Ask question

Ask question

All categories

3 years ago

8

Richard is planning a trip to Italy he thinks he will need $2750 for the trip if the trip is 40 weeks away which is the best est

imate of how much money Richard needs to save each week.

1 answer:

WITCHER [35]3 years ago

7 0

$69 dollars or $68.75

You might be interested in

Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal prob

Novay_Z [31]

Answer:

a) By the Central Limit Theorem, it is approximately normal.

b) The standard error of the distribution of the sample mean is 1.8333.

c) 0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d) 0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours

e) 0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

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 estabilishes 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 36 hours and a standard deviation of 5.5 hours.

This means that $\mu = 36, \sigma = 5.5$

a. What can you say about the shape of the distribution of the sample mean?

By the Central Limit Theorem, it is approximately normal.

b. What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.)

Sample of 9 means that $n = 9$ . So

$s = \frac{\sigma}{\sqrt{n}} = \frac{5.5}{\sqrt{9}} = 1.8333$

The standard error of the distribution of the sample mean is 1.8333.

c. What proportion of the samples will have a mean useful life of more than 38 hours?

This is 1 subtracted by the pvalue of Z when X = 38. So

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

By the Central Limit Theorem

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

$Z = \frac{38 - 36}{1.8333}$

$Z = 1.09$

$Z = 1.09$ has a pvalue of 0.8621

1 - 0.8621 = 0.1379

0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d. What proportion of the sample will have a mean useful life greater than 34.5 hours?

This is 1 subtracted by the pvalue of Z when X = 34.5. So

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

$Z = \frac{34.5 - 36}{1.8333}$

$Z = -0.82$

$Z = -0.82$ has a pvalue of 0.2061.

1 - 0.2061 = 0.7939

0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours.

e. What proportion of the sample will have a mean useful life between 34.5 and 38 hours?

pvalue of Z when X = 38 subtracted by the pvalue of Z when X = 34.5. So

0.8621 - 0.2061 = 0.656

0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

4 0

3 years ago

Two boats left a base camp at bearing of S60°E and S50°W at speeds of 50km/h and 70km/h respectively. Find the distance between

ivolga24 [154]

Answer:

197.93 km

Step-by-step explanation:

Let the base camp is point A and boats' locations after two hours are points B and C.

By connecting the three points together we get a triangle ABC with sides:

AB = 50*2 = 100 km
AC = 70*2 = 140 km

The angle between AB and AC is:

60 + 50 = 110 degrees (opposite directions from south)

We are looking for the distance BC, which can be found by using the law of cosines:

BC² = AB² + AC² - 2AB*AC*cos ∠BAC

BC² = 100² + 140² - 2*100*140*cos 110°
BC² = 39176.56 (rounded)
BC = √39176.56 = 197.93 km (rounded)

The distance between the boats is 197.93 km.

7 0

2 years ago

Can someone plz help with math?!

Flura [38]

Answer:

1. <em>L = 3D +59</em>

2. 158 miles

Step-by-step explanation:

For the second part:

Since each day 3 miles of the road is being added, we can simply multiply the number 33 by 3, getting 99. If the road started with 59 miles, we need to add 59 to 99, getting 158 miles.

5 0

2 years ago

Explain how you know 21/100 is greater than 1/5

Nataly_w [17]

This problem can be solved using equivalent fractions. The first step in resolving this problem is to realize that fractions are best compared when they both have the same denominator. In this case, I will choose to make that common denominator 100. There is no need to rewrite the fraction $\frac{21}{100}$ as the denominator for this fraction is already 100. The fraction $\frac{1}{5} =\frac{20}{100}$ . This is achieved by multiplying both the numerator and denominator by 20. Now that the two fractions have the same denominator, we can easily see that $\frac{21}{100}$ is greater than $\frac{1}{5}$ because it is greater than its equivalent fraction.

5 0

3 years ago

Find the domain and range

tresset_1 [31]

Domain: all real numbers
Range: y>-3

4 0

2 years ago