The sum of the digits of a 5 digit positive integer =2 how many five digits integers exist?

last week, ezra's sold 1/3 of tangerines and Anthony sold 3/4 of tangerines as ezra's did. how many boxes of tangerines did Anth

Bond [772]

Answer:

¼ box

Step-by-step explanation:

Ezra sold ⅓ box of tangerines.

Anthony sold ¾ as many as Ezra = ¾ × ⅓ box

Cancel the 3s = ¼ × ¹/₁ box

Multiply numerators and denominators = ¼ box

Anthony sold ¼ box of tangerines.

6 0

3 years ago

You roll four six-sided die. Find the probability that exactly three of the <br> dice are odd.

s344n2d4d5 [400]

Answer:

3/8

Step-by-step explanation:

There is a 1/2 probability one die will be odd because 1, 3, and 5 are the only possible odd numbers.

Since you want 3/4 of the dice to be odd, you multiply 1/2 by 3/4 to get 3/8.

I believe it is 3/8.

3 0

3 years ago

Read 2 more answers

In a given year, the average annual salary of a NFL football player was $189,000 with a standard deviation of $20,500. If a samp

nika2105 [10]

Answer:

15.15% probability that the sample mean will be $192,000 or more.

Step-by-step explanation:

To solve this problem, 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 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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 189000, \sigma = 20500, n = 50, s = \frac{20500}{\sqrt{50}} = 2899.14$

The probability that the sample mean will be $192,000 or more is

This is 1 subtracted by the pvalue of z when X = 192000. So

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

By the Central Limit Theorem

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

$Z = \frac{192000 - 189000}{2899.14}$

$Z = 1.03$

$Z = 1.03$ has a pvalue of 0.8485.

1-0.8485 = 0.1515

15.15% probability that the sample mean will be $192,000 or more.

7 0

2 years ago

If the average speed of a garden snail is 0.025 miles per hour and one inch is approximately 2.54 centimeters, how many minutes

Katen [24]

Answer:

Option B 1.5 min

Step-by-step explanation:

we know that

1 mi=1,609.36 m

1 h=60 min

so

$0.025\ mi/h=0.025(1,609.36)/60=0.67\ m/min$

using proportion

$\frac{0.67}{1}\frac{m}{min} = \frac{1}{x}\frac{m}{min}\\\\x=1/0.67\\\\x=1.5\ min$

8 0

3 years ago

A manufacturing company produces x units this week and wants to increase its weekly output by 135 units by next week. The compan

masya89 [10]

Answer:

x + 135 = 1450

Step-by-step explanation:

It says that they are trying to increase the output by 135 so you have the variable to show how much they have to make to reach there goal of 1450 next week

I hope this helped you!

4 0

3 years ago