A marine sales dealer Önds that the average price of a previously owned boat is $6492. He decides to sell boats that will appeal to the middle 66% of the market in terms of price. Find the maximum and minimum prices of the boats the dealer will sell. The standard deviation is $1025, and the variable is normally distributed.

Answer:

5x-3

Step-by-step explanation:

x+4+7x-1-4x

5x-3

Answer:

7 points total

Step-by-step explanation:

So basically the x represents the total amount of times you landed on that color so since its one of each you just replace the x with one and solve normally. 1+4=5 points for green, 7(1)-1= 6 points for blue -4(1)= -4 points for red  

Answer:

Option D) 0.65    

Step-by-step explanation:

We are given the following in the question:

Percentage of executives who read Time magazine = 35%

[tex]P(M) = 0.35[/tex]

Percentage of executives who read U.S. News & World Report = 40%

[tex]P(N) = 0.4[/tex]

Percentage of executives who read both Time magazine and U.S. News & World Report = 10%

[tex]P(M\cap N) = 0.1[/tex]

We have to find the probability that a particular top executive reads either Time or U.S. News & World Report regularly.

Thus, we have to evaluate,

[tex]P(M\cup N) = P(M) + P(N) -P(M\cap N)[/tex]

Putting values, we get,

[tex]P(M\cup N) = 0.35 + 0.4 - 0.1=0.65[/tex]

0.65 is the probability that a particular top executive reads either Time or U.S. News & World Report regularly.

Thus, the correct answer is

Option D) 0.65

Answer:

The answers are for option (a) 0.2070  (b)0.3798  (c) 0.3938

Step-by-step explanation:

Given:

Here Section 1 students = 20

Section 2 students = 30

Here there are 15 graded exam papers.

(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070

(b) Here if x is the number of students copies of section 2 out of 15 exam papers.

 here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15

Then,

Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798

(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)

so,

Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938

Note : Here the given distribution is Hyper-geometric distribution

where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.

Answer:

5(8-x)

Step-by-step explanation:

Fact using the GCF (Greatest Common Factor) of 40 and 5

Factors of 5:

1,5

Factors of 40:

1,2,4,5,8,10,20,40

The greatest factor they both have is 5

Factor by distributing out a 5

40-5x

5(8-x)

The GCF (Greatest Common Factor) of 40 and 5 is 5!

Divide both numbers by 5.

40 / 5 = 8

-5x / 5 = -x

Rewrite the expression using the distributive property.

5(8 - x)

Best of Luck!

Answer:

4

Step-by-step explanation:

AC = AE = 4 cm

Answer: after driving 330 miles, the remaining fuel in each tank be the same.

Step-by-step explanation:

Let x represent the number of miles it will take for the the remaining fuel in each tank to be the same.

Jerry has a large car which holds 22 gallons of fuel and gets 20 miles per gallon. It means that the number of gallons needed to drive 1 mile is 1/20. Then the number of gallons needed to drive x miles is

1/20 × x = x/20

If the tank is full, then the number of gallons of fuel left after driving x miles is

22 - x/20

Kate has a smaller car which holds 16.5 gallons of fuel and gets 30 miles per gallon. It means that the number of gallons needed to drive 1 mile is 1/30. Then the number of gallons needed to drive x miles is

1/30 × x = x/30

If the tank is full, then the number of gallons of fuel left after driving x miles is

16.5 - x/30

For the remaining fuel in each tank to be the same, it means that

22 - x/20 = 16.5 - x/30

Multiplying both sides of the equation by 60(LCM), it becomes

1320 - 3x = 990 - 2x

- 2x + 3x = 1320 - 990

x = 330 miles

Answer:

[tex]\theta_{1} = \frac{\pi}{3} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]

[tex]\theta_{2} = \frac{5\pi}{3} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]

Step-by-step explanation:

The critical numbers are found by the First Derivative Test, which consists in differentiating the function, equalizing it to zero and solving it:

[tex]g'(\theta) = 16 - 4\cdot \sec^{2} \theta[/tex]

Following equation needs to be solved:

[tex]16 - 4\cdot \sec^{2}\theta = 0[/tex]

[tex]\sec^{2}\theta = 4[/tex]

[tex]\cos^{2}\theta = \frac{1}{4}[/tex]

[tex]\cos \theta = \frac{1}{2}[/tex]

The solution is:

[tex]\theta = \cos^{-1} \frac{1}{2}[/tex]

Given that cosine is a periodical function, there are two subsets of solution:

[tex]\theta_{1} = \frac{\pi}{3} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]

[tex]\theta_{2} = \frac{5\pi}{3} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]

The critical numbers of the function g(θ) = 16θ − 4 tan(θ) can be found by determining where the derivative of the function is zero or undefined. However, specific critical numbers can't be determined from the provided prompt due to complexity of the derivative.

The critical numbers of a function occur when the derivative of the function is equal to zero or undefined. Given the function g(θ) = 16θ − 4 tan(θ), the first step to be done is finding the derivative (g'(θ)) of the function.

This involves applying the rules of differentiation to each term: the coefficient rule (b), and chain rule for the tangent part. After calculating the derivative, we set g'(θ) equal to zero and solve for θ to determine the critical numbers. For the term involving tan(θ), it is undefined at θ = π/2 + πn, where n is an integer. Therefore, these are also considered as critical numbers.

However, due to the complexity of the derivative, finding critical numbers usually requires using algebraic and trigonometric techniques or potentially numerical methods if the equation cannot be solved analytically. As the specific steps for this complex derivative calculation are not provided in the prompt, we can't provide the specific critical numbers.

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Final answer:

To test whether the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer, we can use a one-sample proportion test. Using a significance level of 0.05, we find that the test statistic is -2.09. Comparing this to the critical value from the standard normal distribution (-1.645), we reject the null hypothesis and conclude that there is evidence to suggest that the percentage of correctly guessed penalty shots is less than 50%.

Explanation:

To test whether there is evidence that the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer, we can use a one-sample proportion test. We will assume that the null hypothesis is true and that the goalkeeper's correct guesses are no better than random chance. The alternative hypothesis would be that the goalkeeper's correct guesses are significantly less than 50%. Using a significance level of 0.05, we can calculate the test statistic and compare it to the critical value from the standard normal distribution.

Null hypothesis (H0): The percentage of correctly guessed penalty shots is 50%.

Alternative hypothesis (Ha): The percentage of correctly guessed penalty shots is less than 50%.

Test statistic: We can use the z-test statistic since the sample size is large enough. The formula for the z-test statistic is z = (p - P0) / SE, where p is the sample proportion, P0 is the hypothesized proportion, and SE is the standard error. In this case, since the standard error is given as 0.043, we can plug in the values to calculate the test statistic.

Calculate the z-test statistic: z = (0.41 - 0.5) / 0.043 = -2.09

Find the critical value: Since our alternative hypothesis is that the percentage is less than 50%, we will use a one-tailed test. With a significance level of 0.05, the critical value from the standard normal distribution is -1.645.

Compare the test statistic to the critical value: Since the test statistic (-2.09) is less than the critical value (-1.645), we can reject the null hypothesis. There is evidence to suggest that the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer.

Answer:

(a) The outlier in the data is $810.

(b) The distribution of money spent on textbooks for the 34 students is right skewed.

Step-by-step explanation:

The data provided for the amount of money 34 college students spent on books is:

S = {120, 130, 130, 140, 150, 150, 160, 170, 180, 210, 220, 230, 240, 250, 260, 280, 280, 290, 290, 290, 310, 320, 320, 370, 380, 390, 410, 440, 450, 470, 510, 530, 620, 810}

(a)

An outlier of a data set is a value that is very different from the other values of a data set. It is either too large or too small.

The most common way to determine whether a data set consists of any outliers of not is,

Here

Q₁ = first quartile

Q₃ = third quartile

IQR = Inter-quartile range = Q₃ - Q₁.

The first quartile is the value that is more than 25% of the data values. The first quartile is the median of the first half of the data.

Compute the value of first quartile as follows:

First half of data: {120, 130, 130, 140, 150, 150, 160, 170, 180, 210, 220, 230, 240, 250, 260, 280, 280}

There are 17 values.

The median of an odd data set is the middle value.

The middle value is: 180

The first quartile is Q₁ = 180.

The third quartile is the value that is more than 75% of the data values.

Compute the value of first quartile as follows:

Second half of data: {290, 290, 290, 310, 320, 320, 370, 380, 390, 410, 440, 450, 470, 510, 530, 620, 810}

There are 17 values.

The median of an odd data set is the middle value.

The middle value is: 390

The third quartile is Q₃ = 390.

Compute the inter-quartile range as follows:

IQR = Q₃ - Q₁

      = 390 - 180

      = 210

Compute the value of [Q₁ - 1.5 IQR] as follows:

[tex]Q_{1}-1.5\ IQR =180-(1.5\times 210)=-135[/tex]

Compute the value of [Q₃ + 1.5 IQR] as follows:

[tex]Q_{3}+1.5\ IQR =390+(1.5\times 210)=705[/tex]

There are no values that are less than [Q₁ - 1.5 IQR]. But there is one value that is more than [Q₃ + 1.5 IQR].

X = 810 > [Q₃ + 1.5 IQR] = 705

Thus, the outlier in the data is $810.

(b)

A distribution is known as to be skewed to the right, or positively skewed, when maximum of the data are collected on the left of the distribution.

In the stem plot above, it is shown that maximum of the data values are collected on the left of the chart. This implies that the distribution is positively skewed.

Thus, the distribution of money spent on textbooks for the 34 students is right skewed.

Outliers are data elements that are relatively far from other data elements

(a) The outlier

From the plot, the stems are given as:

Stems: 1 2 3 4 5 6 7 8

However, stem 7 does not have any leaf

Using the above highlight, the next entry after 620 is 810

810 is relatively far from the other datasets.

Hence, 810 is an outlier

(b) The distribution

The data on the stem-plot is more concentrated at the top, and it reduces as the stem increases.

When there are more data elements at the top or left, then the distribution is right skewed.

Hence, the distribution is right skewed.

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Answer:

D. 12.5

Step-by-step explanation:

You can notice that ∠JMK and ∠JML are similar triangles.

That means that JM (from triangle on the left) is similar to KM (from triangle of the right)- the ratio between these must be:

[tex]\frac{6}{8}[/tex], that means that we can multiple KM (from triangle on the right) by this ratio to get ML (from triangle on the right).

[tex]6 * \frac{6}{8}=\frac{36}{8}=4.5[/tex]

Now we add JM to ML to to get JL - so 8 + 4.5 = 12.5

Answer:

12.5

Step-by-step explanation:

x less-than-or-equal-to 6 is the inequality which is represented by the graph.

The problem can be solved by following steps

It is given that

According to above statement a graph is drawn below

The shaded part of the left side describes the value of x which are less than 6 or equal to 6

Hence , The inequality that is represented in the graph is x is less than or equal to 6

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Answer: B

Step-by-step explanatii got it right on edge

This question is incomplete, in that the Excel File: data07-11.xlsx a. was not provided, but I was able to get the information on the Excel File: data07-11.xlsx a. from google as below:

57 61 86 74 72 73

20 57 80 79 83 74

The image of the Excel File: data07-11.xlsx a. is also attached below.

Answer:

a) Point estimate of sample mean  = 68

b) Point estimate of standard deviation (4 decimals) = 17.8122

Step-by-step explanation:

a) Point estimate of sample mean, \bar{x}  =  ∑Xi / n = (57 + 61 + 85 + 74 + 73 + 72 + 20 + 58 + 81 + 78 + 84 + 73)/12 = 68

b) Point estimate of standard deviation = sqrt ∑ Xi² - n\bar{x}² / n-1)

= sqrt(((57 - 68)^2 + (61 - 68)^2 + (85 - 68)^2 + (74 - 68)^2 + (73 - 68)^2 + (72 - 68)^2 + (20 - 68)^2 + (58 - 68)^2 + (81 - 68)^2 + (78 - 68)^2 + (84 - 68)^2 + (73 - 68)^2)/11) = 17.8122

The value is 50

A fraction represents a part of a whole or, more generally, any number of equal parts.

Given:

p/4 + pq

As, p= 8, q= 6

We have,

8/4 + 8*6

= 2 + 48

= 50

Hence, p/4 + pq= 50

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The equation results to get 50.

To evaluate the expression p/4 + pq when p = 8 and q = 6, follow these steps:

Therefore, the value of the expression when p = 8 and q = 6 is 50.

The complete question is:

Evaluate p/4 +pq when p=8 and q=6

Answer:

The slope is -1

Step-by-step explanation:

Let's find the slope between your two points.

(1,4);(6,−1)

(x1,y1)=(1,4)

(x2,y2)=(6,−1)

Use the slope formula:

m= y2−y1/x2−x1  = −1−4/6−1

= −5/5

= −1

Hope this is a better explanation :)

Hope it helps!!!!!!!!!

Please provide us with an attachment of image A and B so we can answer your question correctly.

8 (y - 7) = - 16

Divide by 8 on both sides

y - 7 = - 2

Add 7 to both sides

y = 5

Answer:

y = 5

Step-by-step explanation:

8(y-7) = -16

Divide each side by 8

8(y-7)/8 = -16/8

y-7 = -2

Add 7 to each side

y -7+7 = -2+7

y = 5

Answer: a) 1.029, b) (-5.318, -1.282).

Step-by-step explanation:

Since we have given that

[tex]n_1=13\\\\n_2=17\\\\\bar{x_1}=1.9\\\\\bar{x_2}=5.2[/tex]

and

[tex]s_1=2.1\\\\s_2=3.5[/tex]

So, the standard error for comparing the means :

[tex]SE=\sqrt{\dfrac{s^2_1}{n_1}+\dfrac{s^2_2}{n_2}}\\\\SE=\sqrt{\dfrac{2.1^2}{13}+\dfrac{3.5^2}{17}}\\\\SE=\sqrt{1.0598}\\\\SE=1.029[/tex]

At 95% confidence interval, z = 1.96

So, Confidence interval would be

[tex]\bar{x_1}-\bar{x_2}\pm z\times SE\\\\=(1.9-5.2)\pm 1.96\times 1.0294\\\\=-3.3\pm 2.017624\\\\=(-3.3-2.018,-3.3+2.018)\\\\=(-5.318,-1.282)[/tex]

Hence, a) 1.029, b) (-5.318, -1.282).

Answer:

[tex]\frac{33}{36}[/tex]

Step-by-step explanation:

Combinations greater than a 10

5 - 5

5 - 6

6 - 5

There are total 36 combinations (6 * 6).

3 of these combinations are higher than 10.

So 36 - 3 combinations are less than 10.

Answer:

7.64% probability that they spend less than $160 on back-to-college electronics

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 237, \sigma = 54[/tex]

Probability that they spend less than $160 on back-to-college electronics

This is the pvalue of Z when X = 160. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{160 - 237}{54}[/tex]

[tex]Z = -1.43[/tex]

[tex]Z = -1.43[/tex] has a pvalue of 0.0763

7.64% probability that they spend less than $160 on back-to-college electronics

The capability of the process is determined using the process capability index (Cp). In this case, the process has a capability of 0.833, indicating that it is not meeting the specifications well.

To calculate the capability of the process, we need to use the process capability index (Cp). Cp is calculated by dividing the tolerance range by six times the standard deviation. The tolerance range in this case is 65 - 55 = 10 psi. So, Cp = 10 / (6 * 2) = 10 / 12 = 0.833. Since Cp is a measure of how well the process meets the specifications, a value closer to 1 indicates a better capability.

In this case, the process has a capability of 0.833, which means it is not meeting the specifications very well.

Keywords: capability, process capability index, tolerance range, standard deviation, specifications

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The process capability index (Cpk) for the grease aerosol can pressurization process is 0.67, indicating that the process is not capable of meeting the design specifications as it is less than the acceptable limit of 1.33.

The capability of the process in question relates to its ability to meet design specifications, which can be quantified using statistical measures like the process capability index (Cpk). To compute the Cpk, you need to determine the worst-case process capability scenario by comparing the distance of the process mean to the nearest specification limit in terms of standard deviations. The Cpk can be calculated using the formula:

Cpk = minimum [(USL - x) / (3σ), (x - LSL) / (3σ)]

Where USL is the upper specification limit, LSL is the lower specification limit, x is the process mean, and σ is the standard deviation.

For this process with an average of 61 psi and a standard deviation of 2 psi:

You would calculate two separate indices:

Thus, the Cpk would be the smaller of these two indices which is 0.67. A Cpk of 0.67 indicates that the process is not capable of meeting the design specifications since it is less than the acceptable limit of 1.33 for most industries.

Answer:

1. Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex]  70 mph

  Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 70 mph

2. Value of test statistics is 2.652.

Step-by-step explanation:

We are given that Kansas Highway Patrol recorded the speed of 450 interstate vehicles and found the mean speed of 70.2 mph with standard deviation 1.6 mph.

We have to conduct a hypothesis test at significance level 0.01 to decide whether or not to reduce the money sent to Kansas.

Let [tex]\mu[/tex] = true mean speed of all vehicles on the Kansas interstate highway system.

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex]  70 mph   {means that the federal government will not reduce the amount of funding it provides as the speed limit is less than or equal to 70 mph}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 70 mph   {means that the federal government will reduce the amount of funding it provides as the speed limit exceed 70 mph}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                        T.S.  = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean speed limit of 450 interstate vehicles = 70.2 mph

             s = sample standard deviation = 1.6 mph

             n = sample of vehicles = 450

So, test statistics  =   [tex]\frac{70.2-70}{\frac{1.6}{\sqrt{450} } }[/tex]  ~ [tex]t_4_4_9[/tex]

                               =  2.652

Hence, the value of test statistics is 2.652.

Answer:

For this case we want to test if there is a higher incidence of smoking among women than among men in a neighborhood (alternative hypothesis). And we define p1 for men and p2 for women, so for this case the best system of hypothesis are:

Null hypothesis:[tex] p_1- p_2 \geq 0[/tex]

Alternative hypothesis: [tex]p_1 -p_2 <0 [/tex]

And the best option would be:

a) h0: p1-p2 ≥ 0 h1: p1-p2 < 0

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Solution to the problem

For this case we want to test if there is a higher incidence of smoking among women than among men in a neighborhood (alternative hypothesis). And we define p1 for men and p2 for women, so for this case the best system of hypothesis are:

Null hypothesis:[tex] p_1- p_2 \geq 0[/tex]

Alternative hypothesis: [tex]p_1 -p_2 <0 [/tex]

And the best option would be:

a) h0: p1-p2 ≥ 0 h1: p1-p2 < 0

Answer:

The range of the p-value is: 0.050 < p-value < 0.100.

Step-by-step explanation:

For checking the equivalence of two population variances of independent samples, we use the f-test.

The test statistic is given by:

[tex]F=\frac{S_{1}^{2}}{S_{2}^{2}}\sim F_{\alpha, (n_{1}-1)(n_{2}-1)}[/tex]

It is provided that the hypothesis test is one-tailed.

The computed value of the test statistic is:

F = 4.23.

The degrees of freedom of the numerator and denominator are:

[tex]df_{1}=4\\df_{2}=5[/tex]

Use MS-Excel to compute the p-value as follows:

Step 1: Select function fX → F.DIST.RT.

Step 2: A dialog box will open. Enter the values of f-statistic and the two degrees of freedom.

*See the attachment below.

Step 3: Press OK.

The p-value is, 0.0728.

The range of the p-value is:

0.050 < p-value < 0.100

Answer:

[tex]\large \boxed{z = \dfrac{83}{6 - C - T}}[/tex]

Step-by-step explanation:

[tex]\begin{array}{rccl}-Cz + 6z & = & Tz + 83 & \\6z -Cz - Tz &= & 83 & \text{Subtracted Tz from each side}\\z(6 - C - T) & = & 83 & \text{Removed the common factor}\\z& = & \mathbf{\dfrac{83}{6 - C - T}} &\text{Divided each side by 6 - C - T}\\\end{array}\\\\\large \boxed{\mathbf{z = \dfrac{83}{6 - C - T}}}[/tex]

Answer:A

Go from the dark blue to light green flips on y and down 1

Step-by-step explanation:

Answer:

63 intervals would be expected to contain the true population average.

Step-by-step explanation:

90% confidence interval:

Mean that we are 90% sure that the interval contains the true population mean. That is, 90% of the intervals are expected to contain the true population average.

Of these 70 intervals, how many of these intervals would you expect to contain the true population average?

Following the explained logic

0.9*70 = 63

63 intervals would be expected to contain the true population average.

Answer:

so the common multiple of 4 and 6 is 2  i hope this helps!   :)

Step-by-step explanation:

well the common multiples of 4 and 6 is 2

2 * 2 = 4

2 * 3 = 6

so there has to be a certain number that goes into both of the numbers you are working with

Answer:

Common multiples of 4 and 6 are 12, 24, 36, 48, 60, 72, 84, 96...

Answer:

At first we should know that:

The properties of the square are:

The properties of Rhombus

So, Wade is incorrect because the quadrilateral may be Rhombus

And the quadrilateral to be a square, she needs to show that It has two pairs of perpendicular lines.

Answer: The answer is 1. Substitute (2,1) into x+3y=5

Step-by-step explanation: Remember, (2,1) 2 is substituted for x and 1 is used as the y substitute

Answer:

1. 2+3(1)=5

2. 5=5 is true

3. 1=-(2)+3

4. 1=1 is true

Answer:

Step-by-step explanation:

The null hypothesis is the hypothesis that is assumed to be true. It is an expression that is the opposite of what the researcher predicts.

The alternative hypothesis what the researcher expects or predicts. It is the statement that is believed to be true if the null hypothesis is rejected.

From the given situation,

The mean number of pounds of fruit on this plot of land with the old fertilizer was 416 pounds. This is the null hypothesis.

H0 : µ = 416

Agriculture scientists believe that the new fertilizer may decrease the yield. This is the alternative hypothesis.

H0 : µ < 416