Round off 80.6 to the nearest whole number

algol13

What do we know about these angles? Immediately, you might notice that (4y-8)° and (16x-4)° share a line. The same is true of (16x-4)° and (14x+4)°. Any straight line forms what's called a straight angle, which measures 180°, so we know that, since they add up to form a straight angle, (14x+4)° and (16x-4)° must add up to 180°. We can use that fact to set up an equation to solve for x:

(14x+4)+(16x-4)=180

After you solve for x, you should look to solve for y. How can we figure out what y is? If you're familiar with the vertical angle theorem, you'll know that all vertical angles (angles that are directly across from each other diagonally) are equal. So we know that 14x+4=4y-8. You can use the value of x you solved for before to solve this one fairly easily, and then you'll have both values.

8 0

3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

4 years ago

Find two rational numbers between 5/6 and 83/ 100 . Explain how you know (without using decimals)

kolbaska11 [484]

Answer:

998/1200 and 997/1200

Step-by-step explanation:

It’s hard to compare 5/6 and 83/100 in their current states. Let’s find a common denominator. Take 600. We know that this will work because both 6 and 100 go into it. Multiplying 5 by 100 and 83 by 6 to find the common denominator, we can compare them at 498/600 and 500/600. However, there is only one number (499/600) between them. Thus, we will multiply each number by 2/2, which is 1. Our numbers are then 996/1200 and 1000/1200. Two numbers between them are then 997/1200 and 998/1200

8 0

3 years ago

What does 2+3.5k=18.43

Mars2501 [29]

Is about 4.6942...

2 + 3.5k = 18.43
-2 -2

3.5k = 16.43
/3.5

4 0

4 years ago

The drama club sold 1500 tickets for the end-of-the-year performance. Admission prices were $12 for adults and $6 for students.

Zepler [3.9K]

The system of equations to determine the number of adult tickets, a, and the number of student tickets, s, the drama club sold is a + s = 1500; 12a + 6s = 16,200. 300 students attended the play.

<h3>Simultaneous equation</h3>

number of adult tickets = a
number of student tickets = s

The system of equation:

a + s = 1500

a + s = 150012a + 6s = 16,200

From equation (1)

a = 1500 - s

Substitute into (2)

12a + 6s = 16,200

12(1500 - s) + 6s = 16,200

18,000 - 12s + 6s = 16200

- 12s + 6s = 16200 - 18,000

-6s = -1800

s = -1800 / -6

s = 300

Substitute s = 300 into (1)

a + s = 1500

a + 300 = 1500

a = 1500 - 300

a = 1200

Therefore, there are 300 students and 1200 adults at the play respectively.

Learn more about simultaneous equation:

brainly.com/question/16863577

#SPJ1

8 0

2 years ago