All categories

3 years ago

How much ounces 5 1/8 pounds? (show all work)

Mathematics

1 answer:

Alekssandra [29.7K]3 years ago

8 0

It’s definitely 5.125

You might be interested in

According to 2013 report from Population Reference Bureau, the mean travel time to work of workers ages 16 and older who did not

gregori [183]

Answer:

a) 48.80% probability that his travel time to work is less than 30 minutes

b) The mean is 30.7 minutes and the standard deviation is of 3.83 minutes.

c) 13.13% probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 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 normally distributed samples are 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.

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.

In this question, we have that:

$\mu = 30.7, \sigma = 23$

a. If a worker is selected at random, what is the probability that his travel time to work is less than 30 minutes?

This is the pvlaue of Z when X = 30. So

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

$Z = \frac{30 - 30.7}{23}$

$Z = -0.03$

$Z = -0.03$ has a pvalue of 0.4880.

48.80% probability that his travel time to work is less than 30 minutes

b. Specify the mean and the standard deviation of the sampling distribution of the sample means, for samples of size 36.

$n = 36$

Applying the Central Limit Theorem, the mean is 30.7 minutes and the standard deviation is $s = \frac{23}{\sqrt{36}} = 3.83$

c. What is the probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes?

This is 1 subtracted by the pvalue of Z when X = 35. So

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

By the Central Limit Theorem

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

$Z = \frac{35 - 30.7}{3.83}$

$Z = 1.12$

$Z = 1.12$ has a pvalue of 0.8687

1 - 0.8687 = 0.1313

13.13% probability that in a random sample of 36 NJ workers commuting to work, the mean travel time to work is above 35 minutes

8 0

3 years ago

I need help asap )/)/)/)

Andrej [43]

Answer:

$49.20

Step-by-step explanation:

6 0

3 years ago

The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

Alexandra [31]

Answer:

$\bar x = 260.1615$

$\sigma = 70.69$

The confidence interval of standard deviation is: $53.76$ to $103.25$

Step-by-step explanation:

Given

$n =20$

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

$\bar x = 260.1615$

$\bar x = 260.16$

Solving (b): The standard deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

So, the confidence interval of the standard deviation is:

$\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} }$ to $\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }$

$70.69 * \sqrt{\frac{20 - 1}{32.852}$ to $70.69 * \sqrt{\frac{20 - 1}{8.907}$

$70.69 * \sqrt{\frac{19}{32.852}$ to $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

8 0

2 years ago

The sum of 5 consecutive integers is 205. What is the third number?

Elza [17]

If 5 consecutive integers is 205,

then a + b + c + d + e = 205
but also, each integer is separated by a difference of 1

⇒ a + (a + 1) + (a + 2) + (a + 3) + (a + 4) = 205
⇒ 5a + 10 = 205
⇒ 5a = 195
⇒ a = 39

∴ third term = 39 + 2
= 41

7 0

2 years ago

£6846 is shared equally between 7 people. How much does each person get?

jok3333 [9.3K]

Answer:

£978

Step-by-step explanation:

i did use a calculator so i don't know if this is wrong or not, sorry if it is.

8 0

2 years ago

Read 2 more answers