There are 12 girls and 16 boys and Javier's math class or 26 girls and 14 boys and Javier's core class is the race ratio of girls to boys in the two classes equivalent

Answer:

288 hours.

Step-by-step explanation:

You exercise 24 hours in 1 month.

There are 12 months in one year.

24 * 12 = 288

Therefore, you exercised 288 hours in one year.

Answer:

Calculation with Distribution

24 (1+1+1+1+1+1+1+1+1+1+1+1)

24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1

24+24+24+24+24+24+24+24+24+24+24+24

288 hours

You did 288 hours of exercise.

Calculation without Distribution

24(1+1+1+1+1+1+1+1+1+1+1+1)

24(12)

288 hours

You did 288 hours of exercise.

Step-by-step explanation:

The growth factor of g is given by [tex]A = e^{0.6t}.[/tex]

The function exhibits exponential growth.

We have,

(a)

Let's denote the growth factor as A(t), where t is the independent variable representing time.

The rate of change of g at any time t is proportional to g itself, with a constant of proportionality of 0.6.

This can be expressed as:

dA/dt = 0.6A

This is a separable differential equation, which we can solve by separating the variables and integrating them:

1/A dA = 0.6 dt

Integrating both sides,

ln|A| = 0.6t + C

Here, C is the constant of integration.

Exponentiating both sides:

[tex]|A| = e^{0.6t + C}[/tex]

Since the growth factor A is always positive, we can drop the absolute value:

[tex]A = e^{0.6t + C}[/tex]

We can do this by considering the initial condition.

Let's assume that at t = 0, the growth factor A is 1 (no growth). Substituting these values into the equation:

[tex]1 = e^{0.6(0) + C}\\1 = e^C[/tex]

Taking the natural logarithm of both sides:

[tex]ln(1) = ln{e^C}[/tex]

0 = C

So,

[tex]A = e^{0.6t}[/tex]

(b)

The function g exhibits exponential growth, as the value of g increases exponentially with time.

Thus,

The growth factor of g is given by [tex]A = e^{0.6t}.[/tex]

The function exhibits exponential growth.

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The growth factor of gg is 0.6. The function exhibits exponential growth.

(a) The growth factor of an exponential function is the base of the exponent. In this case, the growth factor is 0.6.

(b) The function gg exhibits exponential growth, meaning that as the input increases, the output increases at a proportional rate.

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Answer:

The value of A is 4.

Step-by-step explanation:

Consider the provided quadratic equation.

[tex]4x^2-20x+3=0[/tex]

Subtract 3 from both sides.

[tex]4x^2-20x+3-3=-3[/tex]

[tex]4x^2-20x=-3[/tex]

Now take 4 common from the left side of the above equation.

[tex]4(x^2-5x)=-3[/tex]

Compare the above equation with [tex]A(x^2 - 5x) = -3[/tex]

By the comparison it is clear the value of A is 4.

Hence, the value of A is 4.

To solve this problem, we use the z statistic.  The z formula is:

z = (x – u) / s

where x is lakers score = 101.7, u is Celtics score = 99.2, s is standard dev = 10.5

z = (101.7 – 99.2) / 10.5 = 0.24

using the normal distribution table at z = 0.24, the P value is:

P (z = 0.24) = 0.5984 = 59.84% 

To find the proportion of games in which the Boston Celtics scored more than the Los Angeles Lakers' mean score, we need to standardize the Lakers' mean score, find the z-score, and then find the proportion from the standard normal distribution table.

To find the proportion of games in which the Boston Celtics scored more than the Los Angeles Lakers' mean score, we need to find the area under the normal distribution curve to the right of the Lakers' mean. First, we need to standardize the Lakers' mean score by finding the z-score using the formula:

z = (x - μ) / σ

where x is the mean score, μ is the mean of the Celtics' scores, and σ is the standard deviation of the Celtics' scores. Substituting the given values, we have:

z = (101.7 - 99.2) / 10.5 = 0.2381

Next, we find the proportion of scores greater than the Lakers' mean in the standard normal distribution table or using a calculator. Looking up the z-score of 0.2381 in the table, we find the proportion to be approximately 0.4069, or 40.69%.

Answer:(-2.7,2.1)

Step-by-step explanation:

False, the given equation of hyperbola is vertical hyperbola.

What is hyperbola?

Hyperbola is defined as a conic section, two-branched open curve, produced by the intersection of a circular cone.

[tex]\frac{(y-2)^2}{16} -\frac{(x+1)^2}{144}=1[/tex]

Simplify each term in the equation in order to set the right side equal to 1

The standard form of an ellipse or hyperbola requires the right side of the equation be 1

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

[tex]\frac{(y-k)^2}{a^2} -\frac{(x-h)^2}{b^2}=1[/tex]

Match the values in this hyperbola to those of the standard form. The variable

h represents the x-offset from the origin,

k represents the y-offset from origin, a.

a=4

b=12

k=2

h=−1

c, the distance from the center to a focus  4√10

The foci is (−1,2+4√10),(−1,2−4√10)

Hence, the hyperbola is vertical.

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"To create a box and whisker plot and calculate the Q1, Q3, and IQR for the given set of quiz grades, we must first organize the data in ascending order:

58, 60, 69, 71, 78, 78, 85, 85, 86, 87, 88, 91, 93, 95, 95, 95, 96, 98, 100, 100

Next, we find the median (Q2), which is the middle value in the ordered list. Since there are 20 scores, the median will be the average of the 10th and 11th scores:

Q2 = (87 + 88) / 2 = 175 / 2 = 87.5

Now, we need to find the first quartile (Q1), which is the median of the first half of the data (excluding the median if the number of data points is odd). Since we have an even number of scores, we take the first 10 scores and find their median:

Q1 = (71 + 78) / 2 = 149 / 2 = 74.5

For the third quartile (Q3), we take the median of the second half of the data (excluding the median if the number of data points is odd). Again, we have an even number of scores, so we take the second set of 10 scores and find their median:

Q3 = (95 + 96) / 2 = 191 / 2 = 95.5

Finally, the interquartile range (IQR) is calculated by subtracting Q1 from Q3:

IQR = Q3 - Q1 = 95.5 - 74.5 = 21

Therefore, the Q1, Q3, and IQR for the fifth period Coordinate Algebra class quiz grades are:

Q1 = 74.5

Q3 = 95.5

IQR = 21

These values are essential for constructing the box and whisker plot and for understanding the spread of the middle 50% of the quiz scores."

Answer-

The best-fit quadratic equation is  [tex]y=14.9786x^2+3.1057x+56.2771[/tex] and the game cost in 2010 will be $1586

Solution-

Plotting a table taking year as input variable and cost as output variable.

X= year - 2000    

Y= cost in dollar.

Quadratic equation formula,

[tex]y=ax^2+bx+c[/tex]

[tex]a=\frac{(\sum x^2y\sum xx)-(\sum xy\sum xx^2)}{(\sum xx\sum x^2x^2)-({\sum xx^2)}^2}[/tex]

[tex]b=\frac{(\sum xy\sum x^2x^2)-(\sum x^2y\sum xx^2)}{(\sum xx\sum x^2x^2)-({\sum xx^2)}^2}[/tex]

[tex]c=\frac{\sum y}{n}-b\frac{\sum x}{n}-a\frac{\sum x^2}{n}[/tex]

Where,

[tex]\sum xx=\sum x^2-\frac{(\sum x)^2}{n}[/tex]

[tex]\sum xy=\sum xy-\frac{\sum x\sum y}{n}[/tex]

[tex]\sum xx^2=\sum x^3-\frac{\sum x\sum x^2}{n}[/tex]

[tex]\sum x^2y=\sum x^2y-\frac{\sum x^2\sum y}{n}[/tex]

[tex]\sum x^2x^2=\sum x^4-\frac{(\sum x^2)^2}{n}[/tex]

Putting the values, we get

[tex]a=14.9786,b=3.1057,c=56.2771[/tex]

Putting these in the quadratic equation,

[tex]y=14.9786x^2+3.1057x+56.2771[/tex]

In order to  get the cost of game in 2010, we can put x=10 to get the value of y or the cost of game.

[tex]y(10)=14.9786(10)^2+3.1057(10)+56.2771=1585.1941 \approx 1586[/tex]

Answer:

(m-n)(r-s)

Step-by-step explanation:

We will factor this by grouping.  First we rearrange the terms so the m's are together and the n's are together:

mr-ms+ns-nr

Next we group the first two and the last two:

(mr-ms)+(ns-nr)

The GCF of the first group is m; factor this out:

m(r-s)+(ns-nr)

The GCF of the second group is n; factor this out:

m(r-s)+n(s-r)

We want to have the same thing in both sets of parentheses; to do this, we will multiply the second factor by -1:

m(r-s)+-n(r-s)

Factoring out the term now in common, r-s, we have

(r-s)(m-n) = (m-n)(r-s)

Answer:

PANCKACKES

Step-by-step explanation:

The solution to the first algebraic equation is x = 7, and the solution to the second equation is x = -3. Both problems involve setting up and solving algebraic equations to find the value of the unknown number.

To solve the problems stated, we need to set up algebraic equations and solve for the unknown number x.

For the first one:

Fourteen less than five times a number is three times the number: 5x - 14 = 3x.

After simplifying, we get 2x = 14.

Divide both sides by 2 to find x = 7.

For the second one:

Twelve more than seven times a number equals the number less six: 7x + 12 = x - 6.

Rearrange to find 6x = -18.

Divide both sides by 6 to find x = -3.

We have found that the unknown number in the first equation is 7 and in the second equation, it is -3.

0-9 = 10 numbers

 1st digit can be 1 0f 10

2nd digit can be 1 of 9

3rd digit can be 1 of 8

4th digit can be 1 of 7

10 * 9 *8*7 = 5040 different combinations

The number of different possible passwords are possible

Given that each digit can be used once, then:

So, the number of passwords is:

n = 10 * 9 * 8 * 7

Evaluate the products

n = 5040

Hence, there are 5040 different possible passwords

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[tex] p\%=\dfrac{p}{100}=p:100\\\\21\dfrac{1}{4}\%=\dfrac{21\cdot4+1}{4}=\dfrac{85}{4}\%=\dfrac{85}{4}:100=\dfrac{85}{4}\cdot\dfrac{1}{100}=\dfrac{85}{400}=\dfrac{85:5}{400:5}=\dfrac{17}{80} [/tex]

Answer:

the answer is 17/80 hope this helps