The hypotenuse of an isosceles right triangle is 14 centimeters longer than either of its legs. Find the exact length of each side.​ (Hint: An isosceles right triangle is a right triangle whose legs are the same​ length.)

Answer:

Square Pyramid

Step-by-step explanation:

Thats what a Square Pyramid is

The area of the sector of the circle is: A. 50/3π in.².

The area of a sector that is bounded by two radii of a circle is calculated using the formula, ∅/360 × πr², where we have the following parameters:

Thus, we are given the following regarding the sector of the circle:

Central angle (∅) = 60 degrees

Radius (r) = 10 inches.

Plug in the values into ∅/360 × πr²:

Area of sector = 60/360 × π(10²)

Area of sector = 1/6 × π(100)

Area of sector = 100/6 × π

Area of sector = 50/3 × π

Area of sector = 50/3π in.²

Thus, the area of the sector of the circle is: A. 50/3π in.².

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Step-by-step explanation:

the total sale price for the magazines were

2×4.95 = $9.90

so,

100% = 9.9

1% = 100%/100 = 9.9/100 = 0.099

6.75% = 6.75×1% = 6.75×0.099 = 0.66825 ≈ $0.67

so, he paid in total

$9.90 + $0.67 = $10.57

The area below the mean compares to the area above the mean in a normal distribution as the areas are always equal regardless of the mean. Option A This is further explained below.

Generally, The normal distribution, also known as the Gaussian distribution, is a kind of probability distribution that is symmetric around the mean. This means that it demonstrates that data that are closer to the mean are more likely to occur than data that are farther away from the mean. When represented graphically, the normal distribution takes the shape of a "bell curve."

In conclusion, In a normal distribution, the area below the mean is compared to the area above the mean since the areas are always equal regardless of the mean. This is true even if the mean is different.

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Complete Question

How does the area below the mean compare to the area above the mean in a normal distribution?

A. the areas are always equal regardless of the mean

B. the areas are sometimes equal depending upon the standard deviation of the distribution

C. the area above the mean is larger since the values are larger as you move above the mean

D. the areas are sometimes equal depending upon the value of the mean

19 trees should be planted to maximize the total

From the question, we have the following parameters:

Number of apples, x = 18

Yield, f(x) = 80 per tree

When the number of apple trees is increased (say by x).

We have:

Trees = 18 + x

The yield decreases by four apples per tree.

So, we have

Yield = 80 - 4x

So, the profit function is

P(x) = Apples * Yield

This gives

P(x) = (18 + x) *(80 - 4x)

Expand the bracket

P(x) = 1440 - 72x + 80x - 4x^2

Differentiate the function

P'(x) = 0 - 72 + 80 - 8x

Evaluate the like terms

P'(x) = 8 - 8x

Set P'(x) to 0

8 - 8x = 0

Divide through by 8

1 - x = 0

Solve for x

x = 1

Recall that:

Trees = 18 + x

So, we have

Trees = 18 + 1

Evaluate

Trees = 19

Hence, 19 trees should be planted to maximize the total

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The ski club has 44 members in all

What is population?

Population refers to the total number of items or individuals that form a group.

In this case, the population means the total  number of all members in the  Mogul Runners ski club.

What is sample?

Sample means a fraction of the total population, in other words, the 25% of the members, whose number is 11, that signed up to go for the planned trip.

We can convert 25% to 100% by equation 25% to 11 members that signed up to go for the planned trip.

25% of club members=11 members

divide both sides by 25%

25%/25% of club members=11/25% members

club members=44 members

The ski club has 44 members altogether as computed above.

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Answer:

bro it had been decreased

The option A is correct because the value of the statement is 6+3/n-8.

According to the statement

we have given that the statement and we have to write the statement in the numerical form.

So, For this purpose

we know that the given statement is

"Six more than the quotient of three and a number, decreased by eight".

In this statement "Six more than" indicates the sum between the quotient and 6.

And "The quotient of three and a number" indicates a fraction where the numerator is 3 and the denominator is , a number.

"Decreased by eight" indicates a subtraction between the quotient an 8 units".

If we unit all parts of the statements, we would have

6+3/n-8

So, From the given statement, The option A is correct because the value of the statement is 6+3/n-8.

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Answer:

4 inches

Step-by-step explanation:

large rectangle - 18:12 ratio = 3:2

small rectangle - 6:4 = 3:2

Answer:

a company powder large group of people to assess the company's reputation in the surrounding

Answer: The given integral is impossible to solve.

Step-by-step explanation: I tried solving it right now with integration by parts and trigonometric substitution, and I was unable to solve it.

So I looked at two math solvers and tried inputting the question, and it either says "There is nothing more you can do with this problem," or "We are unable to solve this problem."

First, let's substitute [tex]x=\frac w4[/tex] and [tex]dx=\frac{dw}4[/tex] to get rid of the fraction.

[tex]\displaystyle \int \csc^6\left(\dfrac w4\right) \cot^4\left(\dfrac w4\right) \, dw = 4 \int \csc^6(x) \cot^4(x) \, dx[/tex]

Recall that

[tex]\cot^2(x) + 1 = \csc^2(x)[/tex]

[tex]\dfrac{d}{dx} \cot(x) = -\csc^2(x)[/tex]

We can the rewrite the integrand as

[tex]\displaystyle 4 \int \csc^6(x) \cot^4(x) \, dx = 4 \int \left(\cot^2(x) + 1\right)^2 \cot^4(x) \csc^2(x) \, dx[/tex]

then substitute [tex]y=\cot(x)[/tex] and [tex]dy=-\csc^2(x)\,dx[/tex] to get

[tex]\displaystyle -4 \int (y^2 + 1)^2 y^4 \, dy[/tex]

Expand the integrand.

[tex]\displaystyle -4 \int (y^4 + 2y^2 + 1) y^4 \, dy = -4 \int (y^8 + 2y^6 + y^4) \, dy[/tex]

Now integrate with the power rule.

[tex]\displaystyle -4 \left(\frac{y^9}9 + \frac{2y^7}7 + \frac{y^5}5\right) + C[/tex]

[tex]\displaystyle -\frac{4y^9}9 - \frac{8y^7}7 - \frac{4y^5}5 + C[/tex]

Put everything back in terms of [tex]x[/tex], then [tex]w[/tex].

[tex]\displaystyle -\frac{4\cot^9(x)}9 - \frac{8\cot^7(x)}7 - \frac{4\cot^5(x)}5 + C[/tex]

[tex]\displaystyle \boxed{-\frac49 \cot^9\left(\dfrac w4\right) - \frac87 \cot^7\left(\dfrac w4\right) - \frac45 \cot^5\left(\dfrac w4\right) + C}[/tex]

The solution or the value of x = 14 (Option C). See the explanation below.

Where x (bar) is the mean.

x = (12 + 13 + 14 + 15 + 16)/5

= 14

Hence option C is correct.

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Option A is correct. The gross profit that has been calculated for this year is given as 17545

22050 × 97.5%(100-2.5%) = 21498.75

= 21498.75 × 0.975 = 20961.28

= 20961.28 x 0.975 = 20437.24

= 20437.24 x  0.975  = 19926.31

= 19926.31 ×  0.975  = 19428.15

= 19428.15 x  0.975  = 18942.45

= 18942.45 x  0.975  = 18468.80

= 18468.89 ×  0.975  = 18017

= 18007 × 0.975  = $17545

Hence we can see at the end of the solution that the value of the gross profit in 2028 = $17545

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Answer:

The statement "q + 5 is odd" is FALSE.

Step-by-step explanation:

Let p = 2 and q = 3.

A. p + q is odd ... 2 + 3 = 5   TRUE

B. pq is even ... 2*3 = 6   TRUE

C. q + 1 is even ... 3 + 1 = 4   TRUE

D. q + 5 is odd ... 3 + 5 = 8   FALSE

The area of this regular polygon is 46.8 square units

The given parameters are:

Perimeter, p = 54 m

Apothem, a = √3

The area of this regular polygon is then calculated as:

Area = 0.5 * Apothem * Perimeter

Substitute the known values in the above equation

Area = 0.5 * √3 *  54

Evaluate the product

Area = 46.8

Hence, the area of this regular polygon is 46.8 square units

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Answer:

80

Step-by-step explanation:

Let the number of adult tickets sold be a and the number of student tickets sold be s.

Substituting the first equation into the second,

[tex]2a+a-20=220 \\ \\ 3a-20=220 \\ \\ 3a=240 \\ \\ a=80[/tex]

The interpretation of the p-value is that there is a 0.006 = 0.6% probability of the sample means differing by the amount that they differed on the sample.

At the null hypothesis, it is tested if the means are equal, that is:

[tex]H_0: \mu_1 = \mu_2[/tex]

At the alternative hypothesis, it is tested if the means are different, that is:

[tex]H_1: \mu_1 \neq \mu_2[/tex]

Since we are testing if the means are different, we have a two-tailed test, which means that the p-value is the probability of the means differing by the amount they different on the sample, or a greater amount.

Hence, the interpretation is that there is a 0.006 = 0.6% probability of the sample means differing by the amount that they differed on the sample.

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The zeros which are the roots of the polynomial expressions are:

x = -4; x = - 2;  x = 4 - 2√5;  x = 2(2 + √5)

The zeros of a polynomial f(x) are really the values of x that answer the equation f(x) = 0. Thus, f(x) is a function of x, and the polynomial zeros refer to the values of x, which makes the f(x) value equal to zero. The digit of zeros in a polynomial is decided by the degree of equation f(x) = 0.

For a polynomial, there may be specific values of the variable for which the polynomial is zero. These are known as polynomial zeros. They are also known as polynomial roots. In essence, we look for the zeros of quadratic equations to discover the solutions to the given equation.

From the given equation, the zeros of x^4-2x^3-44x^2-88x-32 can be computed as follows:

Using the rational root theorem:

[tex]\mathbf{= (x+2)\dfrac{x^4-2x^3-44x^2-88x-32}{x+2} }[/tex]

[tex]\mathbf{=(x+2) (x^3-4x^2-36x-32) }[/tex]

By factorization, we have:

= (x + 2) (x + 4) (x² - 8x - 4)

The zeros which are the roots of the polynomial expressions are:

x = -4; x = - 2;  x = 4 - 2√5;  x = 2(2 + √5)

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Step-by-step explanation:

The value of b will be 43° as it is vertically opposite angle.. !!

The information given regarding the computation of the profit or loss shows that there's a loss of RS 500,000.

From the information given, it was stated that the real estate agent bought two triangular Kittas' having a common side as shown in the figure from two people for Rs 2000,000 altogether.

It was further depicted that he sold the land in the shape of a quadrilateral at Rs 1,500,000. In their case, the sale is less than the cost and therefore, there's a loss.

The loss will be:

= 2,000,000 - 1,500,000

= 500,000

Therefore, the information given regarding the computation of the profit or loss shows that there's a loss of RS 500,000.

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The effect of  Edible a portion Price on your profit margin , if we only use the as purchased price to determine our cost and selling price is that it will maximize the profit because it will account for every part of the production.

The portion cost can be calculated by multiplying the cost of a usable kg  with the  portion size.

This can be represented as : portion cost = (portion size x cost of usable kg)

It should be noted that Edible portion (EP)  serves as the  portion of food  which will be given top the   customer after  the preparation.

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Answer:

Step-by-step explanation:

well, the regular price is really "x", which oddly enough is 100%, but we also know that $358 is really the discounted price by 20%, namely 100% - 20% = 80%, so we know that 358 is really just 80% of "x", so

[tex]\begin{array}{ccll} amount&\%\\ \cline{1-2} x & 100\\ 358& 80 \end{array} \implies \cfrac{x}{358}~~=~~\cfrac{100}{80} \\\\\\ \cfrac{x}{358}=\cfrac{5}{4}\implies 4x=1790\implies x=\cfrac{1790}{4}\implies x\approx 448[/tex]

Answer:

6840

Step-by-step explanation:

multiply combinations

n!/(r!(n − r)!)

n choose r

n (objects) = |n

r (sample) = r

(4 teachers, CHOOSE 2)*(20 students, CHOOSE 3)

C(4,2) times C(20,3) =

C(n,r)=?

C(n,r) = C(4,2)

4! / (2!(4-2)!)

4! /2! x 2!

= 6

C(n,r)=?

C(n,r) = C(20,3)

20! / (3!(20-3)!)

20! /  3! x 17!

= 1140

6 * 1140 = 6840

alegbracom Edwin McCravy

Answer: q = -2/3

Step-by-step explanation:

First, we need to apply the Distributive Law in order to proceed with the problem.

3(q + 4/3) = 2

3q + 3(4/3) = 2

We already know that 3(4/3) = 4, since the 3's cancel out. Or, 3(4/3) = 12/3 = 4.

Our new equation is,

3q + 4 = 2

Through transposition and isolation of variables, we get,

3q = -2

Dividing by 3 on both sides...

q = -2/3

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

According to given figure,

Angle 11, Angle 20 and Angle 16 forms angles of a triangle,

so, by angle sum property of a triangle :

[tex]\qquad❖ \: \sf \: \angle11 + \angle20 + \angle16 = 180 \degree[/tex]

[tex]\qquad❖ \: \sf \: \angle11 +66 + 43 = 180 \degree[/tex]

[tex]\qquad❖ \: \sf \: \angle11 +109 = 180 \degree[/tex]

[tex]\qquad❖ \: \sf \: \angle11 = 180 \degree - 109[/tex]

[tex]\qquad❖ \: \sf \: \angle11 =71 \degree [/tex]

[tex] \qquad \large \sf {Conclusion} : [/tex]

Answer: 288

Step-by-step explanation:

(16x9)x2

Answer:

True

Step-by-step explanation:

This is true in an equilateral triangle.

The number of times Ken visits, given the £3.6 pool charge, and 1/3 savings per visit is 5 times.

The entry fee = £3.60

Amount saved on entry fee by having a membership card = 1/3 of the entry fee

Amount Ken spends on the membership card and the reduced entry fee = £5

Therefore;

Reduced entry fee = £3.60×(1 - 1/3) = £2.4

Amount saved per visit = £3.6 - £2.4 = £1.2

The number of visits, n, before he gets back his £5 is therefore;

Ken has to visit the swimming pool more than 4 times to get his £5 back.

Rounding to the next larger whole, therefore;

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See below for the values of the probabilities

The probabilities of the whole events

For event A, the probability of the whole event is calculated using

P(A) = n(A)/Total

Using the table of values, we have:

P(F) = 130/200 = 0.65

P(M) = 70/200 = 0.35

P(L) = 60/200 = 0.30

P(S) = 40/200 = 0.20

The probabilities of the intersection events

For events A and B, the probability of the intersection events is calculated using

P(A n B) = n(A n B)/Total

Using the table of values, we have:

P(F n E) = P(E n F) = 60/200 = 0.30

P(F n L) = P(L n F) = 40/200 = 0.20

P(F n S) = P(S n F) = 30/200 = 0.15

P(M n E) = P(E n M) = 40/200 = 0.20

P(M n L) = P(L n M) = 20/200 = 0.10

P(M n S) = P(S n M) = 10/200 = 0.05

The probabilities of the Union (OR) disjoint events

For events A and B, the probability of the union (OR) disjoint events is calculated using

P(A u B) = P(A) + P(B)

Using the table of values, we have:

P(E u S) = P(E) + P(S) = 100/200 + 40/200 = 0.70

P(E u L) = P(E) + P(L) = 100/200 + 60/200 = 0.80

P(E u L u S) = P(E) + P(L) + P(S) = 100/200 + 60/200 + 40/100 = 1

P(F u M) = P(F) + P(M) = 130/200 + 70/200 = 1

The probabilities of the Union (OR) joint events

For events A and B, the probability of the union (OR) joint events is calculated using

P(A u B) = P(A) + P(B) - P(A n B)

Using the table of values, we have:

P(E u F) = P(E) + P(F) - P(E n F) = 100/200 + 130/200 - 60/100 = 0.85

P(L u M) = P(L) + P(M) - P(L n M) = 60/200 + 70/200 - 20/100 = 0.55

The probabilities of the conditional probabilities

For events A and B, the conditional probability is calculated using

P(A/B) = P(A n B)/P(B)

Using the table of values, we have:

P(F/S) = P(F n S)/P(S) = 0.15/0.20 = 0.75

P(F/E) = P(F n E)/P(E) = 0.30/0.50 = 0.60

P(M/S) = P(M n S)/P(S) = 0.05/0.20 = 0.25

P(M/L) = P(M n L)/P(L) = 0.10/0.30 = 0.33

P(S/F) = P(F n S)/P(F) = 0.15/0.65 = 0.23

P(E/F) = P(F n E)/P(F) = 0.30/0.65 = 0.46

P(S/M) = P(M n S)/P(M) = 0.05/0.35 = 0.14

P(L/M) = P(M n L)/P(M) = 0.10/0.35 = 0.29

The multiplication of dependent events

For events A and B, the conditional probability is calculated using

P(A n B) = P(A) * P(B/A)

Using the table of values, we have:

P(F n L) = P(F) * P(L/F)

This gives

P(F n L) = 0.65 * (0.20/0.30)

Evaluate

P(F n L) = 0.43

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