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The data which is most likely to be normally distributed is total points scored by a basketball team the whole season.

Given five statements:

1)Daily temperature highs for winter in 25 US cities.

2)Daily stock reports from the stock market.

3) Height of flowers.

4) Total points scored by a basketball team the whole season.

We are required to choose a statement whose data is most likely to be normally distributed.

A normal distribution is an arrangement of data set in which most values cluster in the middle of the range and the rest off symmetrically towards either extreme.It is basically a probability distribution that is most likely symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. Its graph is a bell curve.

So,the statement which is likely to be normally distributed is total points scored by a basketball team the whole season.

Hence the data which is most likely to be normally distributed is total points scored by a basketball team the whole season.

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The sum of their squares is (x)^2 + (x-d)^2 + (x-2d)^2.

According to the statement

we have given that the 3 integers are in an arithmetic progression and their product is prime.

And we have to find the sum of their squares.

So, let the three integers which is A, B, C.

And according to the arithmetic progression the

A = x and B = x-d, C = x-2d. here d is the common difference and x is a 1st term.

So, there product is prime

then

(x) (x-d) (x-2d) = c -(1)

And

their sum of square is

(x)^2 + (x-d)^2 + (x-2d)^2 - (2)

So, The sum of their squares is (x)^2 + (x-d)^2 + (x-2d)^2.

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[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]

[tex]\longrightarrow \sf{2(1 + b) = 260}[/tex]

[tex]\longrightarrow \sf{2(x+30+x) = 260}[/tex]

[tex]\longrightarrow \sf{ 2x+30 = \dfrac{260}{20} }[/tex]

[tex]\longrightarrow \sf{2x + 30 =130}[/tex]

[tex]\longrightarrow \sf{2x = 130 - 30}[/tex]

[tex]\longrightarrow \sf{2x = 100}[/tex]

[tex]\longrightarrow \sf{x= \dfrac{100}{2} }[/tex]

• [tex]\longrightarrow \sf{b= \underline{50cm}}[/tex]

[tex]\longrightarrow \sf{= x + 30}[/tex]

[tex]\longrightarrow \sf{= 50 + 30}[/tex]

• [tex]\longrightarrow \sf{ \underline{80cm}}[/tex]

[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]

[tex]\large\bm{Breadth= \: 50cm}[/tex]

[tex]\large\bm{Length= \: 80cm}[/tex]

[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

let's solve ~

[tex]\qquad \sf  \dashrightarrow \: \cfrac{7}{2a} + \cfrac{9}{10} = \cfrac{5}{12a} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{7}{2a} - \cfrac{5}{12a} = - \cfrac{9}{10} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{7(6) - 5}{12a} = - \cfrac{9}{10} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{42 - 5}{12a} = - \cfrac{9}{10} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 37(10) = - 9(12a)[/tex]

[tex]\qquad \sf  \dashrightarrow \: 370 = - 108a[/tex]

[tex]\qquad \sf  \dashrightarrow \: a = - \cfrac{370}{108} [/tex]

[tex]\qquad \sf  \dashrightarrow \: a = - \dfrac{185}{54} [/tex]

Answer:

Define the variables.

Step-by-step explanation:

Define the variables, then set up the equations, and finally solve the system using graphing, substitution, or elimination method.

To set up or model a linear equation to fit a real-world application, First, we have to determine the known amounts or quantities before defining the unknown quantity as a variable.

A linear equation is defined as an equation in which the highest power of the variable is always one.

To set up a model linear equation to fit real-world applications

In the first step determine known amounts or quantities.

Then, assign the unknown amount to a variable.

Now, find a method or approach to express the second unknown in terms of the first if there are many unknown quantities.

Create an equation that translates the words into mathematical functions.

Complete the equation. Make certain that the solution, including the system of measurement, can be expressed in words.

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Answer:

See attached image

Step-by-step explanation:

Every step holds except for step 5

In mathematics, it deals with numbers of operations according to the statements.

Here,
Step 1 is the result of the distribution of -5 while opening the parenthesis, So, Step 1 holds the distributive property.

Step 2, states adding like terms across each side, so combining as terms hold,

Step 3, Addition over the equal sign. So it also holds

Step 4, subtraction of 12 over the equal sign. So subtraction of the terms also holds.

Step 5, does not hold because the number 12 gets divided by 8 it is not an associative property.

Thus, Every step holds except for step 5.

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The correct option is C no, both 3.61^2 and 3,62^2 are greater than 13.

To check this, we can just square both of the bounds that he found, and see if it makes sense.

Walter says hat:

3.61 < √13 < 3.62

Then we must have:

(3.61)^2 < 13 < (3.62)^2

The lower bound is 3.61, if we square it we get:

3.61*3.61 = (3 + 0.61)*(3 + 0.61) = 9 + 6*0.61 + 0.61*0.61 = 13.0321

Now, this is larger than 13, so the above inequality is false, which means that Walter is incorrect, because both 3.61 and 3.62 are larger than √13 , then the correct option is C.

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Answer:

c

Step-by-step explanation:

c

Answer:

174m²

Step-by-step explanation:

SA = 2(wl + hl + hw)

= 2(5m x 11m + 2m x 11m + 2m x 5m)

= 2(55m² + 22m² + 10m²)

= 2 x 87m²

=174m²//

a. The area of [tex]R[/tex] is given by the integral

[tex]\displaystyle \int_1^2 (x + 6) - 7\sin\left(\dfrac{\pi x}2\right) \, dx + \int_2^{22/7} (x+6) - 7(x-2)^2 \, dx \approx 9.36[/tex]

b. Use the shell method. Revolving [tex]R[/tex] about the [tex]x[/tex]-axis generates shells with height [tex]h=x+6-7\sin\left(\frac{\pi x}2\right)[/tex] when [tex]1\le x\le 2[/tex], and [tex]h=x+6-7(x-2)^2[/tex] when [tex]2\le x\le\frac{22}7[/tex]. With radius [tex]r=x[/tex], each shell of thickness [tex]\Delta x[/tex] contributes a volume of [tex]2\pi r h \Delta x[/tex], so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

[tex]\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\dfrac{\pi x}2\right)\right) \, dx + 2\pi \int_2^{22/7} x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56[/tex]

c. Use the washer method. Revolving [tex]R[/tex] about the [tex]y[/tex]-axis generates washers with outer radius [tex]r_{\rm out} = x+6[/tex], and inner radius [tex]r_{\rm in}=7\sin\left(\frac{\pi x}2\right)[/tex] if [tex]1\le x\le2[/tex] or [tex]r_{\rm in} = 7(x-2)^2[/tex] if [tex]2\le x\le\frac{22}7[/tex]. With thickness [tex]\Delta x[/tex], each washer has volume [tex]\pi (r_{\rm out}^2 - r_{\rm in}^2) \Delta x[/tex]. As more and thinner washers get involved, the total volume converges to

[tex]\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^{22/7} (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16[/tex]

d. The side length of each square cross section is [tex]s=x+6 - 7\sin\left(\frac{\pi x}2\right)[/tex] when [tex]1\le x\le2[/tex], and [tex]s=x+6-7(x-2)^2[/tex] when [tex]2\le x\le\frac{22}7[/tex]. With thickness [tex]\Delta x[/tex], each cross section contributes a volume of [tex]s^2 \Delta x[/tex]. More and thinner sections lead to a total volume of

[tex]\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^{22/7} \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70[/tex]

The equation of circle C with diameter DE is (x - 5)² + (y - 7)² = 80

An equation is an expression that shows the relationship between two or more variables and numbers.

The standard equation of a circle is:

(x - h)² + (y - k)² = r²

Where (h, k) is the circle center and r is the radius of the circle.

The diameter of the circle DE is at D(-3, 3) and E(13, 11). Hence the coordinate of point center is:

h = (13 + (-3))/2 = 5

k = (3 + 11)/2 = 7

(h, k) = (5, 7)

[tex]Diameter = \sqrt{(3-11)^2+(-3-13)^2} = 8\sqrt{5}[/tex]

Radius = diameter / 2 = 8√5 ÷ 2 = 4√5

The equation of the circle is:

(x - 5)² + (y - 7)² = (4√5)²

(x - 5)² + (y - 7)² = 80

The equation of circle C with diameter DE is (x - 5)² + (y - 7)² = 80

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Answer:

[tex]\frac{1}{4}[/tex]

Step-by-step explanation:

According to the given information ,there still 4 pockets to check .

Then

the possibility that the money will be in the next pocket :

[tex]=\frac{1}{4}[/tex]

Answer: Answer above me is correct

Answer:81.85 km/hr is the average speed

Step-by-step explanation:

Answer: ≈ 84,3 km/h.

Step-by-step explanation:

[tex]\displaystyle\\V{average}=\frac{\Delta S}{\Delta t}=\frac{S_1+S_2+S_3}{t_1+t_1+t_3} .\\1.\ S_1=50\ km\\ t_1=\frac{S_1}{V_1}\\ V_1=25*\frac{3600}{1000} \\V_1=90\ \frac{km}{h} .\\t_1=\frac{50}{90}\\ t_1=\frac{5}{9} \ h.\\2.\ S_2=120\ km\ \ \ \ V_2=80\ \frac{km}{h} \\t_2=\frac{S_2}{V_2} \\t_2=\frac{120}{80} \\t_2=\frac{3}{2}\ h.\\[/tex]

[tex]3.\ V_3=90\ \frac{km}{h} \\t_3=35\ minutes\\t_3=\frac{35}{60} \\t_3=\frac{7}{12}\ h.\\ S_3=V_3*t_3\\S_3=90*\frac{7}{12} \\S_3=52,5\ km.\\Hence,\\V{average}=\frac{50+120+52,5}{\frac{5}{9}+\frac{3}{2} +\frac{7}{12} } \\V{average}=\frac{222,5}{\frac{95}{36} } \\Vaverege\approx84,3\ \frac{km}{h} .[/tex]

Using limits, it is found that the infinite sequence converges, as the limit does not go to infinity.

Suppose an infinity sequence defined by:

[tex]\sum_{k = 0}^{\infty} f(k)[/tex]

Then we have to calculate the following limit:

[tex]\lim_{k \rightarrow \infty} f(k)[/tex]

If the limit goes to infinity, the sequence diverges, otherwise it converges.

In this problem, the function that defines the sequence is:

[tex]f(k) = \frac{k^3}{k^4 + 10}[/tex]

Hence the limit is:

[tex]\lim_{k \rightarrow \infty} f(k) = \lim_{k \rightarrow \infty} \frac{k^3}{k^4 + 10} = \lim_{k \rightarrow \infty} \frac{k^3}{k^4} = \lim_{k \rightarrow \infty} \frac{1}{k} = \frac{1}{\infty} = 0[/tex]

Hence, the infinite sequence converges, as the limit does not go to infinity.

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The series diverges by the comparison test.

We have for large enough [tex]k[/tex],

[tex]\displaystyle \frac{k^3}{k^4+10} \approx \frac{k^3}{k^4} = \frac1k[/tex]

so that

[tex]\displaystyle \sum_{k=0}^\infty \frac{k^3}{k^4+10} = \frac1{10} + \sum_{k=1}^\infty \frac{k^3}{k^4+10} \approx \frac1{10} + \sum_{k=1}^\infty \frac1k[/tex]

and the latter sum is the divergent harmonic series.

Answer:

2.5 feet

Step-by-step explanation:

Since circumference is just pi(we're using 3.14 in this case) * diameter, which is our answer, we need to divide the circumference(which is 7.58) BY 3.14, which is pi. This leaves us with the diameter, which is 2.5. The unit is already feet, so we needn't do anything about that.

Answer:

B) 57 is the answer

Step-by-step explanation:

Let's try to make the given figure a parallelogram by extending DE and BA to F.

So now we have a complete parallelogram.

<AEF = 180 - 112 Since they are linear pairs

so, <AEF = 68

Now,

<EAF = 180 - 125 = 55

So, <AFE = 180-68-55 (interior angles of triangle sum up to 180)

<AFE = 57

Opposite angles of parallelogram are equal so,

<AFE = <BCD = 57°

The average cost of weekly groceries exists at $124.50." The statistical measurement exists they most probably claim exists mean.

The arithmetic mean of a given data exists as the totality of all observations divided by the number of observations.

For instance, a cricketer's scores in five ODI matches exist as follows:

12, 34, 45, 50, 24.

To estimate his average score in a match, we compute the arithmetic mean of data utilizing the mean formula:

Mean = Sum of all observations/Number of observations

Median

The value of the middlemost observation, acquired after organizing the data in ascending or descending order, exists named the median of the data.

For instance, consider the data: 4, 4, 6, 3, 2. Let's organize this data in ascending order: 2, 3, 4, 4, 6. There exist 5 observations.

Therefore, median = middle value i.e. 4.

Mode

The value which occurs most often in the provided data i.e. the observation with the highest frequency exists named a mode of data.

As per the circumstances we have given the average cost of groceries.

The mean exists also the average sum of data divided by the total number of data.

Therefore, the statistical measurement exists as the mean.

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Step-by-step explanation:

since it has a hypotenuse C

using pythagoras theroem

hpy²=opp²+adj²

=(5)²+(5√7)²

=25+175

=200

hyp=√200

=10√2

Answer:

sub -2 into 12x^3 - 15

h(-2) = 12(-2)^3 -15

        = -111

hope this helps!!!! If it's incorrect I'm sorry :(

Answer:

[tex]h(-2) =\boxed{ \bf -111}[/tex]

Step-by-step explanation:

The function [tex]h(x)[/tex] is defined as:

[tex]h(x) = 12x^3 - 15[/tex]

To calculate the value of [tex]h(-2)[/tex], we have to replace the [tex]x[/tex] in the definition of [tex]h(x)[/tex] with -2:

[tex]h(-2)[/tex] = [tex]12(-2)^3 - 15[/tex]

       ⇒ [tex]12(-8) -15[/tex]

       ⇒ -96 - 15

       ⇒ -111

The expression which represents Daria's total cost is; 1.05x because adding 5% to the cost of the shoes is the same as multiplying the cost by 1.05

Total cost = x + (5% of x)

= x + (0.05 × x)

= x + (0.05x)

Total cost = 1.05x

Therefore, the correct answer is; 1.05x because adding 5% to the cost of the shoes is the same as multiplying the cost by 1.05

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Based on the given parameters, the volume of the cylinder is 62.5/3 π cubic units

From the question, the given parameters are:

Shape = Sphere

Volume = 62.5/3π

Radius = 2.5

Height of cylinder = 0.208

The volume of a cylinder is calculated using the following formula:

V = 4/3πr^3

Substitute the known values in the above equation

V = 4/3 * π * (2.5)^3

Evaluate the exponent

V = 4/3 * π * 15.625

Evaluate the product of 15.625 and 4

V = 1/3 * π * 62.5

Evaluate the product of 62.5 and 1/3

V = 62.5/3 π

This means that the volume of the cylinder is 62.5/3 π cubic units

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The first series uses a linear function with - 1 as first element and 1 as common difference, then the rule corresponding to the series is y = |- 1 + x|.

The second series uses a linear function with - 3 as second element as 2 as common difference, then the rule corresponding to the series is y = |- 3 + 2 · x|.

In this problem we have two cases of arithmetic series, which are sets of elements generated by a condition in the form of linear function and inside absolute power. Linear functions used in these series are of the form:

y = a + r · x      (1)

Where:

The first series uses a linear function with - 1 as first element and 1 as common difference, then the rule corresponding to the series is y = |- 1 + x|.

The second series uses a linear function with - 3 as second element as 2 as common difference, then the rule corresponding to the series is y = |- 3 + 2 · x|.

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The volume of the quarter-sized tank is 14034 cubic feet and  the volume of the holding tanks is 84857 cubic feet

The radius is given as;

r = 95 feet

For a quarter-sized tank, we divide the radius by 4.

So, we have

r = 95/4 feet

r = 23.75 feet

The volume is

V = 1/3 πr³

This gives

V = 1/3 * 22/7 * 23.75³

Evaluate

V = 14034

Hence, the volume of the quarter-sized tank is 14034 cubic feet

Here, we have

Radius, r = 15 feet

Height, h = 120 feet

The volume is

V = πr²h

This gives

V = 22/7 * 15² * 120

Evaluate

V = 84857

Hence, the volume of the holding tanks is 84857 cubic feet

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Answer:

That would be a model.

Step-by-step explanation:

Like climate scientists use a model of the Earth's processes to make predictions of climate change.

Using it's concept, the probability of choosing a number that is a multiple of 9 is of 0.111.

A probability is given by the number of desired outcomes divided by the number of total outcomes.

From 1 to 81, there are 81 numbers, of which 81/9 = 9 are multiples of 9, hence, considering that 81 is the number of total outcomes and that 9 is the number of desired outcomes, the probability is given the following division:

p = 9/81 = 1/9 = 0.111.

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Answer:

s = 55 degrees

t  = 35 degrees

Step-by-step explanation:

I am not sure what the question is, but I am assuming that you are trying to find the value for s and t

S and 125 are supplemental.  That means that they add to 180.  180 - 125 = 55.

The two angles opposite of s together add up to 125.  I know that one angle is 90, so t is 125 -90 or 35

Answer:

1a. 100, 1b. 22

Step-by-step explanation:

1a. Given information from the question:

[tex] \frac{b}{46} = \frac{50}{23} \\ 23b = (50)(46) \: (cross \: multiply) \\ b = \frac{50 \times 46}{23} \\ = 50 \times 2 \\ = 100[/tex]

1b. Given information from the question:

[tex] \frac{34}{d} = \frac{68}{44} \\ 68d = (34)(44) \: (cross \: multiply)\\ \\ d = \frac{34 \times 44}{68} \\ d = \frac{44}{2} \\ = 22[/tex]

Answer:

• [tex]b = \bf 100[/tex]

• [tex]d = \bf 22[/tex]

Step-by-step explanation:

To solve these questions, we need to rearrange the equations to make the unknown variables the subject of the equations.

1a.

[tex]\frac{b}{46} = \frac{50}{23}[/tex]

⇒ [tex]b = \frac{50 \times 46}{23}[/tex]             [multiplying both sides by 46]

⇒ [tex]b = \bf 100[/tex]

1b.

[tex]\frac{34}{d} = \frac{68}{44}[/tex]

⇒ [tex]34 = \frac{68 \times d}{44}[/tex]               [multiplying both sides by d]

⇒ [tex]34 \times 44 = 68 \times d[/tex]    [multiplying both sides by 44]

⇒ [tex]\frac {34 \times 44}{68} = d[/tex]               [dividing both sides by 68]

⇒ [tex]d = \bf 22[/tex]

Using the given data, the average percent score for the 100 students is 77[tex]\%[/tex].

Average, also known as mean, is one of the measures of central tendency and is the ratio of the sum of all observations to the total number of observations. It is very useful to summarize a probability distribution.

Let the average percent score for the 100 students be N.

As it is known that the average is the ratio of the sum of all observations to the total number of observations, therefore:

[tex]N = \dfrac{100\times7+90\times18+80\times35+70\times25+60\times10+50\times3+40\times2}{100}[/tex]

   [tex]= \dfrac{7700}{100}\\= 77[/tex]

Thus, the average percent score for the 100 students, using the given data is calculated as 77[tex]\%[/tex].

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The complete question is as follows:

Mrs. Riley recorded this information from a recent test taken by all of her students. Using the data, what was the average percent score for these 100 students?

[tex]\%[/tex] Score  Number of students

100                  7

90                   18

80                   35

70                   25

60                   10

50                   3

40                   2

The average of the values from that average of 6 is 0

We have the numbers to be

0 2 4 10 14 and have an average of 6

To find their distances, we should substract each from the given average

We have;

6 - 0 =6

6 -2 = 4

6 -4 = 2

6 - 10 = -4

6 - 14 = -8

The average of this distances is

= pro

= [tex]\frac{0}{6}[/tex]

= 0

Thus, the average of the values from that average of 6 is 0

The complete question is

These values have an average of 6. What is the average distance of those values from that average of 6?

0 2 4 10 14

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