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The largest two-digit integer a = 99 satisfies the congruence 3x ≡ 4

To obtain the largest two-digit integer that satisfies the congruence 3x ≡ 4 (mod 7), we need to find an integer value for x that satisfies the congruence equation.

First, we can rewrite the congruence as an equation:

3x = 4 + 7k, where k is an integer.

Next, we can iterate through two-digit integers in descending order to find the largest value of a that satisfies the equation.

Starting with a = 99, we substitute it into the equation:

3(99) = 4 + 7k

297 = 4 + 7k

By trying different values of k, we can see that k = 42 satisfies the equation:

297 = 4 + 7(42)

Therefore, x = 99 is a solution to the congruence equation 3x ≡ 4 (mod 7).

However, we need to find the largest two-digit integer, so we continue the iteration.

Next, we try a = 98:

3(98) = 4 + 7k

294 = 4 + 7k

By trying different values of k, we can see that k = 42 also satisfies the equation:

294 = 4 + 7(42)

Therefore, x = 98 is another solution to the congruence equation 3x ≡ 4 (mod 7).

Since we have found the largest two-digit integer a = 99 that satisfies the congruence, we can conclude that a = 99 is the answer to the problem.

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Answer:

16ft by 16 ft: NO

10ft by 15ft: YES

15ft by 20ft: NO

12ft by 16ft: YES

Step-by-step explanation:

First, we know that the scale used is:

1 in = (1 + 1/2) ft.

This means that each inch on the drawing is equivalent to (1 + 1/2) ft.

We know that Michael uses a paper that has the measures:

width = (8 + 1/2) in

length = 11in

Then the maximum dimensions that can be represented with this paper are:

WIDTH = (8 + 1/2)*(1 + 1/2) ft. = (8 + 8/2 + 1/2 + 1/4) ft

            = (8 + 4 + 2/4 + 1/4) ft

            = (12 + 3/4) ft

LENGTH = 11*(1 + 1/2) ft = (11 + 11/2)ft = (11 + 10/2 + 1/2)ft

               = (11 + 5 + 1/2)ft = (16 + 1/2) ft.

Now let's analyze the options, we can only draw the rooms in the paper if the measures are equal or smaller than the ones we found above:

Where the measures are written as: "width by length".

a) 16ft by 16 ft.

width = 16ft

length = 16ft

We can not draw this, because the maximum width that we can draw is (12 + 3/4) ft, which is smaller than 16ft.

b) 10 ft by 15 ft

width = 10ft

length = 15ft

Both are smaller than the maximum measures we found, then yes, we can draw this room.

c) 15 ft by 20 ft

width = 15ft

length = 20ft

Both are larger than the maximum measures, so no, we can not draw this.

d) 12ft by 16ft

width = 12ft < (12 + 3/4) ft = maximum width

lenth = 16ft < (16 + 1/2) ft = maximum length.

Both measures are smaller than the maximum ones, then we can draw this one

18. Option B is correct, the outcomes of the first five spins tell us nothing about what will happen on the next five spins.

19. Option D is correct,  the overall sample is a simple random sample, biased due to imbalance and stratified sample, option d is correct.

18. Given that a  North American roulette wheel has 38 slots, of which 18 are red, 18 are black, and 2 are green.

Each spin of the roulette wheel is an independent event, and the outcomes of previous spins do not influence the outcomes of future spins.

The wheel has no memory of previous results, so the probability of getting a red outcome on the sixth spin is the same as any other spin – 18 out of 38.

So, the outcomes of the first five spins tell us nothing about what will happen on the next five spins.

19. At a large university, a simple random sample of five female professors is selected, and a simple random sample of 10 male professors is selected.

The overall sample of 15 professors is a combination of two simple random samples, one from the female professors and the other from the male professors.

This makes it a stratified sample because it involves dividing the population (professors) into distinct groups (male and female) and then randomly sampling from each group.

Additionally, the overall sample can also be considered a simple random sample because it was obtained by randomly selecting individuals from the population without any bias

Hence, the overall sample is a simple random sample, biased due to imbalance and stratified sample, option d is correct.

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Answer:

$6.41

Step-by-step explanation:

47.92 - 3.05 = 44.87

44.87/7 = $6.41

Answer:

37

Step-by-step explanation:

This question is incomplete, the complete question is;

Alicia would like to know if there is a difference in the average price between two brands of shoes. She selected and analyzed a random sample of 40 different types of Brand A shoes and 33 different types of Brand B shoes. Alicia observes that the boxplot of the sample of Brand A shoe prices shows two outliers. Alicia wants to construct a confidence interval to estimate the difference in population means.

Is the sampling distribution of the difference in sample means approximately normal?

a) Yes, because Alicia selected a random sample

b) Yes, because for each brand it is reasonable to assume that the population size is greater than ten times its sample size.

c) Yes, because the size of each sample is at least 30

d) No, because the distribution of Brand A shoes has outliers

e) No, because the shape of the population distribution is unknown.

Answer:

the sampling distribution of  the difference in sample means is approximately normal because the size of each sample is al least 30

Option (c) Yes, because the size of each sample is at least 30 ) is the correct answer.

Step-by-step explanation:

Given the data in the question;

sample size of brand A shoes [tex]n_A[/tex] = 40

sample size of brand B shoes [tex]n_B[/tex] = 33

As we can notice,

[tex]n_A[/tex] = 40 > 30

[tex]n_B[/tex] = 33 > 30

Hence, we consider both samples to be large.

Therefore, the sampling distribution of  the difference in sample means is approximately normal because the size of each sample is al least 30.

Option (c) Yes, because the size of each sample is at least 30 ) is the correct answer.

Answer: It would be "290 in.2"

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

its 3, -6,-1... got 100%

Step-by-step explanation:

Answer:

Her results changed because the power increased due to the fact that she reduced variability in her data as a result of using a sample that had lesser variability.

Step-by-step explanation:

She is trying to find if there is difference between people who consumed 1 versus 2 beverages.

Now initially she surveyed 40 people between ages 18 to 38 years and of huge weights and found out there was no difference. However, she decided to run the survey again on only women aged 18 to 22 years with normal body weight.

Since she used a lesser sample size and surveyed only women, it means there will be lesser variability in the result as the sample is smaller and streamlined to only women.

Thus, her result changed because the power increased due to the fact that she reduced variability in her data as a result of using a sample that had lesser variability.

Answer:

v^ 3 = 1000

so v = cube root of 1000

    v = 10

Answer:

bit.[tex]^{}[/tex]ly/3a8Nt8n

Step-by-step explanation:

To find the area using the limit of a sum (a Riemann sum) of the region between the graph of y = f(x) and the x-axis from x = a to x = b, the formula is given by: Area = lim n → ∞ ∑ i = 1 n f(x* i )Δx, where f(x* i ) is the height of the ith rectangle and Δx is the width of the ith rectangle.

To find the area between the graph of y = f(x) and the x-axis from x = a to x = b using the limit of a sum, we need to first divide the interval [a, b] into n equal subintervals of length Δx = (b - a)/n. Then, we can choose any point x* i in the ith subinterval [x i-1 , x i ] and use it to determine the height of the ith rectangle f(x* i ).

Finally, we can take the limit as n approaches infinity to obtain the exact area of the region between the graph of y = f(x) and the x-axis from x = a to x = b.

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Answer:

Try 8 if it's wrong sorry

Step-by-step explanation:

Answer:

Step-by-step explanation:

An absolute value of |x| {modulus of x} is the value of a real number x. The given equation is not true.

An absolute value of |x| {modulus of x} is the value of a real number x, the value we get is always a non-negative number, for example, |-5| will give 5, and also, |5| will give 5 as well.

A hypothesis is a speculative conclusion or assertion that is made without proof. Now, to prove that the given equation is not true substitute the value of both the variables as -1. Therefore,

x = y = -1

Substitute the value on the left side of the equation,

|x - y|

= |-1 - (-1)|

= |-1 + 1|

= |0|

= 0

Substitute the value on the right side of the equation,

|x| - |y|

= |-1| - |-1|

= 1 + 1  

= 2

Since the left and the right side of the equation is not satisfied, therefore, it can be concluded that the given equation is not true.

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Step-by-step explanation:

-.5f -5 < -1

-.5f < 4

f > -8

hope this helps

Answer:

12 cm difference

Step-by-step explanation:

√64 cm² = 8 cm

8 cm x 4 = 32 cm perimeter

√121 cm² = 11 cm

11 cm x 4 = 44 cm perimeter

44 cm - 32 cm = 12 cm difference

Answer:

804 cm3

Step-by-step explanation:

multiply the lenght and the width then multiply by 3.14 which is pi.

Answer:

A- In one simulated SRS of 100 students, the average GPA was  = 3.45.

Step-by-step explanation:

Just took the assignment myself, but here's a bit of insight to help anyone later on!

The x-bar means sample statistic. Since it was one sample taken out of the dot plot, 3.45 is a sample statistic.

Answer:

-4

Step-by-step explanation:

First you add together 75 + 65 to get 140 then you subtract that from 180 because 180° is the total of a triangle. Then you add -4 to 44 to get 40

Answer:

-4

Step-by-step explanation:

75 + 65 = 140

triangle angels equal 180°

then x +44 = 180-140

x +44 = 40

x = 40-44

x = -4

The week se will give a 12-minute speech is week 22 (option A)

The table is a linear table. This is because the variables change by a fixed amount.

Rate of change = change in length of speech / change in week

(180 - 150) / (4 - 3)

= 30 / 1 = 30 seconds  

The next step is to convert minutes to seconds

1 minute = 60 seconds

60 x 12 = 720 seconds

Length of speech in week 2 = 150 - 30 = 120 seconds

Length of speech in week 1 = 120 - 30 = 90

Week the speech would have a length of 720 seconds = 90 +[ 30 x (week number - 1)]

720 = 90 + [30 x (x -1)

720 = 90 + 30x - 30

720 = 60 + 30x

720 - 60 = 30x

660 = 30x

x = 660 / 30

x = 22

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Based on the information given, it should be noted that the critical value is +2.33.

We are given that the sample size is 42, the sample proportion is 0.247, and the significance level is 0.05. We want to test the claim that the population proportion is 0.325.

The critical value is the z-score that separates the rejection region from the non-rejection region. The rejection region is the area under the standard normal curve where we would reject the null hypothesis. The non-rejection region is the area under the standard normal curve where we would fail to reject the null hypothesis.

The z-table shows that the critical value for a two-tailed test with a significance level of 0.05 is +2.33. This means that if the z-score is greater than or equal to +2.33, we would reject the null hypothesis.

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Answer:

Option B

Step-by-step explanation:

→ Find 2 coordinates through which the line passes through

( -2 , -5 ) and ( 0 , -1 )

→ Work out the gradient

[tex]\frac{-1--5}{0--2} =2[/tex]

→ Write into y = mx + c

y = 2x + c

→ Substitute in the coordinates ( 0 , -1 )

-1 = 0 + c so c = -1

→ Write equation

y = 2x - 1

→ Expand out all the equations and simply to see which gives y = 2x - 1

Option B

Answer:

A

Step-by-step explanation:

A secant is a line that goes through a circle at two points.

Answer:

A

Step-by-step explanation:

a straight line that cuts a curve in two or more parts.

Answer:

4 horas para compatir com su familia

The curvature of the function y = x³ at x = 1 is 6

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

y = x³

To start with, we need to differentiate the function

So, we have

y' = 3x²

Next, we differentiate the function for the second time to calculate the curvature of the function

So, we have

y'' = 6x

The value of x is 1

So, we have

y'' = 6(1)

Evaluate

y'' = 6

Hence, the curvature of the function y = x³ at x = 1 is 6

See attachment for the graph

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(a) To prove that there exists a neighborhood W of the origin such that f(x) ≤ y for all x ∈ W, given that f is bounded above on a neighborhood V of the origin, we can use the linearity of f.

Since f is a linear function, it satisfies the following properties:

f(0) = 0

f(rx) = rf(x) for any scalar r and vector x

f(x + y) = f(x) + f(y) for any vectors x and y

Given that f is bounded above on V, there exists a positive number M such that f(x) ≤ M for all x ∈ V. Now, let's consider the neighborhood W defined as follows:

W = {x ∈ X | ||x|| < M}

We claim that for any x ∈ W, f(x) ≤ y.

Let x ∈ W. Since x is in the neighborhood W, we have ||x|| < M. By linearity, we can express x as x = rx' for some scalar r and vector x' with ||x'|| = 1.

Now, consider f(x):

f(x) = f(rx') = rf(x')

Since ||x'|| = 1, we have ||rx'|| = |r| ||x'|| = |r|.

Therefore, ||rx'|| < M implies |r| < M.

Using the fact that f is bounded above on V, we have f(x') ≤ M.

Combining these results, we get:

|f(x)| = |rf(x')| = |r| |f(x')| ≤ M

Since this inequality holds for any x ∈ W and |r| < M, we have shown that f(x) ≤ y for all x ∈ W, where W is a neighborhood of the origin.

(b) To prove that f is continuous, we can show that f is bounded above on any compact set in X. Let K be a compact set in X.

Since K is compact, it is also closed and bounded. By the linearity of f, we have:

f(K) = {f(x) | x ∈ K}

Since K is bounded, there exists a positive number M such that ||x|| ≤ M for all x ∈ K. By the linearity of f, we have:

f(K) = {f(x) | x ∈ K} ⊆ {f(x) | ||x|| ≤ M}

Thus, f(K) is bounded above by M.

By the previous result in part (a), if f is bounded above on a neighborhood of the origin, then it is bounded above on any neighborhood of the origin. Therefore, f is bounded above on the neighborhood V of the origin.

Since K is compact, it can be covered by finitely many neighborhoods of the origin, say V1, V2, ..., Vk. Thus, f is bounded above on each Vi, i = 1, 2, ..., k.

Now, consider the open cover {V1, V2, ..., Vk} of K. By compactness, there exists a finite subcover {V1, V2, ..., Vm}. Therefore, f is bounded above on K.

Since f is bounded above on any compact set K, it follows that f is continuous.

(c) The previous part (b) already proves that if f is bounded above on any compact set, it is continuous. Therefore, if f is bounded above on a set 2 with int(2) ≠ Ø (i.e., the interior of 2 is not empty), then f is continuous.

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Answer:

Step-by-step explanation:

Answer:

I don't see an "x" or an equation-