Solve for missing value

After eating dinner at a restaurant Lucy bill before tak is $48.00 the sales tax rate is 8% Lucy decided to leave a 20% tip for

posledela

Answer:

do you have any answers to choose from?

7 0

3 years ago

**Ms. Gayle is driving to work. She travels 10 miles in 10 minutes. How fast is Ms. Gayle traveling?

Dimas [21]

Answer:

60 mph

Step-by-step explanation:

Ms. Gayle drives 10 miles in just 10 minutes. To find out how fast she was going we need to turn the minutes into hours. There are 60 minutes in an hour, so we multiply the miles and minutes by 60.

Ms. Gayle drives 60 miles in an hour.

This means she is traveling at 60 mph.

I hope that this helps! :)

4 0

3 years ago

Find the point at which the line joining the two points (-3,-5) and (7,5) cuts they y axis

8090 [49]

Answer:

(0,-2)

Step-by-step explanation:

First we find the equation of the line passes through $(-3,-5)$ and $(7,5)$ .

The slope of the line is

$m=\frac{5-(-5)}{7-(-3)}\\=\frac{10}{10}\\=1$

So by the point slope formula the equation of the line is

$y-5=1(x-7)\\y=x-7+5\\y=x-2$

When $x=0$ we have $y=0-2=-2$ .

Therefore the line joining the two points cuts the y-axis at (0,-2).

4 0

2 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that $\mu = 8.9, \sigma = 2.9$

Sample of 37:

This means that $n = 37, s = \frac{2.9}{\sqrt{37}}$

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -0.53$

$Z = -0.53$ has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0

2 years ago

Can somebody help me?

alexira [117]

Answer:

it goes by twelves

Step-by-step explanation:

it goes 12 24 36 48 60 72 84 96 and it keeps going till the end.

7 0

3 years ago