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3 years ago

According to the number line, what is the distance between points A and B?

Mathematics

2 answers:

olga_2 [115]3 years ago

6 0

In the given number line point A is on the left of 0 at number -2 on the number line while point B is on the right of 0 at number 12.

The distance of A to 0 is 2 units .Point B is at a distance of 12 units from 0. The distance between a number's place on the number line and 0 is called the number's [absolute value]

A and B are l-2 l+12= 2+12 =14 units apart from each other.

Distance AB= 12+2=14 units.

exis [7]3 years ago

4 0

14 Units is the Answer.

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Consider the NASCAR race, the Indianapolis 500. Its name refers to the total race distance of 500 miles. There are 44 cars are r

lukranit [14]

Answer: $5289.35

Step-by-step explanation:

Since the cars have an average mileage of 6.78 miles per gallon and the race is 500 miles, the number of gallons each car will take will be:

= 500/6.78

= 73.75 gallons

Since they're 44 cars, the total gallons of fuel will be:

= 44 × 73.75

= 3245 gallons

Since the fuel cost is $1.63 per gallon, the total fuel cost, will be:

= $1.63 × 3245

= $5289.35

8 0

3 years ago

What is the surface area of a cube if it has one side that is 40 cm

belka [17]

The answer to that problem is 9600cm. each side of surface area will be the same multiplied together

5 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

3 years ago

Alana biked 12 miles in 48 minutes. What is Alana's speed in miles per minute?

marusya05 [52]

Question:- Alana biked 12 miles in 48 minutes. What is Alana's speed in miles per minute?

Answer:-

Set up a proportion.

12/48 = 1/x

Cross multiply.

12*x = 12x
48*1 = 48

12x = 48

12x/12 = 48/12

x= 48/12
x= 4

Now, we have our final answer. x = 4

Final answer:- Alana bikes 4 miles per minute.

6 0

4 years ago

explain fractions and, explain if I don't get fractions what do I need to relearn to understand fractions, if you don't understa

Andru [333]

A fraction is a numerical quantity that is not a whole number.What you probably need to relearn is equivalent fractions, Least common denominator and how to multiply and divide fractions.

6 0

3 years ago