i need the answer of this 9th grade question

32 is the number of teams took less than 20minutes to complete the game

Given that Isaac measured the time it took for some teams to complete an escape room game.

The times were recorded in the table

We have to find the number of teams which are  less than 20minutes to complete the game

Number of teams took less than 20 minutes

=8+24

=32

Hence, 32 is the number of teams took less than 20minutes to complete the game.

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Work Shown:

(A'C')/(AC) = 12/5 = 2.4

or you could say

(B'C')/(BC) = 16.8/7 = 2.4

or

(A'B')/(AB) = 9.6/4 = 2.4

The sides of triangle A'B'C' are 2.4 times longer compared to the corresponding sides of triangle ABC.

The equation of proportion will be x/4 = 36/x. Then the value of 'x' will be 12 units.

According to the hypotenuse-leg (HL) principle, a right triangle is harmonious if each of its third side and legs are congruent with its equivalent hypotenuse and legs in another right triangle. The HL theorem proves that these triangles are identical.

The equation is given as,

x/4 = 36/x

Simplify the equation, then we have

x/4 = 36/x

x² = 2² × 6²

x = 2 × 6

x = 12

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The volume of a cylinder can be found using the formula V = πr^2h, where V is the volume, r is the radius, and h is the height.

For Container A, the radius is half the diameter, or 16 feet. So the volume of Container A is:

V_A = π(16 ft)^2(16 ft) = 8,192π cubic feet

For Container B, the radius is half the diameter, or 15 feet. So the volume of Container B is:

V_B = π(15 ft)^2(18 ft) = 12,735π cubic feet

To find the volume of water that was transferred from Container A to Container B, we can subtract the volume of Container A from the volume of both containers combined:

V_water = V_A + V_B - V_A = V_B

V_water = 12,735π cubic feet

To find the volume of water remaining in Container A, we can subtract the volume of water that was transferred from the volume of Container A:

V_remaining = V_A - V_water

V_remaining = 8,192π - 12,735π

V_remaining ≈ -3,543.7 cubic feet

However, a negative volume doesn't make sense in this context, so we know that there must be no water remaining in Container A after the transfer. Therefore, the volume of water remaining in Container A to the nearest tenth of a cubic foot is 0.

Answer:

This graph is not a function.

The domain is -3<x<3.

The range is 1<y<3.

The lengths arranged from shortest to longest is [tex]2\frac{7}{8}[/tex],  [tex]3\frac{1}{4}[/tex], [tex]3\frac{1}{2}[/tex] , [tex]3\frac{5}{6}[/tex], [tex]4\frac{2}{3}[/tex] (option H).

The lengths of the woods are expressed as mixed numbers. Mixed numbers are non-integers that consists of a whole number and a proper fraction. For example, [tex]3\frac{1}{4}[/tex] is a mixed number. 3 is the whole number and [tex]\frac{1}{4}[/tex] is the proper fraction.

In order to arrange the lengths of the woods from shortest to the longest, convert the mixed numbers to decimals for easy comparison.

[tex]3\frac{1}{4}[/tex] becomes 3.25

[tex]2\frac{7}{8}[/tex] becomes 2.875

[tex]3\frac{5}{6}[/tex] becomes 3.833

[tex]4\frac{2}{3}[/tex] becomes 4.667

[tex]3\frac{1}{2}[/tex] becomes 3.5

The lengths arranged from shortest to longest is [tex]2\frac{7}{8}[/tex],  [tex]3\frac{1}{4}[/tex], [tex]3\frac{1}{2}[/tex] , [tex]3\frac{5}{6}[/tex], [tex]4\frac{2}{3}[/tex]

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It is not true that there exists a C > 0 such that for every sample S of n training examples with binary label, which are linearly separable, the hard-SVM and the soft-SVM (with parameter C) solutions will return exactly the same weight vector.

This can be shown by considering the case where n = 2, d = 1 and S = {(x1, yı), (x2, y2)}. Let a > 0 and consider x1 = a, yı = 1, X2 = –a, y2 = -1.

For the hard-SVM, the optimal solution is to find the weight vector w that maximizes the margin between the two classes while ensuring that all training examples are classified correctly. In this case, the margin is 2a and the weight vector is w = (0, 1) or w = (0, -1), depending on the choice of the positive and negative class labels.

For the soft-SVM with parameter C, the optimal solution is to find the weight vector w and the slack variables ξi that minimize the objective function:

min ||w||^2 + CΣξi
s.t. yi(wT xi) ≥ 1 - ξi, ξi ≥ 0

In this case, since the data is linearly separable, the slack variables will be zero for all training examples and the optimal solution will be the same as for the hard-SVM. However, if C is chosen to be a large value, the soft-SVM may allow some misclassifications by penalizing them with the slack variables. In this case, the optimal solution will be different from the hard-SVM.

Therefore, it is not always true that the hard-SVM and soft-SVM solutions will return exactly the same weight vector for linearly separable data, but they may be the same if the parameter C is chosen appropriately.

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Sebastian can choose between 4,320 different sweaters.

to find the total number of different sweaters, we need to take the product between the numbers of options for each of the given selections, we have the numbers:

Color = 10 options.

Pattern = 4 options.

Letter type = 12 options.

Image = 9 options.

Then the total number of different options is:

D = 10*4*12*9 = 4,320

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The predicted amount of debt for a financial education time of 34 hours is $26,350.

The predicted amount of debt for a financial education time of 24 hours is $31,550.

For the given information, we need to know the regression equation that relates financial education time and amount of debt to predict the amount of debt for different time durations.

Assuming a linear relationship between financial education time and amount of debt, we can estimate the regression equation as:

where a and b are the intercept and slope of the regression line, respectively, and Error is the random error term.

Using the given data points, we can estimate the regression line as:

(a) For a financial education time of 34 hours:

Therefore, the predicted amount of debt for a financial education time of 34 hours is $26,350.

(b) For a financial education time of 24 hours:

Therefore, the predicted amount of debt for a financial education time of 24 hours is $31,550.

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Answer:

C

Step-by-step explanation:

It does not belong because like the other ones it is not side ways, as well as it's common sense.

The line of reflection of the given image is:

A reflection across y = -3

A line of reflection is a line that lies between two identical mirror images, so the distance of any point of one figure from the line will equal the distance of the same point of the mirror image (flipped figure).

The definition above tells us that if we are given two images, such as mirror images of each other, the line of reflection can be determined by calculating the midpoint from any two points of the figures.

Looking at the given image, the center point of the mirror images is seen as y = -3

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a) The 99% confidence interval for the proportion of all caterpillars that lived to become a butterfly is between 0.078 and 0.166.

b) About 99% of these confidence intervals will contain the true population proportion of caterpillars that become butterflies and about percent will not contain the true population proportion.

A confidence interval of proportions has the bounds given by the rule presented as follows:

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which the variables used to calculated these bounds are listed as follows:

The confidence level is of 99%, hence the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.575.

The sample size and the estimate for this problem are given as follows:

[tex]n = 369, \pi = \frac{45}{369} = 0.122[/tex]

The lower bound of the interval is given as follows:

0.122 - 2.575 x sqrt(0.122 x 0.878/369) = 0.078.

The upper bound of the interval is given as follows:

0.122 + 2.575 x sqrt(0.122 x 0.878/369) = 0.166.

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The degree of overlap between the two data sets is said to be high degree of overlap

Data overlap is a statistical term that analyzes the extent to which data points or distributions intersect.

It is often employed when comparing two or more datasets, determining how similar or dissimilar they are from each other.

A high degree of overlapping indicates an extensive similarity among shared values between multiple sets of data or distributions. Conversely, if there is a low level of overlapping observed between two datasets, it suggests complete distinctness and no sharing of common quantitative baseline measurements or metrics.

The domain of the overlap according to the plot is [55 85]

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Answer: $1,293.60

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

28.27 CM ^2

Step-by-step explanation:

The formula for area of a circle is pi * r^2 where r = the radius.

Plugging the radius into the formula we get pi3^2, which equals 28.274 cm ^2 (and change)

Answer: C. 2

Step-by-step explanation: you add 2x and x to get 3x. Then add 6 to the other side to get 3x = 6. Then divide 6 by 3 to get x=2

Here are the steps to use the double number line diagram to find a percentage:

It should be noted that the double number line diagram is a visual tool that is used to represent two equivalent sets of values, and it can be useful to find percentages in a given context.

In this case, the steps are given above.

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Based on my experiment, My appropriate conclusion based on the results  is option A: The survey provides strong evidence that the proportion of California residents who agree that police officers should carry video cameras for the purposes of filming their activities while on duty is significantly different from the proportion of New York residents.

This conclusion is based on the truth that the experiment  conducted in California found that 85% of the respondents matches with the view,  one that is diverse from the extent of Modern York inhabitants (88%) who concurred with the same conclusion .

The "strong evidence" based on the experiment shows that the findings are statistically significant and the conclusion will thus be that the extent of California inhabitants who agree is "essentially wide" from the extent of other York inhabitants, showing some differences between the two populaces.

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The probability that exactly 128 flights are on time is 0.0950.

The probability that at least 128 flights are on time is 0.5079.

The probability that fewer than 124 flights are on time is 0.2147.

It should be noted that exactly 128 means P(X=128) = P(127.5 < X < 128.5) (By using continues correlation )

At X = 127.5

Z =(127.5-127.6) / 3.91305 = -0.02

At X = 128.5

Z =(128.5-127.6) / 3.91305 = 0.22

P(127.5 < X < 128.5) = P(-0.02 < Z < 0.22)

P(127.5 < X < 128.5) = P(Z < 0.22) - P(Z < -0.02)

P(127.5 < X < 128.5) =0.58706 - 0.49202

P(127.5 < X < 128.5) =0.0950

(b) at least 128 flights are on time.

P(X > 128) = P(X > 127.5) (by using continues correction)

P(X > 127.5) = P(Z > (127.5-127.6)/3.91305)

P(X > 127.5) = P( Z > -0.02)

P(X > 127.5) = 1- P( Z< -0.02)

P(X > 127.5) = 1- 0.49202

P(X > 127.5) = 0.5079

(c) fewer than 124 flights are on time.

P(X < 124 ) =P(X<124.5) (by using continouse correction)

P(X<124.5) = P(Z < (124.5-127.6)/3.91305)

P(X<124.5) = P(Z < -0.79)

P(X<124.5) = 0.2147

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Answer:

85.84

Step-by-step explanation:

6x4 is the area of 1 rectangle

There r 3 rectangles

6x4 =24

24x3 =72

That’s the area if all the rectangles

There r 2 triangles

(4x3.46)/2 is the area of 1 triangle

=6.92

6.92x2 =13.84

total surface area is all the areas added up

72+13.84

=85.84

The box plot that has the smallest interquartile range would be the fourth box plot.

The bar graph that shows Ellen's steps would be B. Second bar graph.

The interquartile range on a box plot can be found by subtracting the figure where the box starts at, from the figure where the box plot stops.

This means that the figure with the least interquartile range is the fourth box plot with an interquartile range of :

= 8 - 4

= 4

The bar graph that would show Ellen's steps would be the one that has the most steps on the graph, closest to the mean of 10, 000 steps. Only Graph B has steps in the number of days, which are closest to 10, 000 steps consistently so this is the correct graph.

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Answer:

2/3

Step-by-step explanation:

since a is 5 and b is 9.

so, 5+1( according to the expression given)/9

5+1/9 equals 6/9 and taking them down to the lowest terms that makes the answer 2/3

Answer:

Step-by-step explanation:

1

Therefore, for the equation y=x+5 , the slope is 1 , and the y-intercept is 5 .

The surface area of the prism is 70.44 yd²

A prism is a solid shape that is bound on all its sides by plane faces. There are different types of prism, triangular prism, rectangular prism e.t.c

The surface area of a prism is expressed as;

SA = 2B +pH

where B is the base area and p is the perimeter of the base and h is the height of the prism

Base area = 1/2 bh

= 1/2 × 3 × 3 = 4.5 yd²

The hypotenuse of the triangle = √3²+3²

= √9+9 = √18 = 4.24

perimeter of the base = 4.24+3+3 = 10.24yds

Therefore,

SA = 2×4.5 + 10.24 × 6

SA = 9 + 61.44

SA = 70.44 yd²

therefore the surface area of the prism is 70.44yd²

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The measure of the central angle is 0.75 radians

An equation is an expression that can be used to show the relationship between two or more numbers and variables using mathematical operators.

The area of a figure is the amount of space it occupies in its two dimensional state.

The length of an arc with an angle of Ф is given by:

Length of arc = (Ф/360) * (π * diameter)

The diameter is 16 cm and intercepted arc length is 6 cm, hence:

Length of arc = (Ф/360) * (π * diameter)

6 = (Ф/360) * (π * 16)

Ф = 42.97°

Ф = 42.97° * π/180 = 0.75 radian

The measure of the central angle is 0.75 radians

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If u= (-10,-3) and v= (4,8) . Proj vu is  is (640/109, 192/109).

Using this formula to determine the projection of v onto u (proj vu)

proj vu = ((v dot u) / (||u||)^2) × u

Where:

v dot u = dot product of v and u

||u|| = magnitude of u

First step is to find  the dot product of v and u:

v dot u = (-104) + (-38)

v dot u = -40 - 24

v dot u = -64

Second step to to calculate the magnitude of u:

||u|| = √((-10)^2 + (-3)^2)

||u|| = √(100 + 9)

||u|| = √(109)

Let plug in the formula

proj vu = ((-64) / (√(109))^2)× (-10,-3)

proj vu = (-64 / 109) × (-10,-3)

proj vu = (640/109, 192/109)

Therefore  the correct option is A.

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The option that best describes the change in viewers the network is experiencing is B. Each year, the total number of viewers increases to approximately 120% of the previous year's total.

When 't' surpasses 1/ln(5) at 3.5, the expression yields a negative result; connoting the network is witnessing a dwindling viewership.

Consequently, the network initially experiences an intensifying viewership ratio which then fades; in accordance with Option B, "Each year, the total number of viewers increases to roughly 120% of the preceding year's sum". Clearly, this demarcation reveals itself as the accurate answer.

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To show that Sk - ku is a martingale, we need to show that it satisfies the two properties of a martingale:

Sk - ku is integrable for all k.

For any k, E[Sk+1 - ku+1 | E0,...,Ek] = Sk - ku.

First, we show that Sk - ku is integrable for all k:

E[|Sk - ku|] = E[|(X1 + ... + Xk) - k|]

= E[|X1 + ... + Xk| - k] (using linearity of expectation)

≤ E[|X1| + ... + |Xk|] + k (by triangle inequality)

= kE[|X1|] + ... + kE[|Xk|] + k (using linearity of expectation)

= kE[|X1|] + ... + kE[|X1|] + k (since Xi are IID)

= kE[|X1|] + k (since there are k terms)

< ∞ (since E[|X1|] is finite)

Therefore, Sk - ku is integrable for all k.

Next, we show that for any k, E[Sk+1 - ku+1 | E0,...,Ek] = Sk - ku:

E[Sk+1 - ku+1 | E0,...,Ek] = E[(X1 + ... + Xk+1) - (k+1) | E0,...,Ek]

= E[(X1 + ... + Xk) + Xk+1 - (k+1) | E0,...,Ek]

= E[(X1 + ... + Xk) - k | E0,...,Ek] + E[Xk+1 | E0,...,Ek]

= Sk - ku + E[Xk+1] (by the martingale property of Sk - ku)

= Sk - ku + u (since the mean of X1, X2, ... is u)

Therefore, Sk - ku is a martingale.

Now, we can use the optional stopping theorem to prove that E[ST] = uE[T]:

By the optional stopping theorem, we have:

E[ST] = E[(Sk - ku)(T - k) + Sku]

= E[(Sk - ku)(T - k)] + E[Sku]

= E[E[(Sk - ku)(T - k) | E0,...,Ek]] + E[Sku] (using law of total expectation)

= E[(Sk - ku)E[T - k | E0,...,Ek]] + E[Sku] (using linearity of expectation)

= E[(Sk - ku)(E[T | E0,...,Ek] - k)] + E[Sku] (using law of total expectation)

= E[(Sk - ku)(T - k)] + E[Sku] (since T is a bounded stopping time)

= E[SkT - kuT - kS + ku] + E[Sku]

= E[SkT] - uE[KT] - kE[S] + ku + E[Sku] (using linearity of expectation)

= uE[KT] - uE[KT] - kE[S] + ku + E[Sku] (using the optional stopping theorem)

= ku + E[S(T-k)]

= ku + E[Sk - k^2] (since T is a stopping time)

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