Find the product.<br><br> -a 2b 2c 2(a + b

Easy problems I'm having trouble remembering how to do!

beks73 [17]

<h2>Answer:</h2>

1. - 2°C + 4°C

= 2°C.

2. 3/4 × 4

= 3pounds.

7 0

3 years ago

A random sample of 121 automobiles traveling on an interstate showed an average speed of 73 mph. from past information, it is kn

tatiyna

121 is big enough to assume normality and not worry about the t distribution. By the 68-95-99.7 rule a 95% confidence interval includes plus or minus two standard deviations. So 95% of the cars will be in the mph range

$(73 - 2 \cdot 11, 73 + 2 \cdot 11) = (51,95)$

The question is a bit vague, but it seems we're being asked for the 95% confidence interval on the average of 121 cars. The 121 is a hint of course.

The standard deviation of the average is in general the standard deviation of the individual samples divided by the square root of n:

$\sigma = \dfrac{ 11}{\sqrt{121}} = 1$

So repeating our experiment of taking the average 121 cars over and over, we expect 95% of the averages to be in the mph range

$(73 - 2 \cdot 1, 73 + 2 \cdot 1) = (71,75)$

That's probably the answer they're looking for.

4 0

3 years ago

mary lou is twice geoge's and kate is two years younger than george the sum of all of their ages is 46 how old is everyone

saw5 [17]

Answer:

George is 12, Mary Lou is 24 and Kate is 10.

Step-by-step explanation:

To find these, start by setting George's age as x. This means that we can model Mary Lou's age as 2x, since she is twice as old. We can also model Kate's age as x - 2 since she is two years younger. Now we can add these 3 together and set equal to 46

x + 2x + x - 2 = 46

4x - 2 = 46

4x = 48

x = 12

This means that George is 12.

Mary Lou = 2x

Mary Lou = 2(12)

Mary Lou = 24

Kate = x - 2

Kate = 12 - 2

Kate = 10

3 0

3 years ago

Read 2 more answers

Last month, a dwarf lemon tree grew half as much as a semi dwarf lemon tree. A full size lemon tree grew three times as much as

Nadusha1986 [10]

Using a simple guess and check method, i started with the lowest common multiple of 3, and 27 (the total growth) and used 3. if you plug i the numbers, we will say a is the dwarf lemon tree, b is the semi-dwarf lemon tree, and c is the full lemon tree, you know 3times2 equals 6. if you triple 6 you get 18. so if A grew 3 inches , B had to have grown 6 inches, and C had to have grown 18 inches. add them all and you get 27. so the dwarf lemon tree grew 3 inches, semi dwarf grew 6 inches, the full size grew 18 inches.

3 0

3 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3

In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% 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:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that $\mu = 8.3, \sigma = 3.3$ .

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

We are working with a sample mean of 37 jets. So we have that:

$s = \frac{3.3}{\sqrt{37}} = 0.5425$

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

This probability is the pvalue of Z when $X = 8.65$ . So

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

$Z = \frac{8.65 - 8.3}{0.5425}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422. This means that there is a 74.22% 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?

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is subtracted by the pvalue of Z when $X = 7.43$

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

$Z = \frac{7.43 - 8.3}{0.5425}$

$Z = -1.60$

$Z = -1.60$ has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% 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?

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is the pvalue of Z when $X = 8.65$ subtracted by the pvalue of Z when $X = 7.43$ .

So:

From a), we have that for $X = 8.65$ , we have $Z = 0.65$ , that has a pvalue of 0.7422.

From b), we have that for $X = 7.43$ , we have $Z = -1.60$ , that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0

3 years ago

Find the product. -a 2b 2c 2(a + b - c)