Please answer this question

a) The probability of hitting the black circle inside the target is closer to zero.

b) The probability of hitting the white portion inside the target is closer to one.

A probability is calculated as the division of the desired number of outcomes by the total number of outcomes in the context of a problem/experiment.

The total area of the region is given as follows:

10² = 100 units squared.

The black circle has a radius of one unit(as the diameter is of 2 units), hence it's area is given as follows:

A = 3.14 x 1²

A = 3.14 units squared.

Then the probability of hitting the black circle is of:

3.14/100 = 0.0314 -> closer to zero.

The probability of hitting the white circle is given as follows:

1 - 0.0314 = 0.9686 -> closer to one.

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The score located 2 standard deviations below the mean is 100. This score can be found by subtracting 2 standard deviations (20) from the mean (120).

The normal distribution is a bell-shaped curve that is symmetrical around the mean. This means that if you calculate the number of standard deviations away from the mean, you can use the same number to calculate how many standard deviations away from the mean the score is.

For example, in this question, the mean is 120 and the standard deviation is 10. To find the score located 2 standard deviations below the mean, subtract 2 standard deviations from the mean. This means the score is 120 - 20 = 100.

In general, the formula for calculating the score located x standard deviations away from the mean is:

Score = Mean + (x * Standard Deviation)

For example, to find the score located 4 standard deviations away from the mean, the formula is:

Score = Mean + (4 * Standard Deviation)

In this example, the score is 120 + (4 * 10) = 160.

In summary, to find the score located x standard deviations away from the mean, use the formula:
Score = Mean + (x * Standard Deviation)

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Answer:Tham I can’t even se the letters you should have posted each question one by one

Step-by-step explanation:

good luck

P(t) = 5(1 + 0.125)^t

where P(t) is the population of fruit flies after t days.

Answer:   Based on the problem statement, we can define a recurrence relation as follows:

t(n) = t(n-1) + t(n-2) + t(n-4)

This means that the number of ways to construct a tower of height n cm can be obtained by considering the possible heights of the bottom block in the tower. If the bottom block has height 1 cm, then the remaining blocks make up a tower of height (n-1) cm, for which there are t(n-1) ways to construct it. If the bottom block has height 2 cm, then the remaining blocks make up a tower of height (n-2) cm, for which there are t(n-2) ways to construct it. If the bottom block has height 4 cm, then the remaining blocks make up a tower of height (n-4) cm, for which there are t(n-4) ways to construct it.

Since we are assuming an unlimited supply of blocks of each size, we can use these blocks repeatedly to construct towers of different heights. Also, we can use dynamic programming to compute the values of t(n) for each integer n from 1 to 25, by using the recurrence relation above and the base cases:

t(0) = 1 (there is only one way to construct a tower of height 0 cm, which is to not use any blocks)

t(n) = 0 for n < 0 (there is no way to construct a tower of negative height)

Using these, we can compute the values of t(n) for n = 1, 2, ..., 25, as follows:

t(0) = 1

t(1) = t(0) = 1

t(2) = t(1) + t(0) = 2

t(3) = t(2) + t(1) = 3

t(4) = t(3) + t(2) + t(0) = 6

t(5) = t(4) + t(3) + t(1) = 10

t(6) = t(5) + t(4) + t(2) = 19

t(7) = t(6) + t(5) + t(3) = 32

t(8) = t(7) + t(6) + t(4) = 61

t(9) = t(8) + t(7) + t(5) = 104

t(10) = t(9) + t(8) + t(6) = 195

t(11) = t(10) + t(9) + t(7) = 332

t(12) = t(11) + t(10) + t(8) = 626

t(13) = t(12) + t(11) + t(9) = 1065

t(14) = t(13) + t(12) + t(10) = 2002

t(15) = t(14) + t(13) + t(11) = 3405

t(16) = t(15) + t(14) + t(12) = 6403

t(17) = t(16) + t(15) + t(13) = 10946

t(18) = t(17) + t(16) + t(14) = 20618

t(19) = t(18) + t(17) + t(15) = 350

Step-by-step explanation:

A patient's probability of recovering is 0.8, so the likelihood that they won't is 0.2.

The probability that a patient will recover from a stomach disease is 0.8. Assume that 20 persons have reportedly got this illness.

Let X = # of recoveries out of 20 patients who caught the condition while PMF for X and resolve issues. Imagine that a 20-person sample was chosen at random.

Here, Bernoulli trials are applicable.

Recall that the chance of k successes in n trials using the Bernoulli approach is given by:

P(k)=( n/k )p (power k) * (q power n−k)

where q=1-p is the probability that an attempt would fail

where q=1-p is the probability that an attempt would fail

Here, let's make recovery a success. Hence, p = 0.8, q = 0.2, and n = 20

The probability that at least 10 recoveries will occur is

P(X10) = P(10) + P(11) + P(12) +... + P (20)

The complete question will be:

The probability that a patient recovers from a stomach disease is 0.8. Suppose twenty people are known to have contracted this disease. What is the probability that at least ten recover?

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Answer:

biquadratic polynomial

Step-by-step explanation:

it has degree 4

n = ceil(sqrt((2008/4phi)[90 - arccos(k(2 cos(phi/2008)))]))

       2 [cos(alpha) sin(alpha) + cos(4 alpha) sin(2 alpha) + cos(9 alpha) sin(3 alpha) + ... + cos(n^2 alpha) sin(n alpha)]

     = [sin(2 alpha) + sin(8 alpha) + sin(18 alpha) + ... + sin(n^2 alpha)].

        sin(2 alpha) + sin(8 alpha) + sin(18 alpha) + ... + sin(n^2 alpha)

     = (sin(2 alpha) - sin(2n^2 alpha))/(1 - sin(2 alpha))

     = (2 sin(n^2 alpha) cos(n^2 alpha))/(2 cos(alpha) - 1)

     = [sin(2n^2 phi/2008)]/(2 cos(alpha) - 1)

        [sin(2n^2 phi/2008)]/(2 cos(alpha) - 1)

        sin(2n^2 phi/2008) = k(2 cos(alpha) - 1)

        cos(90 - 2n^2 phi/2008) = k(2 cos(alpha) - 1)

       90 - 2n^2 phi/2008 = ±arccos(k(2 cos(alpha) - 1))

       n^2 = (2008/4phi)[90 ± arccos(k(2 cos(alpha) - 1))]

       n = ceil(sqrt((2008/4phi)[90 - arccos(k(2 cos(phi/2008)))]))

We can find the smallest positive integer n by incrementing the value of k starting from 1 until ceil(k^2*alpha) is greater than or equal to n.

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Answer:

Commutative property

The Commutative property is most simply shown with: a x b = b x a. In multiplication, the values can shift or "commute" in any order

The gorilla has a heart rate that is four times faster than the chimpanzee's when they are resting.

Expression in math is a combination of numbers, variables, operations and/or symbols that represent a value, quantity or an equation. It can either be a single term, or several terms connected by mathematical operations such as addition, subtraction, multiplication and division. Expressions are typically used in equations, formulas and problems to help solve them.

To calculate the heart rate of the gorilla, we use the power function given: H(m) = 240m. Substituting in m = 200, we get H(200) = 240(200) = 48,000 beats per minute.

To calculate the heart rate of the chimpanzee, we use the same power function: H(m) = 240m. Substituting in m = 50, we get H(50) = 240(50) = 12,000 beats per minute.

Therefore, the gorilla has a heart rate that is four times faster than the chimpanzees when they are resting.

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Answer:

4

Step-by-step explanation:

Answer:

[tex]x\leq 6[/tex]

Step-by-step explanation:

The length of the hypotenuse is 2 times the square root of 5. The Pythagorean theorem can be used to determine the length of the hypotenuse.

To graph the right triangle with the given points as the hypotenuse, we first plot the points on a coordinate plane . The two points form the endpoints of the hypotenuse, which is the line segment connecting them. We can find the length of this line segment using the distance formula:

distance = [[tex]\sqrt{(x2 - x1)^2 + (y2 - y1)^2]}[/tex]

In this case, we have:

distance = [[tex]\sqrt{(-5 - (-3))^2 + (-6 - (-4))^2}[/tex]]

distance = [tex]\sqrt{(-5 - (-3))^2 + (-6 - (-4))^2}[/tex]]

distance = [[tex]\sqrt{4 + 4}[/tex]]

distance = [[tex]\sqrt{8}[/tex]]

We can simplify [tex]\sqrt{8}[/tex] by factoring out the perfect square factor of 4:

distance = [[tex]\sqrt{4 * 2}[/tex]]

distance = [tex]\sqrt{4} *\sqrt{2}[/tex]

distance = 2 * [tex]\sqrt{2}[/tex]

Thus, the distance between the two points is 2 * [tex]\sqrt{2}[/tex] ] units.

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The answer and workout is provided in the attachment.

The volume of balloon A is [tex]\frac{2}{5}[/tex] gallon and the volume of balloon B is [tex]\frac{1}{5}[/tex] gallon.

What is volume?

Any three-dimensional solid's volume is simply the amount of space it takes up. A cube, cuboid, cone, cylinder, or sphere can be one of these solids.

We are given that two balloons have a total volume of [tex]\frac{3}{5}[/tex] gallon.

Let the volume of balloon A be x and the volume of balloon B be y.

So, we get

x + y = [tex]\frac{3}{5}[/tex]

Also, it is given that the difference of their volumes is [tex]\frac{1}{5}[/tex] gallon and Balloon A has a greater volume than Balloon B.

So, we get another equation as

x - y = [tex]\frac{1}{5}[/tex]

Now, on adding both the equations, we get

⇒ 2x = [tex]\frac{4}{5}[/tex]

Now, on solving we get

⇒ x = [tex]\frac{2}{5}[/tex]

Now, on substituting the value of x in the equation, we get

⇒ [tex]\frac{2}{5}[/tex] + y = [tex]\frac{3}{5}[/tex]

⇒ y = [tex]\frac{1}{5}[/tex]

Hence, the volume of balloon A is [tex]\frac{2}{5}[/tex] gallon and the volume of balloon B is [tex]\frac{1}{5}[/tex] gallon.

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If the number of samples is doubled, the new confidence interval would be 9.61, 10.39 or narrower (smaller) while keeping the same confidence level.

. When calculating the confidence interval, the standard error is used, along with the sample mean and the critical value from the distribution. If we have a larger sample size, we can be more confident in our estimate of the population parameter because the sample mean will more closely resemble the population mean. As a result, the confidence interval can be narrower, indicating a higher degree of precision. For Example:Suppose a sample of 50 was taken, and the mean weight of an object was 10 grams with a standard deviation of 2 grams.

At a 95 percent confidence level, the confidence interval for the mean weight would be 10 ± (1.96)(2/√50) = (9.15, 10.85)Now suppose that the sample size is doubled to 100. The standard error will be cut in half, i.e., 2/√100 = 0.2. As a result, the new confidence interval would be 10 ± (1.96)(0.2) = (9.61, 10.39). Notice that the new confidence interval is narrower, indicating a higher degree of precision while keeping the same confidence level.

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Answer: 41$

Step-by-step explanation:

This is because 5x6=30 (To find how much money is made)

then 11+30=41 (add both amounts)

ANSWER-

Let the nonadhacent anglesof complementary angle be x and y where, x=79 and y =?

WE KNOW,

x+y=90

or,79+y=90

or, y=90-79

:. y=11,,

The nurse should administer 3.5 mL of Lanoxin elixir 0.05 mg/mL to the patient to achieve the desired dose of 0.175 mg.

First, we need to convert the prescribed dose of 0.175 mg to milligrams per milliliter (mg/mL) using the concentration of the Lanoxin elixir, which is 0.05 mg/mL.

To determine how many milliliters of Lanoxin elixir 0.05 mg/mL the nurse should administer to the patient, we need to use a simple formula:

Dose = Desired dose / Stock strength

Where:

Desired dose is 0.175 mg

Stock strength is 0.05 mg/mL

So, substituting the values into the formula:

Dose = 0.175 mg / 0.05 mg/mL

Dose = 3.5 mL

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Step-by-step explanation:

hope this will help u

Answer:

Part A is 8/13 of the whole

Part B is 5/13 of the whole

Step-by-step explanation:

Assuming there are no other parts,

the Whole = A + B is the denominator:

Whole = 40 + 25 = 65

Part A = 40 and Part B = 25 are numerators for each fraction.

The fractions are then:

40/65 and 25/65

Meaning:

Part A is 40/65 of the whole

Part B is 25/65 of the whole

Reducing the fractions, it is also true that:

Part A = 8/13 of the whole

Part B = 5/13 of the whole

The value 1.106 represents the slope of the regression line or the rate of change of y with respect to x.

In the given regression equation y = 3.915(1.106)x:

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Answer:

N*r^n

Step-by-step explanation:

Let the initial number of cases at the start of year 1 be represented by N.

From the given information, we know that the number of cases increases by the same factor each year. Let this factor be represented by r.

Then, at the end of year 1, the number of cases would be N*r, since it has increased by a factor of r.

Similarly, at the end of year 2, the number of cases would be Nrr, or N*r^2.

At the end of year 3, the number of cases would be Nrrr, or Nr^3.

We can use this pattern to write a general expression for the number of cases after n years:

N * r^n

where N is the initial number of cases, r is the common factor by which the number of cases increases each year, and n is the number of years elapsed.

Arranged from smallest to largest, the given values are 0.2, 3/20, 1/8, 11%, and 13%.

To compare these values, we need to convert the percentages to decimals. We can do this by dividing them by 100. So, 11% becomes 0.11 and 13% becomes 0.13.

Next, we can convert 3/20 and 1/8 to decimals by dividing them using a calculator. We get:

3/20 = 0.15

1/8 = 0.125

Now, we can arrange these values in ascending order:

0.2 < 0.125 < 0.15 < 0.11 < 0.13

Therefore, the values arranged in order from smallest to largest are 0.2, 3/20, 1/8, 11%, and 13%.

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The probability that at most one sample is mutated in 10 samples is 0.746.

To calculate this probability, we use the binomial distribution formula.

The binomial distribution formula is used to calculate the probability of a certain number of successes (in this case, samples that are mutated) in a certain number of trials (10 samples). We need to find the probability of 1 success or fewer in 10 trials.

This is equal to P(x<=1) = 1 - P(x>1), where x is the number of successes.

For this calculation, we need the following parameters: n = 10 (number of trials), p = 0.15 (probability of a single sample being mutated), and x = 1 (number of successes). So, P(x<=1) = 1 - P(x>1) = 1 - P(x = 2) - P(x = 3) - P(x = 4) - P(x = 5) - P(x = 6) - P(x = 7) - P(x = 8) - P(x = 9) - P(x = 10).

The probability of at most one sample being mutated in 10 samples is calculated by adding the individual probabilities of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 samples being mutated, which equals 0.746.

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the t-score for a 99.8% confidence interval with a sample size of 5 is 4.604.

The t-score for a 99.8% confidence interval with a sample size of 5 can be found using a t-distribution table or calculator.

Since we have a sample size of 5, the degrees of freedom (df) will be n - 1 = 4.

From the t-distribution table or calculator, the t-score for a 99.8% confidence interval with 4 degrees of freedom is approximately 4.604.

Therefore, the t-score for a 99.8% confidence interval with a sample size of 5 is 4.604.

the complete question is :

what is the value of the t score for a 99.8% confidence interval if we take a sample of size 5?

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Answer:

2.80

Step-by-step explanation:

It asks to find the shipping and handling price.

The first step is take that percentage, 20% and convert into decimal form. Then you need to multiply that by the original price. So it would be 0.2 x 14.

That gives you 2.80 which is the shipping and handling price. That's all you have to do.

triangle C

Step-by-step explanation:

If you draw it out, it looks unique

Answer: what do you mean? I need more info-

Step-by-step explanation:

I can answer it with more info :)

Answer: y - 10.26

Step-by-step explanation: Add 5.4 on both sides and that leaves you with y

Nonparametric tests can be useful in situations where the data may not follow a specific distribution or where the assumptions of a parametric test are not met.

The nonparametric tests mentioned in your question do not assume any specific probability distribution for the data. Hence, they are called nonparametric tests. These tests are used when the assumptions required for parametric tests (e.g., normality) are not met, or when the data is measured on ordinal or nominal scales rather than continuous ones.

The Wilcoxon rank-sum test, sign test, and Wilcoxon signed-rank test are used to compare two independent or dependent samples. The Kruskal-Wallis test and Friedman test are used to compare three or more independent or dependent samples.

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