A submarine is at 756 feet below sea level how many feet will it need to rise to be at the surface?

Marrrta [24]

300567672468.78-200.67=
300567672268.11005

4 0

3 years ago

a spinner is divided into nine equal sections numbered 1 through 9. predict how many times out of 270 spins the spinner is most

evablogger [386]

I'd say around 150 times

5 0

3 years ago

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

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:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

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

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

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

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

6 0

3 years ago

Comparing the two ratios 2/4 and 1/2 2/4 is larger

kumpel [21]

Answer:

3.5 : 7 > 1.2 : 2.5.

Step-by-step explanation:

1. Compare the ratios 4 : 5 and 2 : 3.

Solution:

Express the given ratios as fraction

4 : 5 = 4/5 and 2 : 3 =2/3

Now find the L.C.M (least common multiple) of 5 and 3

The L.C.M (least common multiple) of 5 and 3 is 15.

Making the denominator of each fraction equal to 15, we have

4/5 = (4 ×3)/(5 ×3) = 12/15 and 2/3 = (2 ×5)/(3 ×5) = 10/15

Clearly, 12 > 10

Now, 12/15 > 10/15

Therefore, 4 : 5 > 2 : 3.

2. Compare the ratios 5 : 6 and 7 : 9.

Solution:

Express the given ratios as fraction

5 : 6 = 5/6 and 7 : 9 =7/9

Now find the L.C.M (least common multiple) of 6 and 9

The L.C.M (least common multiple) of 6 and 9 is 18.

Making the denominator of each fraction equal to 18, we have

5/6 = (5 ×3)/(6 ×3) = 15/18 and 7/9 = (7 ×2)/(9 ×2) = 14/18

Clearly, 15 > 14

Now, 15/18 > 14/18

Therefore, 5 : 6 > 7 : 9.

3. Compare the ratios 1.2 : 2.5 and 3.5 : 7.

Solution:

1.2 : 2.5 = 1.2/2.5 and 3.5 : 7 =3.5/7

1.2/2.5 = (1.2 ×10)/(2.5 ×10 ) = 12/25 and 3.5/7 = (3.5 ×10)/(7 ×10) = 35/70 = 1/2

[We removed the decimal point from the ratios now, we will compare the ratio]

Now find the L.C.M (least common multiple) of 25 and 2

The L.C.M (least common multiple) of 25 and 2 is 50.

Making the denominator of each fraction equal to 50, we have

= 12/25 = (12 ×2)/(25 ×2) = 24/50 and 1/2 = (1 ×25)/(2 ×25) = 25/50

Now, 25/50 > 24/50

Therefore, 3.5 : 7 > 1.2 : 2.5.

4 0

3 years ago

An apple farmer has planted 0.3 of his orchard with cherry trees. The orchard is 8.59 acres in size. How many acres are planted

Lynna [10]

0.3(8.59) = 2.577 rounds to 2.58 acres

6 0

3 years ago