Use linear approximation, i.e. the tangent line, to approximate √81.1 as follows:
Let f(x) = √x. The equation of the tangent line to f(x) at x=81 can be written in the form y=mx+b where m is:
and where b is:
Using this, we find our approximation for √81.1 is:_________

Answer:

The boat traveled at 10 miles per hour for 2 hours and at 30 miles per hour for 4 hours.

Step-by-step explanation:

Let x be the time (in hours) that the boat traveled at 10 miles per hour, and let y be the time (in hours) that it traveled at 30 miles per hour.

We know that:

x + y = 6 (the total time the boat traveled)

We also know that the boat traveled a total distance of 140 miles, which can be expressed as:

10x + 30y = 140

We can use the first equation to solve for one of the variables in terms of the other:

y = 6 - x

Substituting this expression for y into the second equation, we get:

10x + 30(6 - x) = 140

Simplifying and solving for x, we get:

10x + 180 - 30x = 140

-20x = -40

x = 2

Therefore, the boat traveled at 10 miles per hour for 2 hours and at 30 miles per hour for 4 hours.

Answer:

[tex]a=\dfrac{1}{2},\;\;b=\dfrac{1}{2}[/tex]

Step-by-step explanation:

A function f is continuous at x = a when:

    [tex]\bullet\quad\textsf{$f(a)$\;is\;d\:\!efined.}[/tex]

    [tex]\bullet\quad\textsf{$\displaystyle \lim_{x \to a} f(x)$\;exists.}[/tex]

    [tex]\bullet\quad\displaystyle \lim_{x \to a} f(x) = f(a)[/tex]

To ensure that f(x) is continuous at x = 2 and x = 3, we need to make sure that the limit of f(x) as x approaches 2 from the left is equal to the limit of f(x) as x approaches 2 from the right, and similarly for x = 3.

First, factor the numerator and simplify the rational function:

[tex]f(x)=\dfrac{x^2-4}{x-2}=\dfrac{(x+2)(x-2)}{x-2}=x+2[/tex]

As x approaches 2 from the left, x < 2. Therefore, to find the limit as x approaches 2 from the left, substitute x = 2 into the first sub-function:

[tex]\displaystyle \lim_{x \to 2^{-}} f(x)=2+2=4[/tex]

As x approaches 2 from the right, x > 2. Therefore, to find the limit as x approaches 2 from the right, substitute x = 2 into the second sub-function:

[tex]\begin{aligned}\displaystyle \lim_{x \to 2^{+}} f(x)&=a(2)^2-b(2)+3\\&=4a-2b+3\end{aligned}[/tex]

To ensure continuity at x = 2, equate the limits:

[tex]4=4a - 2b + 3[/tex]

Solve for b in terms of a:

[tex]\begin{aligned}4&=4a - 2b + 3&\\ 2b+4&=4a+3\\2b&=4a-1\\b&=2a-\dfrac{1}{2}\end{aligned}[/tex]

Now, we need to find a and b so that f(x) is continuous at x = 3.

As x approaches 3 from the left, x < 3. Therefore, to find the limit as x approaches 3 from the left, substitute x = 3 into the second sub-function:

[tex]\begin{aligned}\displaystyle \lim_{x \to 3^{-}} f(x)&=a(3)^2 - b(3) + 3\\&= 9a - 3b + 3\end{aligned}[/tex]

As x approaches 3 from the right, x > 3. Therefore, to find the limit as x approaches 3 from the right, substitute x = 3 into the third sub-function:

[tex]\begin{aligned}\displaystyle \lim_{x \to 3^{+}} f(x)&= 2(3) - a + b\\&= 6 - a + b\end{aligned}[/tex]

To ensure continuity at x = 3, equate the limits:

[tex]9a - 3b + 3 = 6-a+b[/tex]

Solve for b in terms of a:

[tex]\begin{aligned}9a - 3b + 3 &= 6-a+b\\10a-3b+3&=6+b\\10a+3&=6+4b\\10a-3&=4b\\4b&=10a-3\\b&=\dfrac{5}{2}a-\dfrac{3}{4}\end{aligned}[/tex]

Substitute this expression for b into the equation obtained from continuity at x = 2:

[tex]\dfrac{5}{2}a-\dfrac{3}{4}=2a-\dfrac{1}{2}[/tex]

Solve for a:

[tex]\begin{aligned}\dfrac{5}{2}a-\dfrac{3}{4}&=2a-\dfrac{1}{2}\\\\\dfrac{1}{2}a-\dfrac{3}{4}&=-\dfrac{1}{2}\\\\\dfrac{1}{2}a&=\dfrac{1}{4}\\\\a&=\dfrac{1}{2} \end{aligned}[/tex]

Substitute the found value of a into the expression for b and solve for b:

[tex]\begin{aligned}b&=2\left(\dfrac{1}{2}\right)-\dfrac{1}{2}\\b&=1-\dfrac{1}{2}\\b&=\dfrac{1}{2}\end{aligned}[/tex]

Therefore, the values of a and b that make f(x) continuous at all points are:

[tex]a=\dfrac{1}{2},\;\;b=\dfrac{1}{2}[/tex]

So the function f(x) is:

[tex]f(x)=\begin{cases} \dfrac{x^2-4}{x-2}&\text{if}\;\;x < 2\\\\\dfrac{1}{2}x^2-\dfrac{1}{2}x+3\quad&\text{if}\;\;2 \leq x < 3\\\\2x&\text{if}\;\;x \geq 3\end{cases}[/tex]

The total number of people who preferred swimming at the beach is: 24

In survey sampling, the term probability of selection is one that refers to the chance (i.e. the probability from 0 to 1) that a member (element) of a population can be chosen for a given survey.

We are told that two out of 10 people preferred swimming at the beach. We are also told that there was a total of 120 people that the researcher asked about swimming. Thus:

Fraction of people that prefer swimming = 2/10 = 0.2

Thus, number of people that prefer swimming in a sample of 120 people is: 0.2 * 120 = 24 people

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To determine the probability that the roots of the given quadratic equation are both real, we need to find the values of a for which the discriminant of the equation is non-negative.

The discriminant of the quadratic equation ax^2 + bx + c = 0 is b^2 - 4ac. In this case, the discriminant of the given equation is:

a^2 - 4(a+35)

For the roots to be real, this discriminant must be non-negative. That is:

a^2 - 4(a+35) ≥ 0

Simplifying this inequality, we get:

a^2 - 4a - 140 ≥ 0

Factorizing the left-hand side, we get:

(a-14)(a+10) ≥ 0

This inequality is satisfied for a ≤ -10 or a ≥ 14, or when a is in the interval [-12, -10) or (14, 15].

Since a is uniformly distributed over the interval [-12, 15], the probability that lies in the interval [-12, -10) or (14, 15] is:

Probability = Length of the interval [-12, -10) + Length of interval (14, 15] / Total length of the interval [-12, 15]
Probability = (2 + 1) / (15 - (-12))
Probability = 3/27
Probability = 1/9

Therefore, the probability that the roots of the given quadratic equation are both real is 1/9.

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a) The probability that the number of drivers will be at most 19 is approximately 0.411 or 41.1%.

b) The probability that the number of drivers will exceed 29 is approximately 0.004 or 0.4%.

(a) To find the probability that the number of drivers will be at most 19, we need to use the Poisson distribution formula:

P(X ≤ 19) = e^(-20) * (20^0/0!) + e^(-20) * (20^1/1!) + ... + e^(-20) * (20^19/19!)

Using a calculator or statistical software, we get P(X ≤ 19) ≈ 0.088.

(b) To find the probability that the number of drivers will exceed 29, we can use the complement rule:

P(X > 29) = 1 - P(X ≤ 29)

Using the same Poisson distribution formula as in part (a), we can find P(X ≤ 29) ≈ 0.963. So,

P(X > 29) = 1 - 0.963 = 0.037 (rounded to three decimal places).

Note: "Dynamic Ride Sharing" is not directly related to this question and is not necessary for answering it.
Hi! I'd be happy to help you with your question.

(a) To find the probability that the number of drivers will be at most 19, you can use the cumulative distribution function (CDF) of the Poisson distribution. The parameter for this distribution is μ = 20. The formula for the Poisson CDF is:

P(X ≤ k) = Σ (e^(-μ) * (μ^x) / x!) for x = 0 to k

In this case, k = 19. Plugging in the values and calculating the sum, we get:

P(X ≤ 19) ≈ 0.411

Therefore, the probability that the number of drivers will be at most 19 is approximately 0.411 or 41.1%.

(b) To find the probability that the number of drivers will exceed 29, you can use the complementary probability rule. First, find the probability that the number of drivers will be at most 29, and then subtract that from 1.

P(X > 29) = 1 - P(X ≤ 29)

Using the Poisson CDF formula with k = 29 and μ = 20:

P(X ≤ 29) ≈ 0.996

Now, subtract this value from 1:

P(X > 29) = 1 - 0.996 ≈ 0.004

Therefore, the probability that the number of drivers will exceed 29 is approximately 0.004 or 0.4%.

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The mean of the probability distribution of the random variable is 3

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

The probability distribution

From the probability distribution, we can see that the data are normally distributed

This means that the mean, the median and the mode are equal

From the distribution of the random variable, we have the following readings

Hence. the calculated mean of the random variable is 3

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The manager could use random sampling to  find out which type of jelly bean would be popular.

To ensure that every client has an equal chance of being chosen, random sampling entails choosing a sample of consumers at random from the full population of jelly bean customers.

In order to use random sampling,  the manager may create a list of every jelly bean customer before using a random number generator to pick a representative sample of customers.

The manager's desired level of accuracy and confidence in the survey results would determine the sample size.

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The term that represents the typical average speed is 30/3s

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

Expression = 20/s + 30/3s

We understand that

She traveled at 20 miles per second for some time and the rest at her typical speed

This means that

Typical speed = 30/3s

Hence, the term of the average speed is 30/3s

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To calculate the probability of getting at most five successes, we can add up the probabilities of getting 0, 1, 2, 3, 4, or 5 successes

Sure, here are the expressions that can be used to compute each of the probabilities:

a) p(exactly five successes) = (7 choose 5) * p^5 * q^2
This expression uses the binomial coefficient (7 choose 5), which represents the number of ways to choose 5 successes out of 7 trials. We then multiply this by the probability of getting 5 successes (p^5) and the probability of getting 2 failures (q^2).

b) p(at least five successes) = p(five successes) + p(six successes) + p(seven successes)
To calculate the probability of getting at least five successes, we can add up the probabilities of getting exactly 5, 6, or 7 successes. We can use the expression from part a to calculate each of these probabilities.

c) p(at most five successes) = p(zero successes) + p(one success) + p(two successes) + p(three successes) + p(four successes) + p(five successes)
. We can again use the expression from part a to calculate each of these probabilities.

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A Single transformation maps triangle ABC onto triangle DEF.

Given  that Reflection the single  transformation maps ABC onto A'B'C' is reflection .

The rigid transformations can map triangle Δ ABC onto triangle Δ DEF is reflection then translation .

The rigid transformations that will map Δ ABC to Δ DEF is rotation then translation .

However pair of triangles can be proven congruent by the HL theorem is rotation then translation .

Thus, rigid transformation S can map Triangle ABC onto Triangle DEF is reflection then rotation .

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The value corresponding to the percentile rank of 30% is found in the 10-14 class interval, as its cumulative percentage is 30%. The approximate percentile rank of X = 36 is 90%, as the cumulative percentage for the class interval of 35-39 is 90%.

To create a relative frequency table of raw score values, we first need to create class intervals with a width of 5. The class intervals are: 5-9, 10-14, 15-19, 20-24, 25-29, 30-34, and 35-39.

| Raw Scores | Frequency | Cumulative Frequency | Cumulative Percentage |
|------------|-----------|----------------------|-----------------------|
| 5-9        | 0         | 0                    | 0%                    |
| 10-14      | 2         | 2                    | 20%                   |
| 15-19      | 1         | 3                    | 30%                   |
| 20-24      | 1         | 4                    | 40%                   |
| 25-29      | 3         | 7                    | 70%                   |
| 30-34      | 1         | 8                    | 80%                   |
| 35-39      | 1         | 9                    | 90%                   |

To draw a relative frequency histogram, we plot the class intervals on the x-axis and the relative frequencies on the y-axis. The shape of the distribution appears to be positively skewed.

The value corresponding to the percentile rank of 30% can be found by looking at the cumulative percentage column in the frequency table. The cumulative percentage at the end of the 2nd class interval is 20%, so the value corresponding to the 30th percentile is somewhere between 10 and 14. Using linear interpolation, we can estimate that the value corresponding to the 30th percentile is approximately 11.6.

To find the approximate percentile rank of X = 36, we can look at the raw scores column in the frequency table and find the class interval that contains 36, which is 35-39. The cumulative frequency at the end of this interval is 9, and the total number of scores is 10. Therefore, the percentile rank of X = 36 is approximately 90%.
To create a relative frequency table of raw score values using class intervals with a width of 5, we can categorize the data as follows:

Class Intervals (Raw Scores): 10-14, 15-19, 20-24, 25-29, 30-34, 35-39, 40-44

Frequency: 2, 1, 2, 3, 0, 1, 1

Cumulative Frequency: 2, 3, 5, 8, 8, 9, 10

Cumulative Percentage: 20%, 30%, 50%, 80%, 80%, 90%, 100%

Next, draw a relative frequency histogram with the class intervals on the x-axis and the frequency on the y-axis. The shape of the distribution is positively skewed, as most data points are on the lower end of the scale, with a few points in the higher range.

The value corresponding to the percentile rank of 30% is found in the 10-14 class interval, as its cumulative percentage is 30%.

The approximate percentile rank of X = 36 is 90%, as the cumulative percentage for the class interval of 35-39 is 90%.

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The simplified value of the polynomial-expression "√(64m⁸n¹²)" is 8m⁴n⁶..

A "Polynomial-Expression" is an expression which consists of variables, coefficients, and exponents having operations of addition, subtraction, multiplication, and non-negative integer exponents.

To simplify the given polynomial expression √(64m⁸n¹²), we can use the property of square-roots which states that √(a×b) = √a × √b;

So, We have

⇒ √(64m⁸n¹²) = √(64) × √(m⁸) × √(n¹²),

Now, we simplify each of square roots separately:

⇒ √(64) = 8, because 8×8 = 64;

⇒ √(m⁸) = m⁴, because m⁴×m⁴ = (m⁴)² = m⁸;

⇒ √(n¹²) = n⁶, because n⁶×n⁶ = (n⁶)² = n¹²,

Substituting the values,

We get,

⇒ √(64m⁸n¹²) = 8m⁴n⁶

Therefore, the simplified polynomial expression is 8m⁴n⁶.

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The given question is incomplete, the complete question is

Simplify the polynomial expression. √(64m⁸n¹²).

Examining the figure, the missing angles are

angle AOC = 148 degrees

angle OCD =  21 degrees

Line AB and CB are tangents to the circle and hence will make angle 90 degrees at the point of tangent.

OA bisects angle AOC and angles ABC

In triangle AOB

90 + 32/2 + angle AOB = 180 degrees

angle AOB = 180 - 90 - 32 / 2

angle AOB =  74 degrees

angle AOC = 74 x 2 = 148 degrees

Using inscribed angle theorem

angle D = 1/2 x angle AOC

angle D = 74 degrees

In quadrilateral ABCD

143 + 32 + 74 + angle C = 360

angle C = 360 - 143 - 32 - 74

angle C = 111 degrees

angle OCD = 111 - 90

angle OCD = 21 degrees

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Answer: That is correct. In a binary tree, each node has at most two child nodes (hence the name "binary"), and the height of the tree is the length of the longest path from the root node to a leaf node.

If the tree has n leaves, then the number of nodes in the tree is at most 2n-1 (since each node can have at most 2 child nodes and there is only one root node), and the height of the tree is log2(2n) = log2n + 1 (since there are 2n nodes at level log2n, and we need to add 1 for the root node).

However, if the tree is not perfectly balanced, it is possible for the height to be slightly larger than log2n + 1. Nonetheless, log2n is still a tight upper bound on the height of a binary tree with n leaves.

Answer:

Step-by-step explanation:

the standard equation for the circle is:

(x-a)²+(y-b)² = r²

the center is : A(a,b)    and ridus r

you have : a= 2 and b= - 3   r²= (3-2)²+(5+3)²= 1+ 64

r² 65    

the standard equation for the circle is:

(x-2)²+(y²+3 )=65

Answer: The theoretical probability that the number cube will land on five is 1/6. The previous outcomes do not affect the probability of rolling a five on the next roll, as each roll of the number cube is independent of the previous roll. Therefore, the probability of rolling a five on the next roll is the same as the probability of rolling a five on any other roll of the number cube, which is 1/6.

Step-by-step explanation:

Answer:

1/6

Step-by-step explanation:

All the wording makes it confusing, but it is a simple probability of rolling a 5 out of the 6 faces on the die. It only asks about that time, so the probability doesnt increase or decrease at all depending on what was rolled before

Answer: The answer is 3

Step-by-step explanation: I had the same question.

The volume of the 'scooped' pyramid is approximately 26.6667 cubic units.

To find the volume of the 'scooped' pyramid, we first need to determine the area of its base. Since the cross-sectional area of the pyramid is x 4 at a distance x from the tip, we can assume that the area at the tip is zero. This means that the area of the base is 4 times the area at a distance of 5 from the tip (since the distance from tip to base is 5).

Therefore, the area of the base is 4x4 = 16 square units. To find the volume, we can use the formula for the volume of a pyramid, which is:

Volume = (1/3) x Base Area x Height

In this case, the height of the pyramid is 5 units. So, we can substitute the values we have:

Volume = (1/3) x 16 x 5

Volume = 26.6667 cubic units

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Answer:

X - Y and X + Y are uncorrelated. To show that X - Y and X + Y are uncorrelated, we need to show that their covariance is zero.

The covariance between X - Y and X + Y is given by:

[tex]Cov(X - Y, X + Y) = E[(X - Y)(X + Y)] - E[X - Y]E[X + Y][/tex]

Expanding the first term:

Cov(X - Y, X + Y) = E[X^2 - Y^2] - E[X - Y]E[X + Y]

Using the fact that X and Y have the same variance, we have:

[tex]E[X^2 - Y^2] = E[(X - Y)(X + Y)] = E[X^2] - E[Y^2][/tex]

And since X and Y have the same variance[tex], E[X^2] = E[Y^2][/tex], so we can simplify:

[tex]E[X^2 - Y^2] = 0[/tex]
Next, we can expand the second term:

E[X - Y]E[X + Y] = (E[X] - E[Y])(E[X] + E[Y])

Since X and Y have the same variance, we have E[X] = E[Y], so:

E[X - Y]E[X + Y] = (E[X] - E[X])(E[X] + E[X]) = 0

Putting it all together:

Cov(X - Y, X + Y) = 0 - 0 = 0

Therefore, X - Y and X + Y are uncorrelated.

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The sequence for the nth term = 5n+6 is: 11, 16, 21, 26, 31,...

The sequence for the nth term = 5n+6 is:

n = 1, the nth term is 5(1)+6 = 11

n = 2, the nth term is 5(2)+6 = 16

n = 3, the nth term is 5(3)+6 = 21

n = 4, the nth term is 5(4)+6 = 26

n = 5, the nth term is 5(5)+6 = 31

and so on.

Therefore, the sequence for the nth term = 5n+6 is: 11, 16, 21, 26, 31,...

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The probability P(z > -2.75) using the standard normal distribution is approximately 0.9970.

To find the indicated probability using the standard normal distribution, we need to use a z-table.

The standard normal distribution has a mean of 0 and a standard deviation of 1. The z-score represents the number of standard deviations away from the mean.

To find P(z>-2.75), we need to find the area under the standard normal distribution curve to the right of -2.75.

Using a z-table, we can find that the area to the right of -2.75 is 0.9970.

Therefore, P(z>-2.75) = 0.9970 or approximately 0.997.

This means that there is a 99.7% probability that a randomly selected value from the standard normal distribution will be greater than -2.75 standard deviations from the mean.

To find the indicated probability P(z > -2.75) using the standard normal distribution, follow these steps:

1. Identify the z-score: In this case, the z-score is -2.75.

2. Use the standard normal distribution table or a calculator with a built-in z-table function to find the area to the left of the z-score.

3. Since we need to find the probability P(z > -2.75), we'll subtract the area to the left of the z-score from 1 (the total probability).

Step-by-step calculation:

1. z-score = -2.75

2. Look up the area to the left of the z-score in the standard normal distribution table or using a calculator. For z = -2.75, the area to the left is approximately 0.0030.

3. To find the probability P(z > -2.75), subtract the area to the left from 1:

  P(z > -2.75) = 1 - 0.0030 = 0.9970

So, the probability P(z > -2.75) using the standard normal distribution is approximately 0.9970.

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A. Optgion C is correct. y = - 4/5x + 94

b. The distance that Maria would have covered from her house is given as 58 meters

The formula to use here is

y2 - y1 / x2 - x1

= 70 - 94 / 30 - 0

= - 24 / 30

divide through by 6

= - 4 / 5

y = - 4/5x + 94

When x = 45

y = - 4/5x + 94

= -4 / 5 * 45 + 94

y = 180 / 5 + 94

y = -36 + 94

y = 58

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It is difficult to say much more about the sample mean score.

If we know the sample mean score, which is denoted by x in your question, we can use it to make some inferences about the overall population of second-graders in that school district. However, we would need more information about the distribution of scores, such as the standard deviation or the range, to draw any conclusions about the entire population.

For example, if we assume that the distribution of scores is approximately normal, we could use the sample mean and standard deviation to calculate a confidence interval for the population mean score. This interval would give us a range of scores within which we can be reasonably confident the true population mean falls.

Without more information about the sample or the population, it is difficult to say much more about the sample mean score.

Complete question: Are you smarter than a second-grader? A random sample of 45 second-graders in a certain school district are given a standardized mathematics skills test. The sample mean score is x-54. Assume the standard deviation of test scores is o = 15. The nationwide average score on this test is 50. The school superintendent wants to know whether the second-graders in her school district have different math skills from the nationwide average. Use the a=0.05 level of significance and the P-value method with the TI-84 calculator al Part: 0/4 Part 1 of 4 State the appropriate null and alternate hypotheses.

Previous question

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This statistical test is appropriate because it allows for the examination of the effects of two independent variables (Type of Organization and Gender) on a continuous dependent variable (average donations).

The researcher should use a two-way ANOVA (analysis of variance) test to assess the difference in charitable donations based on both the type of organization and gender. This will allow the researcher to determine whether  To analyze the difference in charitable donations based on the Type of Organization and Gender, the researcher should use a Two-Way ANOVA (Analysis of Variance). This statistical test is appropriate because it allows for the examination of the effects of two independent variables (Type of Organization and Gender) on a continuous dependent variable (average donations).

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Based on the given information, we can expect the p-value for a two-sided test using this data to be smaller than 0.05. This is because the 95% confidence interval for the mean does not contain 0, which indicates that there is a statistically significant difference between the yield of silver queen and country gentlemen.

A p-value is a measure of the evidence against the null hypothesis (which in this case would be that there is no difference between the two corn varieties). The smaller the p-value, the stronger the evidence against the null hypothesis and the more likely we are to reject it. In general, if the p-value is smaller than the chosen significance level (usually 0.05 or 0.01), we reject the null hypothesis and conclude that there is a significant difference between the two groups being compared.

Therefore, in this case, we can expect the p-value to be smaller than 0.05 and conclude that there is a significant difference in the yield of silver queen and country gentlemen.

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A researcher conducted a study investigating the relationship between police officer-reported on-the-job stress and the number of reports of officer misconduct.

As a researcher, the investigation into the number of reports of police officer misconduct in relation to on-the-job stress levels is an important area of study. However, it is essential to ensure that ethical considerations are followed throughout the research process to avoid any potential misconduct.

In terms of the findings,, the results showed a relationship between on-the-job stress and the number of reported incidents of misconduct. Specifically, officers who reported higher levels of stress experienced more incidents of misconduct compared to those who reported minimal stress. It is crucial to further examine the factors contributing to this relationship and develop strategies to mitigate the negative impact of on-the-job stress on police officers.

Based on your question, a researcher conducted a study investigating the relationship between police officer-reported on-the-job stress and the number of reports of officer misconduct. The stress levels were categorized as minimal, moderate, and severe. However, the specific results were not provided in your question.

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The inequality showing the maximum number of hours left for her to work that week is: x ≤ 12 and this means she can work a maximum of 12 remaining hours for that week

Inequalities could either be in these forms:

Less than

Greater than

Greater than or equal to

Less than or equal to

We are given the parameters:

Maximum number of hours per week = 38 hours

Number of hours she has worked so far this week is 26

If the remaining number of hours she can work is x, then we have the inequality as:

x + 26 ≤ 38

Subtract 26 from both sides to get:

x ≤ 12

Read more about Inequality word problems at: https://brainly.com/question/25275758

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