HELPPPPPP EASY 15 POINTSSSSSSSS HELP A GIRL OUT GUYS!!!!!

Answer:6

Step-by-step explanation:If jayden made 5% of his free throws and he shot 120 free throws you have to find 5% of 120.

5% of 120 = 6  

So Jayden shot a total of 6 free throws during the season

the height of the antenna is approximately 24 meters.

A triangle is said to be right-angled if one of its angles is exactly 90 degrees. The total of the other two angles is 90 degrees. Perpendicular and the triangle's base are the sides that make up the right angle. The longest of the three sides, the third side is known as the hypotenuse.

From the diagram, we see that we need to find the distance d and the height h. We can use the tangent function to find these values.

First, let's find d:

tan(θ) = h / (d + 1.51)

Rearranging, we get:

d = (h / tan(θ)) - 1.51

Next, let's find h:

tan(θ') = h / d

Substituting the expression for d that we found above, we get:

tan(θ') = h / ((h / tan(θ)) - 1.51)

Multiplying both sides by (h/tan(θ1)) - 1.51 and simplifying, we get:

h = (29 * tan(θ') * tan(θ)) / (tan(θ') - tan(θ))

Now we can plug in the values for the angles and solve for h:

h = (29 * tan(31°) * tan(17°)) / (tan(31°) - tan(17°))

h ≈ 24

So the height of the antenna is approximately 24 meters.

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Answer:

The quotient of 5984 divided by 32 is 187.5

Step-by-step explanation:

Answer:

The quotient of 5984 divided by 32 is 187.

so the rate is per annum, so compounding period is 1 per year, hmmm we could use 1 or 2 years to get "r", hmmm let's use 2.

[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \pounds 5903.13\\ P=\textit{original amount deposited}\dotfill &\pounds 5500\\ r=rate\to r\%\to \frac{r}{100}\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}[/tex]

[tex]5903.13 = 5500\left(1+\frac{\frac{r}{100}}{1}\right)^{1\cdot 2} \implies \cfrac{5903.13}{5500}=\left( 1+\cfrac{r}{100} \right)^2 \\\\\\ \sqrt{\cfrac{5903.13}{5500}}=\cfrac{100+r}{100}\implies 100\sqrt{\cfrac{5903.13}{5500}}=100+r \\\\\\ 100\sqrt{\cfrac{5903.13}{5500}}-100=r\implies \stackrel{ \% ~ ~~ ~ }{\boxed{3.6\approx r}}[/tex]

For the functiοn, the range is deduced tο be [-2, 5].

What is the range of data?

The range οf data is defined as a measure of the difference between the maximum value of data and the minimum value οf data.

The rangeοf a graph refers to the set of all possible output values, or the y-values, οf the function represented by the graph.

The given range of −2 ≤ x ≤ 5 actually represents the domain of the function, or the set οf all possible input values, or the x-values, fοr which the function is defined.

Sο, the range can also be written as [-2, 5] for the given set of data.

Therefore, the range is οbtained as [-2, 5].

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No, the set of vectors x, x + 3x², x³, x² does not span P[x].

To span P[x], a set of vectors must be able to generate any polynomial in P[x] through a linear combination of the vectors in the set. However, the set given only includes polynomials of degree 1, 2, and 3. This means that it cannot generate polynomials of degree 0 (constants) or polynomials of degree 4 or higher.

For example, the polynomial 5 cannot be generated from a linear combination of the vectors in the set, as there are no constants in the set. Similarly, the polynomial x⁴ cannot be generated, as there are no vectors of degree 4 or higher in the set.

Therefore, the set of vectors x, x + 3x², x³, x² does not span P[x].

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The equation that can be used to find the area of the figure below is: (10)(8) + (1/2)(6)(8).

The area of mixed shapes is the area that is covered by any hybrid shape. The composite shape is a shape created by joining a small number of polygons to create the desired shape. These forms or figures can be constructed from a variety of shapes, including triangles, squares, quadrilaterals, etc. To calculate the area of a composite object, divide it into simple shapes such a square, triangle, rectangle, or hexagon.

A composite form is essentially a combination of fundamental shapes. A "composite" or "complex" shape is another name for it.

The area of the rectangle is given as:

A = (l)(w)

A = 10(8)

A = 80 sq. units

The area of the triangle is:

A = 1/2(b)(h)

In the figure:

b = 16 - 10 = 6 and h = 8.

A = 1/2(6)(8)

A = 24 sq. units

The total area of the figure is:

Area = area of rectangle + area of triangle

Area = 80 + 24

Area = 104 sq. units

Hence, the equation that can be used to find the area of the figure below is: (10)(8) + (1/2)(6)(8).

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Perimeter of trapezoid is 35x+2, Perimeter of square is 16x-32, Perimeter of Pentagon is 10x-20 and Perimeter of rhombus is 28x+8.

A polygon is a plane figure made up of line segments connected to form a closed polygonal chain.

Perimeter of trapezoid=9x+4+7x+8+12x-2+7x-8

Add the like terms

35x+2

Perimeter of trapezoid is 35x+2

Perimeter of square= 4(4x-8)

=16x-32

Perimeter of Pentagon=5(2x-4)

=10x-20

Perimeter of rhombus=4(7x+2)

=28x+8

Hence, Perimeter of trapezoid is 35x+2, Perimeter of square is 16x-32, Perimeter of Pentagon is 10x-20 and Perimeter of rhombus is 28x+8.

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The degree 4 polynomial with the given zeroes is P(x) = x^4-(94/5)x^3-(24/5)x^2.

A degree 4 polynomial with the given zeroes can be represented as:


P(x) = (x-4)(x+(6)/(5))(x-0)^2


Simplifying the polynomial gives:


P(x) = (x-4)(x+(6)/(5))(x^2)


P(x) = (5x-20)(x+(6)/(5))(x^2)


P(x) = (5x^2-20x+(6x)/(5)-24/(5))(x^2)


P(x) = (25x^2-100x+6x-24)/(5)(x^2)


P(x) = (25x^2-94x-24)/(5)(x^2)


P(x) = (5x^4-94x^3-24x^2)/(5)


P(x) = x^4-(94/5)x^3-(24/5)x^2


So the degree 4 polynomial with the given zeroes is P(x) = x^4-(94/5)x^3-(24/5)x^2.

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If the matrix A = [a1 a2 a3 a4 a5] has reduced row echelon form [1−10​2​][00031​][00001​][00000​], then a basis for Col(A) is [a1−a2a3a4].

Reduced row echelon form is a way of representing a matrix that has been transformed into a more simplified version. It is commonly used in linear algebra and has many applications.

The columns of a matrix that contain at least one non-zero element in the reduced row echelon form are known as pivot columns. Pivot columns can be used to form a basis for the column space of the matrix.

The number of pivot columns determines the rank of the matrix. Since in the given matrix A has three pivot columns, therefore, the rank of the matrix is 3. Since there are three pivot columns in the reduced row echelon form, a basis for Col(A) must consist of the first three columns of A.

Hence, the required basis for Col(A) is [a1−a2a3a4].

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The length of the arc AB is 44.7 mm

A sector is a part of a given circle which is bounded by two radii and an arc. The length of an arc is simply the path of the incomplete circle.

The length of an arc can be determined by:

length of an arc = θ/ 360 2πr

Where is the measure of the central angle, and r is the radius of the circle.

From the diagram given,

The length of arc AB =  θ/ 360 2πr

                                   = 61/ 360*2*22/7*42

                                   = 44.7333

The length of arc AB is 44.7 mm.

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It appears that Nolan has met the conditions for constructing a confidence interval for the mean birth weight in Somalia. However, he should also check the skewness and presence of outliers in the sample to ensure that the normal approximation is appropriate.

In other words, data with a lower bound are frequently skewed right, and data with an upper bound are typically biased left.

Start-up effects can also cause skewness.

To construct a confidence interval for the mean birth weight in Somalia, we need to ensure that the following conditions are met:

Random Sampling: Nolan used a simple random sample of 100 births, which meets the condition of random sampling.

Independence: Each birth weight in the sample should be independent of the other. This condition is met if the sample size is less than 10% of the total population of births in Somalia.

Sample size: In general, a sample size of at least 30 is recommended to use the normal distribution to approximate the sampling distribution of the sample mean. Since Nolan's sample size is 100, this condition is met.

Skewness and outliers: Nolan mentioned that the sample data was slightly skewed with a few outliers.

Therefore, all the required conditions are given above.

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Answer:

the data is a random sample from the population of interest.

the sampling distribution of x is approximately normal.

individual observations can be considered independent.

Step-by-step explanation:

Answer:

2 and -2

Step-by-step explanation:

For the expression to be equal to 0, it means it has to be undefined, so all we need do, is equate the denominator to 0, since any number divided by 0 = 0

Therefore,

5x² - 20 = 0

5x² = 0 + 20

5x² = 20

Divide both sides by 5,

x² = 20/5

x² = 4

Square root both sides,

x² = √4

x = ±2

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a) The regression prediction for a classroom with 22 students is 651.24.

b) The predicted change in the classroom's average test score for a change from 19 to 23 students is 23.28.

c) The sample average of the test scores across the 100 classrooms is approximately 639.88.

d) The R-squared (R²) measure is 0.08, which indicates a relatively small amount of variation in test scores is explained by variation in class size.

e) The magnitude of the coefficient of correlation (r) is 0.283, indicating a weak positive linear association between class size and test scores.

f) The sample standard deviation of test scores across the 100 classrooms is approximately 11.28.

a) If a classroom has 22 students, then the regression prediction for that classroom's average test score would be:

S = 520.4 + 5.82(22) = 651.24

b) If last year a classroom had 19 students and this year it has 23 students, then the predicted change in the classroom's average test score would be:

ΔS = 5.82(23-19) = 23.28

Note that this calculation assumes that the other factors affecting test scores remain the same, which may not be realistic.

c) The sample average class size across 100 classrooms is 21.4. To find the sample average of the test scores across the 100 classrooms, we simply plug in the average class size into the regression equation:

S = 520.4 + 5.82(21.4) = 639.88

Therefore, the sample average of the test scores across the 100 classrooms is approximately 639.88.

d) The R-squared (R²) measure in this model is 0.08. This is a relatively small value, which means that only 8% of the variation in test scores can be explained by variation in class size. The remaining 92% of the variation is due to other factors that are not captured by the regression model.

e) The magnitude of the coefficient of correlation (r) in this model can be found by taking the square root of R²:

r = √0.08 = 0.283

The interpretation of r is that there is a weak positive linear association between class size and test scores. This means that as class size increases, test scores tend to increase slightly, but the relationship is not very strong.

The difference between R² and r is that R² measures the proportion of variance in the dependent variable (test scores) that is explained by the independent variable (class size) and any other variables in the model, while r measures the strength and direction of the linear association between the two variables.

f) The standard error of the regression (SER) is given as 11.5. This represents the average amount of variation in test scores that is not explained by the regression model. The sample standard deviation of test scores can be estimated by multiplying SER by the square root of (1 - R²) and dividing by the square root of the number of observations:

s = SER × √(1 - R²) / √n

= 11.5 × √(1 - 0.08) / √100

= 11.28

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The quadratic equation with roots 1 and 5 and leading coefficient 2 is 2x^2 - 12x + 10 = 0

To write the quadratic equation with roots 1 and 5 and leading coefficient 2, we can use the fact that the general form of a quadratic equation is:

ax^2 + bx + c = 0

where a, b, and c are constants.

Since the roots are 1 and 5, we know that the factors of the quadratic equation are:

(x - 1) and (x - 5)

Multiplying these factors together, we get:

(x - 1)(x - 5) = x^2 - 6x + 5

To satisfy the condition of having a leading coefficient of 2, we can simply multiply the entire equation by 2

2(x^2 - 6x + 5) = 2x^2 - 12x + 10

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Answer: (a) From the graph, we can see that the function f is increasing on the interval [-5, -1), (-2, 2), and (3, 5]. Therefore, we can write the increasing intervals in interval notation as:

[-5, -1) ∪ (-2, 2) ∪ (3, 5]

(b) From the graph, we can see that the function f is decreasing on the interval (-1, 1) and (2, 3]. Therefore, we can write the decreasing intervals in interval notation as:

(-1, 1) ∪ (2, 3]

(c) From the graph, we can see that the function f is constant on the interval [-4, -3]. Therefore, we can write the constant interval in interval notation as:

[-4, -3]

Step-by-step explanation:

The equation of the line is y = 780x + 150, r² value  is 0.186, the r value is 0.431, and the predicted population in year 12 is 9450.

To find a linear regression that fits this data, we need to use the following formula:
y = mx + b
where y is the population, x is the number of years, m is the slope, and b is the y-intercept.

We can use the data given to find the slope (m) and the y-intercept (b). The slope can be found by calculating the difference in the population between two consecutive years and dividing it by the difference in the number of years. The y-intercept can be found by plugging in the values of x and y into the equation and solving for b.

Using the data given, we can find the slope and y-intercept as follows:

m = (1950 - 1170) / (3 - 2) = 780
b = 150 - (0)(780) = 150

Therefore, the equation of the line is:
y = 780x + 150

To find the r² value, we can use the formula:
r² = 1 - (SSres / SStot)

where SSres is the sum of squares of residuals and SStot is the total sum of squares.

SSres = (150 - 150)² + (620 - 930)² + (1170 - 1710)² + (1950 - 2490)² = 2,484,400
SStot = (150 - 1222.5)² + (620 - 1222.5)² + (1170 - 1222.5)² + (1950 - 1222.5)² = 3,051,875

r² = 1 - (2,484,400 / 3,051,875) = 0.186

The r value can be found by taking the square root of the r²value:
r = √0.186 = 0.431

To predict the population in year 12, we can plug in the value of x into the equation of the line:
y = 780(12) + 150 = 9450

Therefore, the predicted population in year 12 is 9450.

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The price for adults to visit the mammoth cave is $16 and the price for students is $9.

The first step is to form a system of equations that represent the information in the question.

2a + 5s = 77 equation 1

2a + 7s = 95 equation 2

Where:

a = price of one adult

s = price of one student

The elimination method would be used to solve the equations.

Subtract equation 1 from equation 2:

2s = 18

Divide both sides of the equation by 2

s = 18/2

s = $9

Substitute for s in equation 1:

2a + 5(9) = 77

2a + 45 = 77

2a = 77 - 45

2a = 32

a = 32 / 2

a = 16

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Benito and his family will save $1,315 by the move to Oakland.

Savings in house = $1200 - $565

                               = $635

Savings in food = $655 - $545

                          = $110

Saving in health care = $495 - $245

                                   = $250

Saving in taxes  = $625 - $450

                          = $175

Saving in necessities = $495 - $350

                                   = $145

Total saving = $635 + $110 + $250 + $175 + $145

                    = $1,315.

Hence, Benito and his family will save $1,315 by the move to Oakland.

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Your question is incomplete, the complete question is:

Benito's family is thinking of relocating from Los Angeles to Oakland to save money. They set up a budget comparing the cost of living for both cities.

Oakland     Los Angeles  

Cost Housing       $565        $1200

Food                     $545        $655

Health Care         $245        $495

Taxes                    $450         $625

Other Necessities $350        $495

How much money will they save monthly by the move to Oakland? $1315, $1560, $1665, or $1765?

The equation of the line that passes through the point (2,0) and is perpendicular to the line [tex]y-4=-13(x+3)[/tex] is [tex]x - 13y = 2.[/tex]

To find the equation of the line that passes through the point (2,0) and is perpendicular to the line [tex]y-4=-13(x+3)[/tex] we need to find the slope of the perpendicular line and use the point-slope form of the equation of a line.

Step 1: Find the slope of the given line.
The equation of the given line is[tex]y-4=-13(x+3)[/tex]. This is in the form [tex]y - y1 = m(x - x1)[/tex], where m is the slope of the line. Therefore, the slope of the given line is -13.

Step 2: Find the slope of the perpendicular line.
The slope of two perpendicular lines are negative reciprocals of each other. So, the slope of the perpendicular line is 1/13.

Step 3: Use the point-slope form of the equation of a line to find the equation of the perpendicular line.
The point-slope form of the equation of a line is [tex]y - y1 = m(x - x1)[/tex], where m is the slope of the line and [tex](x1, y1)[/tex] is a point on the line. Substituting the values of the slope and the point into the equation, we get:

[tex]y - 0 = (1/13)(x - 2)[/tex]
[tex]y = (1/13)x - (2/13)[/tex]

Step 4: Simplify the equation.
Multiplying both sides of the equation by 13, we get:

[tex]13y = x - 2[/tex]

Step 5: Rearrange the equation to get the standard form of the equation of a line.
The standard form of the equation of a line is Ax + By = C, where A, B, and C are constants. Rearranging the equation, we get:

[tex]x - 13y = 2[/tex]

Therefore, the equation of the line that passes through the point (2,0) and is perpendicular to the line[tex]y-4=-13(x+3)[/tex] is x - [tex]13y = 2[/tex].

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So, by resolving the given, we obtain the result: As a result, the range of standard deviation timings between 72.5 and 87.5 seconds is the middle 99.7% of Aaron's 400 m finishing speeds.

A statistic called standard deviation may be used to represent the variability or variation of a collection of statistics. A low standard deviation means that the values often tend to be closer to the set mean, whereas a large standard deviation shows that the values are widely spread. The standard deviation is an indicator of how far the data are from the mean (or ). The data tend to cluster around the mean when the standard deviation is small, and are more scattered when it is big. Standard deviation is the average degree of variability in the data collection. It displays each score's standard deviation from the mean.  

We can thus get the z-scores that correspond to the bottom and upper bounds of this range using the z-score calculation formula:

Lower bound: z = (x - )/z = (x - 80)/z = 2.5 z = -3

The maximum value is z = (x - ) / z = (x - 80) / 2.5 z = 3.

The area under the curve between z = -3 and z = 3 can be calculated using a calculator or a conventional normal distribution table.

The minimum value is z = -3 (x - 80) / 2.5 = -3 x - 80 = -7.5 x = 72.5.

z = 3 (x - 80) / 2.5 = 3 x - 80 = 7.5 x = 87.5 is the upper limit.

As a result, the range of timings between 72.5 and 87.5 seconds is the middle 99.7% of Aaron's 400 m finishing speeds.

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Answer: 10134.2425

Step-by-step explanation:

Your initial number, 8900, is multiplied by 1+.033 because your percent and your adding money. It all is then raised to the power of 4. Your equation will be 8900(1+.033)^4 Your percentage is always moved two decimal points to the left all multiplied by the amount of years.

Answer:

0.866

Step-by-step explanation:

sin (200) = −0.34202014

cos (80) = 0.17364817

-

cos (200) = −0.93969262

sin (80) = 0.98480775

=

0.86602540338, or 0.866

Answer:

22

Step-by-step explanation:

10+11=21

.05+.05=1

1+21=22

The correct option is (b) i.e. there are 38 students age 15 and above who take a car to school.

What is a linear equation?

A linear equation is one in which the maximum power of the variable is one. Mathematically, an equation of the type axe + b = 0 or axe + by + c = 0, where a, b, and c are real integers and x and y are variables of maximum power one.

We know that,

Students under age 15 who prefer walk/bike + Students age 15 and above who prefer walk/bike = total of column

So, Students under age 15 who prefer walk/bike

   = total of column - Students age 15 and above who prefer walk/bike

   = 152 - 65

   = 87.

Now, total of rows = total of columns

       Row 1 + Row 2 = 360

      Row 1 = 360 - Row 2

                 =360-195

                 = 165.

Now, Students under age 15 who prefer bus + Students under age 15 who prefer walk/bike + Students under age 15 who prefer car = Row table

So, Students under age 15 who prefer bus

    = 165 - 60 - 87

    = 18.

Similarly,

Students under age 15 who prefer bus + Students 15 and above who prefer bus  = column total

So, Students 15 and above who prefer bus  = 110 - 18

                                                                         = 92.

Finally, students age 15 and above take a car to school

           = 195 - 92 - 65

           = 38.

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