac{3}{4}\right)[/tex]

P(Ball is red in the second box) = 3/48 + 3/16 + 3/56

P(Ball is red in the second box) = (21 + 63 + 18)/336

P(Ball is red in the second box) = 102/336

P(Ball is red in the second box) = 17/56

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The 95% of regular grade gasoline sells for between $3.23 & $3.63 and per gallon (to decimal) and 85% of regular grade gasoline sells for between $3.23 & $3.53 and per gallon (to decimal) and 16% percentage of regular grade gasoline sells for less than $3.53 per gallon (to decimal).

Suppose that mean retail price per gallon of regular grade gasoline 3.43.

empirical rule explains that  approximately 68%,95% and 99.7% of data lies between (µ ± 1*σ) ,(µ ± 2*σ) and (µ ± 3*σ) respectively.

⇒If some data {x} are normally distributed, the corresponding {z} will be normal with mean 0 and standard deviation 1 where the correspondence between the {x} and {z} is given by

Such z's are called z-scores.

(µ ± 1*σ)=68%

(µ ± 2*σ)=95%

(µ ± 3*σ)=99.7%

a)

P(3.23<x<3.63) =[tex]P[ \frac{(x1- \mu)}{\sigma} < z < \frac{(x2- \mu)}{\sigma}} ][/tex]

=P[( [tex]\frac{3.23-3.43}{0.10}[/tex] < z < [tex]\frac{3.63-3.43}{0.10}[/tex] ]

=P[-2 < z < 2 ]

=0.95

=95.0 %

b)

P(3.23<x<3.53) =[tex]P[ \frac{(x1- \mu)}{\sigma} < z < \frac{(x2- \mu)}{\sigma}} ][/tex]

=P[( [tex]\frac{3.23-3.43}{0.10}[/tex] < z < ([tex]\frac{3.53-3.43}{0.10}[/tex] ]

=P[-2 < z < 1 ]

=0.815

=81.5 %

c)

P(x>3.53) =[tex]P[ z > \frac{x-\mu}{\sigma} ][/tex]

=P[ z > [tex]\frac{3.53-3.43}{0.10}[/tex] ]

=P[ z > 1 ]

=0.16

=16.0 %.

Therefore, 95% of regular grade gasoline sells for between $3.23 & $3.63 and per gallon (to decimal) and 85% of regular grade gasoline sells for between $3.23 & $3.53 and per gallon (to decimal) and 16% percentage of regular grade gasoline sells for less than $3.53 per gallon (to decimal).

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The sum of 9/10 and 5/6 is 26/15.

The sum of two fractions can be calculated only if the denominator (the bottom number) is the same. Here, in 9/10, 10 is the denominator and in 5/6, 6 is the denominator. To equate two denominators LCM(least common multiple) should be done.

LCM of 10 and 6 is 30.

To convert is to multiply each by the bottom number of the other over the other

9/10 + 5/6

= 9/10×3/3 + 5/6×6/6

=27/30 + 25/30

=52/30

=26/15 or 1 11/15

Therefore, the sum of 9/10 and 5/6 is 26/15 or 1 11/15.

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The sum of 9/10 and 5/6 equals 26/15.

Only if the denominator (the bottom number) is the same can two fractions' total be calculated. In this case, the denominator in 9/10 is 10 and in 5/6 is 6. The LCM (least common multiple) method should be used to equate two denominators.

30 is the LCM of 10 and 6.:

By multiplying each by the bottom number of the other over the other, you may convert.

9/10 + 5/6

= 9/10×3/3 + 5/6×6/6

=27/30 + 25/30

=52/30

=26/15 or 1 11/15

As a result, 9/10 plus 5/6 equals 26/15.

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If you join Gym A, it will cost $20 to join for 1 month plus $3 per visit. If you visit the gym x times, your total cost will be 20 + 3x dollars.

If you join Gym B, it will cost a flat rate of $40 for 1 month.

To determine which membership is more cost-effective, you would need to compare the costs for different number of visits. For example, if you plan to visit Gym A 10 times in a month, your total cost would be 20 + 3(10) = $50. If you plan to visit Gym B only 2 times in a month, membership B is more cost-effective. If you plan to visit more than 13 times, membership A is more cost-effective.

It would be helpful to make a table with different number of visits to compare the cost of both gym memberships.

Answer:

249.6 in^2

Step-by-step explanation:

Since it's an equilateral triangle as the base, all of the other faces will be the same equilateral triangle, so 62.4 x 4 gives you the answer.

The interest rate is -17.72%

What is the interest rate?

The proportion of a loan that is charged as interest to the borrower is typically expressed as an annual percentage of the loan outstanding.

The formula for finding the interest rate is:

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

Where A is the end balance, P is the starting principal, r is the interest rate, n is the number of compounding periods per year, and t is the number of years.

Since the interest rate is compounded semi-annually, n = 2. And the starting principal is $4,848, t = 6 years. So the equation becomes:

$6,520.02 = $4,848 * (1 + r/2)^(2 * 6)

Now we'll isolate r by taking the natural logarithm of both sides:

ln(6520.02 / 4848) = ln((1 + r/2)^(2 * 6))

ln(6520.02 / 4848) = 12 * ln(1 + r/2)

Now we'll solve for r by dividing both sides by 12 and taking the exponential of the result:

r/2 = (ln(6520.02 / 4848) / 12)^(1/12) - 1

r = 2 * ((ln(6520.02 / 4848) / 12)^(1/12) - 1)

Finally, we'll convert the interest rate to a percentage by multiplying by 100:

r = (2 * ((ln(6520.02 / 4848) / 12)^(1/12) - 1)) * 100%

= (2 * ((ln(1.34814) / 12)^(1/12) - 1)) * 100%

= (2 * (0.113774 - 1)) * 100%

= (2 * -0.886226) * 100%

= -17.72452%

hence, the interest rate is -17.72%

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To minimize the time it takes us to get to the town, we should swim to a point that is √(1/14) miles from the town.

We can use the Pythagorean theorem to solve this problem. Let's assume the point where we swim to shore is x miles from the town.

The distance we need to swim is √(1^2 + x^2) = √(x^2 + 1)

The distance we need to walk is 3-x

The total time it takes us to get to the town is the sum of the time it takes us to swim and walk.

So, Time = (distance to swim)/(swimming speed) + (distance to walk)/(walking speed)

Time = √([tex]x^2[/tex] + 1)/2 + (3-x)/4

To minimize the time, we need to find the value of x that minimizes the expression above. To do this, we can take the derivative of the expression with respect to x and set it equal to zero:

d/dx (√( [tex]x^2[/tex]  + 1)/2 + (3-x)/4) = 0

Expanding the derivative:

(x/(2√(  [tex]x^2[/tex]  + 1))) - 1/4 = 0

Solving for x, we get,

x/(2√([tex]x^2[/tex] + 1)) = 1/4

x = √([tex]x^2[/tex] + 1)/4

Squaring both sides, we get,

[tex]x^2[/tex]   = ([tex]x^2[/tex]  + 1)/16

15[tex]x^2[/tex]  = x^2 + 1

14[tex]x^2[/tex]   = 1

[tex]x^2[/tex]  = 1/14

Taking square root, we get,

x = ± √(1/14)

Since x must be positive, we have x = √(1/14)

So, to minimize the time it takes us to get to the town, we should swim to a point that is √(1/14) miles from the town.

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The complete question is -

I am 1 mile in the ocean and wish to get to a town 3 miles down the coast. I need to swim to the shore and then walk along the shore and I can swim at 2 mph and walk at 4 mph. To what point on the shore should I swim to minimize the time it takes me to get to the town?

Answer:

D)  64.25

Step-by-step explanation:

Break the figure up into 2 shapes:  half-circle with radius of 5;  triangle with base of 10 and height of 5

Area-half-circle = πr² = (π(5)²) / 2 = 25π /2 = 39.27

Area-triangle = 1/2bh = 1/2(10)(5) = 25

Atotal = 25 + 39.27 = 64.27

closest answer is D)  64.25.  They used π=3.14 to get this answer.

With the money they have left, the students in M. Houton's fifth grade class of 20 students can buy 32 foils.

This is calculated by subtracting the cost of the transportation from the total amount raised ($128 - $5*20 = $28) and then dividing that amount by the cost of each foil ($4). Therefore, the students can buy $28 / $4 = 32 foils.

To calculate how many foils the students in M. Houton's fifth grade class of 20 students can buy with the money they have left, you must subtract the cost of their transportation from the total amount raised ($128 - $5*20 = $28).

Then, divide the remaining amount by the cost of each foil ($4) to get the number of foils they can buy with the money they have left ($28 / $4 = 32 foils).

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Answer:

0.35 mL

Step-by-step explanation:

There are 1000 mL in 1 Liter. 350 of 1000 is 0.35

The border must have a width of 10 meters.

We first have a rectangular lawn of 10m by 5m, so if we have a border with a width of x meters, the area of the border is given by:

A = 10m*x + 5m*x

Where we compute the area in that way because the flowers will be only on two adjacent sides of the rectangle.

And the area is 150 m^2, then we will get:

150m^2 = 10m*x + 5m*x

150m^2 = 15m*x

150m^2/15m = x

10m = x

The border has a width of 10 meters.

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The answers to each part is given below.

In statistics, a quartile is a type of quantile which divides the number of data points into four parts, or quarters, of more-or-less equal size.

The data must be ordered from smallest to largest to compute quartiles; as such, quartiles are a form of order statistic

Given is the list of prices for a used original radio for a 1955 Thunderbird. The prices vary depending on the condition of the radio.

$210, $210, $320, $200, $300, $10, $340, $300, $245, $325, $700, $250, $240, $200

We can write the sorted data as -

10, 200, 200, 210, 210, 240, 245, 250, 300, 300, 320, 325, 340, 700

{ b } - Median calculation

{n} = 14

Median = (x{n/2} + x{n/2 + 1})/2 = (245 + 250)/2 = 247.5

{ c } - Mode calculation

The mode is the value or values that occur most frequently in the data set. Count the number of times each value in a data set occurs. The mode is the data value with the highest count. We can write the mode as follows -

210, 200, 300

{ d } - Quartiles calculation

Quartiles mark each 25% of a set of data -

The first quartile Q1 is the 25th percentile : 210

The second quartile Q2 is the 50th percentile : 247.5

The third quartile Q3 is the 75th percentile : 320

{ e } - Interquartile range calculation

The interquartile range IQR is the range in values from the first quartile Q1 to the third quartile Q3. Find the IQR by subtracting Q1 from Q3.

IQR = Q3 - Q1

IQR = 320 - 210

IQR = 110

{ f } -

Lower Outlier Boundary : 45

{ e } -

Upper Outlier Boundary : 485

Therefore, the answers to each part is given above.

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Answer:

Step-by-step explanation:

an ant travels from the point $a (0,-63)$ to the point $b (0,74)$ as follows. it first crawls straight to $(x,0)$ with $x \ge 0$, moving at a constant speed of $\sqrt{2}$ units per second. it is then instantly teleported to the point $(x,x)$. finally, it heads directly to $b$ at 2 units per second. what value of $x$ should the ant choose to minimize the time it takes to travel from $a$ to $b$?

Area of a rhombus is half the product of its diagonals:

Area of a trapezoid is the product of the height and half the sum of its bases: