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3 years ago

23 yd

Mathematics

2 answers:

stealth61 [152]3 years ago

4 0

72.22 I think I’m sorry if I’m wrong

lubasha [3.4K]3 years ago

4 0

Answer:

415.47 or 132.25π

Step-by-step explanation:

First what we must do is find the radius. This can be easily found because we know that the radius is always half of the diameter so 23/2. This gives us a radius of 11.5 yards. Now, we put it through the standard formula for a circle π*r*r or π*r^2 (r being the radius). Here is the equation you now have, π*11.5^2 which gives us a total area of 415.47 or 132.25π.

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sveta [45]

I think the answer is B.
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4 0

3 years ago

Read 2 more answers

The GCD(a, b) = 9, the LCM(a, b)=378. Find the least possible value of a+b.

denis-greek [22]

$\mathrm{gcd}(a,b)=9\implies9\mid a\text{ and }9\mid b\implies9\mid a+b$

which means there is some integer $k$ for which $a+b=9k$ .

Because $9\mid a$ and $9\mid b$ , there are integers $n_1,n_2$ such that $a=9n_1$ and $b=9n_2$ , and

$\mathrm{lcm}(a,b)=\mathrm{lcm}(9n_1,9n_2)=9\mathrm{lcm}(n_1,n_2)=378\implies\mathrm{lcm}(n_1,n_2)=42$

We have $42=2\cdot3\cdot7$ , which means there are four possible choices of $n_1,n_2$ :

1, 42
2, 21
3, 14
6, 7

which is to say there are also four corresponding choices for $a,b$ :

9, 378
18, 189
27, 126
54, 63

whose sums are:

387
207
153
117

So the least possible value of $a+b$ is 117.

6 0

3 years ago

What is an example of when you would want consistent data and, therefore, a small standard deviation?

steposvetlana [31]

Answer:

12.1, 12.3,12.4,12.5,12.3,12.1,12.2

$\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271$

And for the standard deviation we can use the following formula:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And after replace we got:

$s = 0.1496$

And as we can ee we got a small value for the deviation <1 on this case.

Step-by-step explanation:

For example if we have the following data:

12.1, 12.3,12.4,12.5,12.3,12.1,12.2

We see that the data are similar for all the observations so we would expect a small standard deviation

If we calculate the sample mean we can use the following formula:

$\bar X=\frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271$

And for the standard deviation we can use the following formula:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And after replace we got:

$s = 0.1496$

And as we can ee we got a small value for the deviation <1 on this case.

8 0

3 years ago

A concert hall has 8,000 seats and two categories of ticket prices, $29 and $34. Assume that all seats in each category can be s

Greeley [361]

Using a system of equations, it is found that:

For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.
For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.
For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.

--------------------

The variables of the system are:

x, which is the number of $29 tickets sold.
y, which is the number of $34 tickets sold.

Total of 8,000 seats, all can be sold, thus:

$x + y = 8000 \rightarrow x = 8000 - y$

--------------------

For a profit of $255,000, we have that:

$29x + 34y = 255000$

$29(8000 - y) + 34y = 255000$

$5y = 23000$

$y = \frac{23000}{5}$

$y = 4600$

$x = 8000 - 4600 = 3400$

For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.

--------------------

For a profit of $271,000, we have that:

$29x + 34y = 271000$

$29(8000 - y) + 34y = 271000$

$5y = 39000$

$y = \frac{39000}{5}$

$y = 7800$

$x = 8000 - 7800= 200$

For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.

--------------------

For a profit of $235,000, we have that:

$29x + 34y = 235000$

$29(8000 - y) + 34y = 235000$

$5y = 3000$

$y = \frac{3000}{5}$

$y = 600$

$x = 8000 - 600 = 7400$

For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.

A similar problem is given at brainly.com/question/22826010

3 0

3 years ago

The service department of a luxury car dealership conducted research on the amount of time its service technicians spend on each

mart [117]

Answer:

Probability that the mean service time is between 1 and 2 hours is 0.96764.

Step-by-step explanation:

We are given that a systematic random sample of 100 service appointments has been collected.

The 100 appointments showed an average preparation time of 90 minutes with a standard deviation of 140 minutes.

Let $\bar X$ = sample mean service time

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = average preparation time = 90 minutes

$\sigma$ = standard deviation = 140 minutes

n = sample of appointments = 100

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, probability that the mean service time is between 60 and 120 minutes is given by = P(60 minutes < $\bar X$ < 120 minutes)

P(60 minutes < $\bar X$ < 120 minutes) = P( $\bar X$ < 120 min) - P( $\bar X$ $\leq$ 60 min)

P( $\bar X$ < 120 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{120-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z < 2.14) = 0.98382

P( $\bar X$ $\leq$ 60 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{60-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z $\leq$ -2.14) = 1 - P(Z < 2.14)

= 1 - 0.98382 = 0.01618

The above probability is calculated by looking at the value of x = 2.14 in the z table which has an area of 0.98382.

Therefore, P(60 min < $\bar X$ < 120 min) = 0.98382 - 0.01618 = 0.96764

7 0

4 years ago