What is a radius example?

Answer:

23

Step-by-step explanation:

a) Lower limit on the 99% confidence interval is 0.085

b) Upper limit on the 99% confidence interval is 0.215

c) Correct option is = a.

Yes, because 0.2 is inside the interval

d) Sample size = n = 4148

Probabilities are mathematical explanations of the chance that an event will occur or that a proposition is true. The chance of an occurrence is represented by a number between 0 and 1, with 0 often signifying impossibility and 1 typically signifying certainty.

Given that,

n = 200

x = 30

Point estimate = sample proportion = \hat p = x / n = 30 / 200 = 0.15

1 - \hat p = 1 - 0.15 = 0.85

At 99% confidence level the z is,

\alpha  = 1 - 99%

\alpha  = 1 - 0.99 = 0.01

\alpha /2 = 0.005

 Z\alpha/2 = 2.576

Margin of error = E = Z\alpha / 2 * \sqrt((\hat p * (1 - \hat p)) / n)

                              = 2.576 * (\sqrt((0.15 * 0.85 ) / 200 )

                              = 0.065

A 99% confidence interval for population proportion p is ,

\hat p - E < p < \hat p + E

0.15 - 0.065 < p < 0.15 + 0.065

(0.085 < p < 0.215 )

a) lower limit = 0.085

b) upper limit = 0.215

c) correct option is = a.

Yes, because 0.2 is inside the interval

d) \hat p = 1 - \hat p = 0.5

margin of error = E = 0.04 / 2 = 0.02

 sample size = n = (Z\alpha / 2 / E )2 * \hat p * (1 - \hat p)

                              = ( 2.576 / 0.02 )2 * 0.5 * 0.5

                              = 4147.36

sample size = n = 4148

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The function y = 2sin has a phase shift of pi/2 to the right (2x - pi).

To find the function which has a phase shift to the right:

The function has a phase shift of pi/2 to the right.

By definition, you have the phase shift is:

When you substitute the values from the function \(y=2sin(2x-\pi )\), where  \(c=-\pi\)  and \(b=2\), you obtain:

Therefore, the function y = 2sin has a phase shift of pi/2 to the right (2x-pi).

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The complete question is given below:
Which function has a phase shift of pi/2 to the right?

Answer:

$1,281.00

Step-by-step explanation:

We start by calculating the value $50 added each month after the first month

= $50 × 11

= $550

Calculation:

First, converting R percent to r a decimal

r = R/100 = 5.5%/100 = 0.055 per year.

P = Principal = 500 + 550

= $1050

Calculation:

First, converting R percent to r a decimal

r = R/100 = 5.5%/100 = 0.055 per year.

Solving our equation:

A = 1050(1 + (0.055 × 4)) = 1281

A = $1,281.00

Therefore, there would be $1,281.00 after 4 years.

The length of the other leg of the triangle is 16.76 inches.

The Pythagorean theorem states that in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. Written as an equation, this is: \(a^2 + b^2 = c^2\). In this case, the two shorter sides are 11 and b and the hypotenuse is c = 20. Solving for b, we get\(b^2 = c^2 – a^2 = 20^2 – 11^2 = 400 – 121 = 279\). Taking the square root of both sides, we get b = √279 = 16.76. So the length of the other leg of the triangle is 16.76 inches.

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Answer: D

D because if everything is distributed and then divided added and everything would make sense then.

Answer:

d. -8

Step-by-step explanation:

so, well, remember the priorities of mathematical operations :

1. brackets

2. exponents

3. multiplications and divisions

4. additions and subtractions

and remember, when 2 signs or a sign and a + or - operation combine, the result is

+ + = +

+ - = -

- + = -

- - = +

so,

-3 + (9 + 6) ÷ (-3)

first brackets

-3 + 15 ÷ -3

then multiplications or divisions

-3 + -5

then additions and subtractions

-3 - 5 = -8

Answer:

a but

Step-by-step explanation:

i think a

The measure of the angle Y is approximately 32.683°.

In this problem we find the lengths of the three sides of the triangle and all angles are unknown, then we can determine the measure of an angle in terms of all known lengths:

cos Y = (ZX² - XY² - YZ²) / (- 2 · XY · YZ)

If we know that ZX = 55 ft, XY = 50 ft and YZ = 90 ft, then the measure of the angle Y is:

cos Y = (55² - 50² - 90²) / (- 2 · 50 · 90)

Y ≈ 32.683°

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Answer:

8

Step-by-step explanation:

David=18

charlie pours 1/2 which is 18 because that is how many david has and 18 is half of charlies thing. So charli has 36 cause 18 times 2 is 36

CHARLI=36

36 is 3/5 of barbra's so 3 divided by 36 is 12 and ther  are 2 more halfs so 12 times 2 is 24

BARBRA=24

24 is 3/4 of Amandas so you have to do 24 divided by 3 which is 8. so each 4th is 8 and there is one more 4th so it is 8

By: M

Division is defined as breaking down of larger quantity into smaller quantities equally. Vice versa, if we desire to calculate larger quantity on the basis of smaller quantities, we need to multiply the smaller quantity with their weights.

The quantity of water Amanda still have is \(20\: \rm ounces\).

Given:

Quantity of water David has is \(18\: \rm ounces\).

Quantity of water with David is 1/2 of water in Charlie's bottle.

Therefore quantity of water Charlie had is \(\rm 18 \times 2 \:ounces = 36\:ounces\)

Quantity of water Charlie had is 3/5 of water in Barbara's bottle.

Therefore quantity of water Barbara had is \(36\times \dfrac{5}{3} = 60\rm \:ounces\)

Quantity of water Barbara had is 3/4 of water in Amanda's bottle.

Therefore quantity of water Amanda had is \(60\times \dfrac{4}{3} \rm \:ounces = 80\: ounces\).

Water left with Amanda:

\(1 - \dfrac{3}{4}=\dfrac{1}{4}\)   of total water i.e.

\(80\times \dfrac{1}{4}= 20\rm \:ounces\)

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Answer:

33 and 18

Step-by-step explanation:

x+y=51

x-y=15

adding,

x+y=51

x-y=15

______

2x=66

x=33

33+y=51

y=18

Answer:

Mixed number form 4 4/5 exact form 24/5

Step-by-step explanation:

hope this helps

Answer:

Step-by-step explanation:

8/15

The probability value of (m, n) is (1, 2^1005).

Let n be a positive odd integer. We are asked to find the probability that its reciprocal gives a terminating decimal. This is equivalent to saying that the only prime factors of n are 2 and 5 because any other prime factor will yield a repeating decimal.

If n is less than 2010, then its only possible prime factors are 3, 7, 11, ..., 2009, since all primes greater than 2009 are greater than n. We want n to have no prime factors other than 2 and 5. There are 1005 odd integers less than 2010.

We want to count how many of these have no odd prime factors other than 3, 7, 11, ..., 2009. This is equivalent to counting how many subsets there are of {3, 7, 11, ..., 2009}. There are 1004 primes greater than 2 and less than 2010. Each of these primes is either in a subset or not in a subset. Thus, there are 2^1004 subsets of {3, 7, 11, ..., 2009}, including the empty set.

Thus, the probability is:

P = (number of subsets with no odd primes other than 3, 7, 11, ..., 2009) / 2^1004

We can count this number using the inclusion-exclusion principle. Let S be the set of odd integers less than 2010. Let Pi be the set of odd integers in S that are divisible by the prime pi, where pi is a prime greater than 2 and less than 2010. Let Pi,j be the set of odd integers in S that are divisible by both pi and pj, where i < j.

Then, the number of odd integers in S that have no prime factors other than 2 and 5 is:

|S - ⋃ Pi + ⋃ Pi,j - ⋃ Pi,j,k + ...|

where the union is taken over all sets of primes with at least one element and less than or equal to 1005 elements.

By the inclusion-exclusion principle:

|S - ⋃ Pi + ⋃ Pi,j - ⋃ Pi,j,k + ...| = ∑ (-1)^k ⋅ (∑ |Pi1,i2,...,ik|)

where the outer summation is from k = 0 to 1005, and the inner summation is taken over all combinations of primes with k elements.

This simplifies to:

(1/2) ⋅ (2^1004 + (-1)^1005)

Thus, the probability is: P = (1/2^1004) ⋅ (1/2) ⋅ (2^1004 + (-1)^1005) = 1/2 + 1/2^1005. Hence, (m, n) = (1, 2^1005).

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Complete question:

If m/n is the probability that the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal, with m and n being relatively prime positive integers. what is probability value of m and n?

The probability value of (m, n) is (1, 2^1005).This is equivalent to saying that the only prime factors of n are 2 and 5 because any other prime factor will yield a repeating decimal.

Let n be a positive odd integer. We are asked to find the probability that its reciprocal gives a terminating decimal. This is equivalent to saying that the only prime factors of n are 2 and 5 because any other prime factor will yield a repeating decimal.

If n is less than 2010, then its only possible prime factors are 3, 7, 11, ..., 2009, since all primes greater than 2009 are greater than n. We want n to have no prime factors other than 2 and 5. There are 1005 odd integers less than 2010.

We want to count how many of these have no odd prime factors other than 3, 7, 11, ..., 2009. This is equivalent to counting how many subsets there are of {3, 7, 11, ..., 2009}. There are 1004 primes greater than 2 and less than 2010. Each of these primes is either in a subset or not in a subset. Thus, there are 2^1004 subsets of {3, 7, 11, ..., 2009}, including the empty set.

Thus, the probability is:

P = (number of subsets with no odd primes other than 3, 7, 11, ..., 2009) / 2^1004

We can count this number using the inclusion-exclusion principle. Let S be the set of odd integers less than 2010. Let Pi be the set of odd integers in S that are divisible by the prime pi, where pi is a prime greater than 2 and less than 2010. Let Pi,j be the set of odd integers in S that are divisible by both pi and pj, where i < j.

Then, the number of odd integers in S that have no prime factors other than 2 and 5 is:

|S - ⋃ Pi + ⋃ Pi,j - ⋃ Pi,j,k + ...|

where the union is taken over all sets of primes with at least one element and less than or equal to 1005 elements.

By the inclusion-exclusion principle:

|S - ⋃ Pi + ⋃ Pi,j - ⋃ Pi,j,k + ...| = ∑ (-1)^k ⋅ (∑ |Pi1,i2,...,ik|)

where the outer summation is from k = 0 to 1005, and the inner summation is taken over all combinations of primes with k elements.

This simplifies to:

\((1/2) * (2^{1004} + (-1)^1005)\)

Thus, the probability is: P = \((1/2)^{1004}* (1/2) *(2^{1004} + (-1)^{1005}) = 1/2 + 1/2^{1005}.\)

Hence, (m, n) = (\(1, 2^{1005\)).

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Answer:

241 mm^2.

Step-by-step explanation:

There are 10 mm in a cm so there are 10*10 = 100 mm^2 in  a cm^2.

So the answer is 2.41 * 100 = 241 mm^2.

Answer:

241mm²

Step-by-step explanation:

If 10mm = 1cm,

then 100mm² = 1cm²

2.41 × 100 = 241mm²

Answer:

See Below

Step-by-step explanation:

y= 2/3x - 1

y= -1/3x +2

Use substitution

2/3x -1 = -1/3 +2

+1/3         +1/3

x-1 = 2

+1   +1

x = 3

Plug that in

y = 2/3(3/1) -1

3 cancel out leaving it with

y = 2-1

y = 1

(3,1)

Answer:

1 5/8

Step-by-step explanation:

3 1/2 = 3 4/8

3 (from 3/8) x 5 = 15/8

3 4/8 - 15/8 = 1 5/8

Answer:

A=151

B= 29

Step-by-step explanation:

Y=3X/2-Z for the equation X=2/3(Y+Z)

Two or more expressions with an Equal sign is called as Equation.

The given equation is

X=2/3(Y+Z)

x equal to two by three times of y plus z

Apply distributive property on RHS

X=2Y/3+2/3Z

Now Subtract 2/3Z on both sides

X-2/3Z=2/3Y

Divide both sides by 2/3

3/2X-Z=Y

Hence Y=3X/2-Z for the equation X=2/3(Y+Z)

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Answer:

y = -4x - 6

Step-by-step explanation:

12x + 3y = -18

Solve for y.

3y = -12x - 18

y = -12/3 x - 18/3

y = -4x - 6

The probability that the child will be color blind is 25%.

In this scenario, the father and mother are carriers of the color-blindness allele. They have one dominant normal vision allele and one recessive color-blindness allele, so they themselves do not have color blindness. Both of their fathers, however, had color blindness. This means that the fathers had two recessive color-blindness alleles, and so they had the condition.

The probability of the child being color-blind can be calculated by a Punnett square. Let's call the normal vision allele "N" and the color-blindness allele "n". The father's genotype is Nn, and the mother's genotype is also Nn. When these are crossed, there are four possible offspring genotypes:NN (normal vision)Nn (normal vision, but carrier of color-blindness allele)nn (color-blind)NN Nn Nn nnNN Nn Nn nnN N N N Nn Nn N n Nn Nn nn nnWhen we count up the possible offspring, we see that there is one chance of an nn genotype out of four possible genotypes, so the probability is 1/4 or 25%.

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Answer:  First, we need to find the composite function (fg)(x):

(fg)(x) = f(g(x))

= f(-4x-3)

= (-4x-3)^2 + (-4x-3) + 9 (substituting g(x) into f(x))

= 16x^2 + 24x + 18

Therefore, (fg)(x) = 16x^2 + 24x + 18.

Next, we need to find the quotient function (f/g)(x):

(f/g)(x) = f(x) / g(x)

= (x^2 + x + 9) / (-4x - 3) (substituting f(x) and g(x))

To simplify this expression, we can use polynomial long division or synthetic division. Using synthetic division, we get:

  -4 |   1    1    9

      |_____-4__ 12

      |  1  -3   21

Therefore, (f/g)(x) = -4x + 3 - 21 / (-4x - 3)

Simplifying further, we get:

(f/g)(x) = -4x + 3 + (21/4)(1/(x + 3/4))

Therefore, (f/g)(x) = -4x + 3 + (21/4)(1/(x + 3/4)).

Step-by-step explanation:

Answer:

no real roots

Step-by-step explanation:

0 has one square root which is 0. Negative numbers have no square roots since a square is either positive or 0.

The bottleneck time is 10, as this is the longest time for any station in the assembly line.

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Answer: A.

Step-by-step explanation:

The ratio of boy to girl who play kickball at race is 6 to 2. There are 18 girl on the team. the number of boys who play kickball at race is 12 boys.

The ratio of boy to girl who play kickball at race is 6 to 2

6 boys: 2 girls

Multiply the number of girls by the ratio:

18 girls x (6 boys / 2 girls) = 18 x 3 = 54

Subtract the number of girls from the total to get the number of boys:

54 - 18 = 36

Therefore, there are 12 boys who play kickball at race.

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Answer:

Grade 8 Worktext-Science Quarter 3

The correct choice below the elements AU(BNC) = b.{d, f, g, b}

To find the set AU(BNC), to determine the individual sets BNC and then take the union with set A.

Set BNC consists of elements that are in both B and C but not in A.

BNC = (B ∩ C) - A

= ({b, f, g} ∩ {a, b, c, e}) - {d, f, g}

= {b} - {d}

= {b}

find the union of set A and set BNC.

AU(BNC) = A ∪ BNC

= {d, f, g} ∪ {b}

= {d, f, g, b}

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Complete question:

Let U= (a, b, c, d, e, f, g), A=(d, f, g), B = {d, f, g, b), and C= {a, b, c, e}. Find the following set. AU (BNC) Select the correct choice below and fill in any answer boxes within your choice. A. AU (BOC) = {} B. AU (BNC) =Ø

Answer:

8.6666666

or

8.7

Step-by-step explanation:

Answer:

29

Step-by-step explanation:

Answer:

29

Step-by-step explanation:

29-28=1

1+28=29

29+11=40

40-11=29

Thank you for this challenge that was half the challenge of a challenge........

Better hope you stay up long enough to see the animatronics. :)

Answer:

1.   752/47=16

2.  375/25=15

3.  576/24=24

hope this helps!!:)

Step-by-step explanation:

1.

1. Find the Greatest Common Factor (GCF) of 752 which is 47 .

2. Divide both the numerator and the denominator by the GCF.

752/47

47/47

3. simplify

16

hope this helps!!:)

Among the given candidates, the operations in (b) and (c) are binary operations on the respective sets. However, the operations in (a) and (d) fail to be binary operations due to closure violation or reliance on external criteria.

(a) The operation x# = \(x^{2}\) is not a binary operation on set S = {1, 2, 3}. To be a binary operation, it must take two elements from the set and produce another element in the set. However, when x = 2, the result \(x^{2}\) = 4 is not an element in set S, violating the requirement for closure. Therefore, this operation fails to be a binary operation on S.

(b) The operation x + 1 if x is odd, x if x is even is a binary operation on set S = {1, 2, 3}. It takes two elements from the set and produces another element in the set, satisfying the closure property. For example, 2 + 1 = 3, which is an element in S. Hence, this operation is a binary operation on S.

(c) The operation x + y = that fraction, x or y, with the smaller denominator is a binary operation on the set of all fractions. It takes two fractions and produces another fraction, satisfying closure. The result is determined by selecting the fraction with the smaller denominator. Therefore, this operation is a binary operation on the set of all fractions.

(d) The operation x + y = that person, x or y, whose name appears first in an alphabetical sort is not a binary operation on the set of 10 people with different names. This operation does not combine two elements from the set to produce another element in the set, violating closure. It relies on external sorting criteria (alphabetical order) and does not operate solely on the elements of the set itself. Thus, this operation fails to be a binary operation on the given set.

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