Please help me with question four thank you !!

Answer:

14+14 + 4pi= 28 + 4pi

\(\pi\)

Answer:

E

Step-by-step explanation:

Here, we want to find the area of the shaded portion on the graph.

That would be P(-1.2<z<0.85)

we complete this by checking these values on the standard score table as follows;

P( z< 0.85) - P(z<-1.2) = 0.68727

Answer:

\(\displaystyle 1\frac{2}{5}=\frac{7}{5}\)

Step-by-step explanation:

\(\displaystyle a\frac{b}{c}=\frac{ac+b}{c}\\\\1\frac{2}{5}=\frac{(1)(5)+2}{5}=\frac{5+2}{5}=\frac{7}{5}\)

Answer:

Step-by-step explanation:

(4-3)/(2-1= 1/1 = slope = 1

y =(1)x + b,  plug in either point to solve for b

3 = 1 +b

b = 2

A)  The 95% confidence interval is:

     0.58 ± 0.062

B) At the 0.01 probability value, there is neither substantial evidence of a home-field  advantage in professional football (they won and over half of the games).

Now, According to the question:

A) Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × \(\frac{\sqrt{pq} }{n}\)

Where:

z = The z score corresponds to the amount of confidence.

p = sample proportion.

q = probability of failure

q = 1 - p

p = x/n

Where

n = the number of samples

x = the number of success

From the information given,

n = 240

x = 138

p = 138/240 = 0.58

q = 1 - 0.58 = 0.42

To determine the z score,  The confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

Thus,

1 - 0.025 = 0.975

The z -score associated with the area just on z table approximately 1.96. Therefore, the z score with a 95% confidence level is 1.96.

As a result, the 95% confidence interval becomes

0.58 ± 1.96√(0.58)(0.42)/240

Confidence interval is

0.58 ± 0.062

B) Earning more than half of the games equates to winning 120 games or more.

p = 120/240 = 0.5

The hypothesis test will be

For the null hypothesis,

P ≥ 0.5

For the alternative hypothesis,

P < 0.5

Probability of success, p = 0.5

q = probability of failure = 1 - p

q = 1 - 0.5 = 0.5

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 138

n = number of samples = 240

P = 138/240 = 0.58

We need to find the values of  the test statistic which will be the z score

z = (P - p)/√pq/n

z = (0.58 - 0.5)/√(0.5 × 0.5)/240 = 2.48

Remember that this is a two-tailed test. We would use the normal distribution table to calculate the probability potential of the property to the right of both the z score.

P value will be = 1 - 0.9934 = 0.0066

Since alpha, 0.01 > the p value, 0.0066, then we would reject the null hypothesis.

The given question is incomplete, The complete question is this:

__"During the 2000 season, the home team won 138 out of 240 regular season National Football League games. (15 points) a) Construct a 95% confidence interval for the winning proportion of the home team during this season. b) At the 0.01 significance level, is there strong evidence of a home field advantage (they win more than half of the games) in professional football? State hypotheses, calculate the test statistic and p-value, and make a conclusion in context"__

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Answer:

22×4=88÷2=44+121=165

Answer:

±1

±1/2

±1/11

±1/22

±11

±11/2

±121

±121/2

Step-by-step explanation:

b=books

3b\(\leq\)23

\(b\leq 23/3\)

\(b\leq 7.67\)

L=lesson

25L\(\leq\)150

L\(\leq\)6

Hope that helps :)

Answer: a)7, B)8

Step-by-step explanation:

Problem a) If Kai wants to buy just one poster that costs 2 dollars, he has 25-2=23 dollars left. If each book is 3 dollars, then he  can buy 23/3 books. The inequality that results is that if b=#ofbooks, then b< or = (25-2)/3, because you cant buy a book for 2 and a half dollars, hence the less than. 23/3 is 7 r2. We get that b< or = to 7 2/3. However, he can't buy 2/3 of a book, so 7. The final inequality we get is that 2+3b<=25

Problem 2) She spends 50 initially, so 200-50=150 dollars left. Thus the number of lessons, or n, cover the rest of the money. Using the same thory as above, the final inequality is 50+25n<=200. n=8.

After 4000 years, only about 52.5 mg of Radium-226 remains in the original sample containing 300 mg of Radium-226

So if a sample contains 300mg of Radium-226 and 4000 years have passed since its creation, we can calculate the remaining amount of Radium-226 using the following formula:

Remainder= Initial value * (1/2) ^ (time / half-life)

where the Initial amount is the initial amount of Radium-226 in milligrams, time is the time in years, and the half-life is the half-life of Radium-226 in years.

Putting the values given into the formula, we get

Amount remaining = 300 * (1/2)^(4000/1590)

= 300 * (1/2)^2.5

= 300 * 0.1755

≈ 52.5 mg

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To find the critical numbers of a function, take its derivative, set it equal to zero, and solve for x.

To find the critical numbers of the function f(x) = x-1 * x²-x+1, we need to take its derivative and set it equal to zero. Let's differentiate the function first. The derivative of f(x) is given by f'(x) = 3x² - 3x - 1. Now, we set f'(x) equal to zero and solve for x.

Solving the equation 3x² - 3x - 1 = 0 can be done using various methods like factoring, quadratic formula, or graphing. Upon solving, we find that there are two critical numbers, approximately x ≈ -0.347 and x ≈ 1.347.

These are the values of x where the function f(x) may have local extrema or points of inflection.


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Answer:

70cm

Step-by-step explanation:

Just like a rectangle, the formula for finding the area of a parallelogram is basexheight. \(14*5=70\)

Hey there!

ASSUMING:

p = 2l + 2w

IF SO:

• We already have our perimeter (p) & our width (w), now you have to find our length (l)

160 = 2l + 2(34)

CONVERT TO:

2l + 2(34) = 160

SIMPLIFY IT:

2l + 68 = 160

SUBTRACT 68 to BOTH SIDES:

2l + 68 - 68 = 160 - 68

SIMPLIFY IT:

2l = 160 - 68

2l = 92

DIVIDE 2 to BOTH SIDES:

2l/2 = 92/2

SIMPLIFY IT:

l = 92/2

l = 46

Therefore, your answer should be:

l = 46

Good luck on your assignment & enjoy your day!

~Amphitrite1040:)

106 bags needed to pack all the brownies.

The unitary technique involves first determining the value of a single unit, followed by the value of the necessary number of units.

For example, Let's say Ram spends 36 Rs. for a dozen (12) bananas.

12 bananas will set you back 36 Rs. 1 banana costs 36 x 12 = 3 Rupees.

As a result, one banana costs three rupees. Let's say we need to calculate the price of 15 bananas.

This may be done as follows: 15 bananas cost 3 rupees each; 15 units cost 45 rupees.

Given:

Jack and Dwayne made 318 Brownies.

Each small bag contain 3 Brownies.

So, the number of bags required

= 318 / 3

= 106 bags

Hence, 106 bags needed.

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The 98% confidence interval is (433.5866,854.6934)

What is a confidence interval?

A confidence interval (CI) for an unknown parameter in frequentist statistics is a range of estimations. The most popular confidence level is 95%, but other levels, such 90% or 99%, are occasionally used for computing confidence intervals. The fraction of related CIs over the long run that actually contains the parameter's true value is what is meant by the confidence level.

Given n = 14

Mean x = 644.14

standard deviation is s = 297.29

here α= 0.02

t(α</2,-1) = 2.65

The 98% confidence interval for mean is given by

z = x'- u / α(/√n) = (23-24/5.5/√5) =

x'-u / α(/√n) ≤ µ ≤ x'+u / α(/√n)

⇒(433.5866≤µ≤854.6934)

Hence, The 98% confidence interval is (433.5866,854.6934)

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Answer:

42 minutes

Step-by-step explanation:

Answer:

$196,797

Step-by-step explanation:

Answer:

255,000 is the value of them home

The number of prime factors of 3×5×7+7 is 3.

To find the number of prime factors, we need to calculate the given expression:

3×5×7+7 = 105+7 = 112.

The number 112 can be factored as 2^4 × 7.

In the first step, we factor out the common prime factor of 7 from both terms in the expression. This gives us 7(3×5+1). Next, we simplify the expression within the parentheses to get 7(15+1). This further simplifies to 7×16 = 112.

So, the prime factorization of 112 is 2^4 × 7. The prime factors are 2 and 7. Therefore, the number of prime factors of 3×5×7+7 is 3.

In summary, the expression 3×5×7+7 simplifies to 112, which has three prime factors: 2, 2, and 7. The factor of 2 appears four times in the prime factorization, but we count each unique prime factor only once. Thus, the number of prime factors is 3.

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Step-by-step explanation:

To find the Highest Common Factor (H.C.F.) of 567 and 255 using Euclid's division lemma, we can follow these steps:

Step 1: Apply Euclid's division lemma:

Divide the larger number, 567, by the smaller number, 255, and find the remainder.

567 ÷ 255 = 2 remainder 57

Step 2: Apply Euclid's division lemma again:

Now, divide the previous divisor, 255, by the remainder, 57, and find the new remainder.

255 ÷ 57 = 4 remainder 27

Step 3: Repeat the process:

Next, divide the previous divisor, 57, by the remainder, 27, and find the new remainder.

57 ÷ 27 = 2 remainder 3

Step 4: Continue until we obtain a remainder of 0:

Now, divide the previous divisor, 27, by the remainder, 3, and find the new remainder.

27 ÷ 3 = 9 remainder 0

Since we have obtained a remainder of 0, the process ends here.

Step 5: The H.C.F. is the last non-zero remainder:

The H.C.F. of 567 and 255 is the last non-zero remainder obtained in the previous step, which is 3.

Therefore, the H.C.F. of 567 and 255 is 3.

The recurrence interval for the second largest flood on the Bow River in 1932 is approximately 1.0106 years. The probability of a flood with a discharge of 1,520 m3/s occurring next year is roughly 98.95%. When examining the 100-year trend of peak discharges, excluding major floods, there is likely a general pattern of fluctuations but with overall stability in typical peak discharge values.

Using the provided data on the highest discharge per year on the Bow River at Calgary, Canada, we can calculate the recurrence interval (R) for specific flood magnitudes and determine the probability of such floods occurring in the future. Additionally, we can examine the 100-year trend for floods on the Bow River, excluding major floods, to identify the general trend of peak discharges over time.

1) Calculating the Recurrence Interval for the Second Largest Flood (1,520 m3/s in 1932):

To calculate the recurrence interval (R) for the second largest flood, we need to determine the rank of that flood. Since there are 95 data points in total, the rank of the second largest flood would be 94 (as the largest flood, in 2013, is excluded). Applying the formula R = (n + 1) / r, we have:

R = (95 + 1) / 94 = 1.0106 years

Therefore, the recurrence interval for the second largest flood (1,520 m3/s in 1932) is approximately 1.0106 years.

2) Probability of a Flood of 1,520 m3/s Occurring Next Year:

The probability of a flood of 1,520 m3/s happening next year can be calculated by taking the reciprocal of the recurrence interval for that flood. Using the previously calculated recurrence interval of 1.0106 years, we can determine the probability:

Probability = 1 / R = 1 / 1.0106 = 0.9895 or 98.95%

Thus, the probability of a flood of 1,520 m3/s occurring next year is approximately 98.95%.

3) Examination of the 100-Year Trend for Floods on the Bow River:

To analyze the 100-year trend for floods on the Bow River while excluding major floods, we focus on the peak discharges over time. Without considering the labeled major floods, we can observe the general trend of peak discharges.

Unfortunately, without specific data on the peak discharges for each year, we cannot provide a detailed analysis of the 100-year trend. However, by excluding major floods, it is likely that the general trend of peak discharges over time would show fluctuations and variations but with a relatively stable pattern. This implies that while individual flood events may vary, there might be an underlying consistency in terms of typical peak discharges over the 100-year period.

In summary, the recurrence interval for the second largest flood on the Bow River in 1932 is approximately 1.0106 years. The probability of a flood with a discharge of 1,520 m3/s occurring next year is roughly 98.95%. When examining the 100-year trend of peak discharges, excluding major floods, there is likely a general pattern of fluctuations but with overall stability in typical peak discharge values.

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Answer:

42 cannot be divided into 926 .. however 926 can be divided by 42 and the answer will be 22.04 rounded of to 22

so your answer is: 22

Answer:

4 or 7

Step-by-step explanation:

First job: 1 penny

Second job: Double the first one, 2*1, so 2

Third job: Double the second job, 2*2, so 4

On the third job you make 4 pennies. Overall you make 1+2+4=7. I'm not sure if you mean just the third job, or after the third job. I put both answers just in case.

To determine the payment for the third job, we need to find the payment for the second job, as the payment for the third job will be double that of the second job.

The payment for the first job is 1 penny. The payment for the second job will be double the payment for the first job, which is 2 pennies.

Therefore, the payment for the third job will be double the payment for the second job, which is 4 pennies.

This situation is an example of exponential growth, where the quantity being measured increases at a constant percentage rate with respect to time or some other independent variable.

Exponential growth is commonly seen in many real-world situations, such as population growth, compound interest, and the spread of diseases.

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The degree measure of the largest angle is 150 degrees

It is important to note that the sum of the angles in a quadrilateral is 360.

This is so because the general formula is;

(n -2)1180

Substitute n = 4

(2)180

360 degrees

From the information given, we have that the ratio of the angles are;

1:2:4:5

The largest is 5, then;

5/ 1 + 2 + 4 + 5 × 360

Add the denominators

5/12 × 360

Multiply the values

1800/12

Divide the values

150 degrees

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Given equation is

The quadratic formula for the quadratic equation

is

Here, a=1,b=-8,c=11.

So, the solutions are

So, the correct options are C and E.

Answer:

28/4=7

Step-by-step explanation:

Divide 4 by 28 and you get 7

9514 1404 393

Explanation:

It is often useful to put everything in terms of sine and cosine.

  sin(x)·(tan(x) +cot(x))

  = sin(x)·(sin(x)/cos(x) + cos(x)/sin(x))

  = sin(x)²/cos(x) +cos(x) . . . . . . use the distributive property*

  = (sin(x)² +cos(x)²)/cos(x) . . . . . combine terms over a common denominator

  = 1/cos(x)

  = sec(x)

_____

* The expression inside parentheses could have been combined to ...

  (sin(x)² +cos(x)²)/(sin(x)·cos(x))

Then the sin/sin factors cancel, leaving the expression on line 4.

The given function is:

At x = 0:

Therefore, one point is (0, 3)

At x = 12:

Another point is (12, 0)

The graph of g(x) considering the points (0, 3) and (12, 0) is plotted below

Step-by-step explanation:

f(x) =14x+4

h(x)=7x

f(x)-h(x)=14x+4-7x

=7x+4

Answer:

Step-by-step explanation:

You are given two points whose y values are the same. There is no need to use the distance formula. The x values alone will give you the length of the line.

to get to the y axis you have to go 12 units to the right.

To get from the y axis to 9, you have to go 9 more units to the right.

The total is 9 + 12 = 21

If you insist to see this done with the formula, it would look like this.

d = sqrt( y2 - y1)^2 + (x2 - x1)^2 )

y2 = 6

y1 = 6

x2 = -12

x1 = 9

d = sqrt ( (6 - 6)^2 + (-12 - 9)^2 )

d = sqrt( 0 + ( -21 )^2 )

d = sqrt( 441)

d = 21 the same thing you got just be reasoning it out.

The first method works only if either both xs are the same or both ys are the same.

Therefore, The equation (x-7)^2 y^2 = 49 in polar coordinates is r^2 - 14r cos(theta) + 49 = 0.

To convert the equation (x-7)^2 y^2 = 49 to polar coordinates, we replace x with r cos(theta) and y with r sin(theta). This gives us (r cos(theta) - 7)^2 (r sin(theta))^2 = 49. Simplifying this equation, we get r^2 - 14r cos(theta) + 49 = 0. This is the equation in polar coordinates.

Therefore, The equation (x-7)^2 y^2 = 49 in polar coordinates is r^2 - 14r cos(theta) + 49 = 0.

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Abby can make 5 complete groups with 9 extra bonus coupons.

The division is one of the basic arithmetic operations in math in which a larger number is broken down into smaller groups having the same number of items.

Here, Abby has total number of coupons = 159

        Number of coupons in a group = 30

Total number of groups = Total Coupons / coupons in a group

                                         = 159 / 30

                                         = 5.3

Thus, Abby can make 5 complete groups with 9 extra bonus coupons.

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