Hannah made 0.7 of her free throws in a basketball game. Abra made 9/10 of her free throws. Dena made 3/4 of her free throws.Who was the best shooter? explain

We need to find the future value of annuity due with the following given values :-

Payment, Pm = 295 dollars.

N=4 (for quarterly)

Rate at 10%, r = 0.10/4 = .025

Time for 6 years, T = 6x4 = 24.

Future Value formula is :-

[tex]FV_{ad}=P_m*(1+r)*[\frac{(1+r)^T-1}{r} ] \\\\ FV_{ad}=295*(1+0.025)*[\frac{(1+0.025)^{24}-1}{0.025} ] \\\\ FV_{ad}=295*(1.025)*[\frac{(1.025)^{24}-1}{0.025} ] \\\\ FV_{ad}=295*(1.025)*[\frac{(1.80872595)-1}{0.025} ] \\\\ FV_{ad}=295*(1.025)*[\frac{0.80872595}{0.025} ] \\\\ FV_{ad}=295*(1.025)*(32.34903798) \\\\ FV_{ad}=9,781.54 \;dollars[/tex]

Hence, the final answer is 9,781.54 dollars.

In a calendrical system where leap years occur every four years without any exceptions, there would be 37 leap years in a 150-year period.

In a calendrical system where leap years occur every four years without exception, the number of leap years in a 150-year period can be calculated with a simple division operation. This is because every 4 years there is a leap year consistent with the rules of this calendar system.

To calculate the maximum number of leap years in a 150-year period, you would divide 150 by 4 as this represents the frequency of leap years. This results in a total of 37.5. Given that a half year cannot be a leap year, we eliminate the fraction and we have 37 leap years.

While the Gregorian calendar also has a leap year every four years, there are exceptions for century years not divisible by 400. But since the question specified a system without such rule, there would be more leap years in this system compared to the Gregorian calendar for the same period of time.

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We have been given that prices increase at a monthly rate of 0.5%.

Number of months in a year = 12

Let us suppose the item cost $1, then the percentage is given by

[tex]P=P_0(1+r)^n\\ \\ P=1(1+0.005)^{12}\\ \\ P=1.062[/tex]\\

Therefore, increase in a year is

[tex]1.062-1\\ =0.062[/tex]

Therefore, the required percentage in a year is 6.2%

Answer:

The slope of the line is [tex]-\frac{1}{2}[/tex]

Step-by-step explanation:

To know the slope of a line we can have the equation or to know points of the line. In this case from the graph we can get to points (-4,-1) and (0,-3). Having this and using the equation to calculate the slope, we have:

[tex]m=\frac{y2-y1}{x2-x1}[/tex]

Defining:

[tex](x1=-4,y1=-1)[/tex] and [tex](x2=0,y2=-3)[/tex]

Replacing this values in the initial equation, this is:

[tex]m=\frac{-3-(-1)}{0-(-4)}[/tex]

[tex]m=\frac{-2}{4} =-\frac{1}{2}[/tex]

The slope of the line is [tex]-\frac{1}{2}[/tex]

Answer:

the third one its 100% verified

Step-by-step explanation:

Answer:

x = 9 ±√23

Step-by-step explanation:

Answer:

4.063 x 10*4

Step-by-step explanation:

take the big number and move it from the exponent then bwa

Final answer:

The visible portion of the longitude line under the plane from an altitude of 5 miles is approximately 320.38 kilometers, using the formula D = 112.88 km √h where h is the altitude in kilometers.

Explanation:

The question asks how much of the longitude line directly under the plane is visible from an altitude of 5 miles high. To find this distance, we can use the formula given for horizon distance: D = 112.88 km √h, where h is the altitude in kilometers. Since 5 miles is approximately 8.05 kilometers, we can calculate the distance to the horizon by taking the square root of the altitude and multiplying by 112.88 kilometers.

First, calculate the square root of the altitude:
√8.05 ≈ 2.838.

Then, we multiply by the constant:
D ≈ 112.88 km × 2.838 ≈ 320.38 km.

Therefore, from an altitude of 5 miles, the visible portion of the longitude line directly under the plane is approximately 320.38 kilometers long.

Answer:

Option D [tex]\frac{b}{15}[/tex]

Step-by-step explanation:

Let

b-----> the number of bagels

we know that

To find the algebraic expression of "the bagels were divided equally into fifteen bags" , divided the number of bagels by fifteen

so

[tex]\frac{b}{15}[/tex]

The question requires solving a system of linear equations which results from given conditions on three unknown numbers. The unknowns can be represented by the letters x, y, and z which need to be solved for by leveraging the knowledge of algebra.

The question given is a classic example of a system of linear equations. Let's represent the first number as x, the second number as y, and the third number as z. From the question, we have the following equations:

[tex]x + y + z = 12 \\2x + 5y + 6z = 59 \\8x - y = 23[/tex]

Solving this system, we can find the values for x, y, and z which represent the three numbers. One of the methods for solving such a system is substitution or elimination method. Nevertheless, the solving such system requires a strong understanding of basic algebra including addition, subtraction, multiplication, and division of numbers and algebraic expressions.

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The probability that worker 1 will outproduce worker 2 in a single day is approximately 0.352 or 35.2%.

To find the probability that worker 1 will outproduce worker 2 in a single day, we need to compare their production rates using their respective means and standard deviations.

Let X be the production of worker 1 and Y be the production of worker 2. We are interested in finding P(X > Y).

Given:

Worker 1: X ~ N(77, 22^2)

Worker 2: Y ~ N(66, 19^2)

First, we standardize the variables:

Z1 = (X - 77) / 22

Z2 = (Y - 66) / 19

Now, we find P(X > Y) = P(X - Y > 0):

= P((X - 77) - (Y - 66) > 0)

= P((X - Y) > (77 - 66))

Now, we find the mean and standard deviation of the difference:

Mean of (X - Y) = 77 - 66 = 11

Standard deviation of (X - Y) = sqrt(22^2 + 19^2) = sqrt(484 + 361) = sqrt(845) ≈ 29.07

Now we standardize the difference:

Z = (11 - 0) / 29.07 ≈ 0.378

Finally, we find the probability using the standardized normal distribution table:

P(Z > 0.378) ≈ 0.352

So, the probability that worker 1 will outproduce worker 2 in a single day is approximately 0.352 or 35.2%.

the probability that Worker 1 will outproduce Worker 2 in a single day is approximately 0.7202 or 72.02%.

To find the probability that Worker 1 will outproduce Worker 2 in a single day, we can use the z-score formula for normal distributions.

The z-score formula is given by:

[tex]\[ z = \frac{{X - \mu}}{{\sigma}} \][/tex]

Where:

- ( X ) is the value we want to find the z-score for (in this case, Worker 1's productivity).

- [tex]\( \mu \)[/tex]is the mean (average) productivity.

- [tex]\( \sigma \)[/tex] is the standard deviation.

For Worker 1:

- [tex]\( X_1 = 77 \)[/tex] units per day (average productivity)

- [tex]\( \mu_1 = 77 \)[/tex]units per day (average productivity)

- [tex]\( \sigma_1 = 22 \)[/tex] units per day (standard deviation)

For Worker 2:

-[tex]\( X_2 = 66 \)[/tex]units per day (average productivity)

- [tex]\( \mu_2 = 66 \)[/tex]units per day (average productivity)

- [tex]\( \sigma_2 = 19 \)[/tex] units per day (standard deviation)

First, we find the z-score for Worker 1's productivity relative to Worker 2's productivity:

[tex]\[ z = \frac{{X_1 - \mu_2}}{{\sigma_2}} = \frac{{77 - 66}}{{19}} = \frac{{11}}{{19}} \approx 0.579 \][/tex]

Next, we look up the z-score in the standard normal distribution table or use a calculator/tool that can calculate probabilities from the standard normal distribution. The z-score of 0.579 corresponds to a probability of approximately 0.7202.

Therefore, the probability that Worker 1 will outproduce Worker 2 in a single day is approximately 0.7202 or 72.02%.