What is the solution f/4 = 12/5

Plzz hellp :( i will mark you the brainlist if some on helps (A rectangular prism has a length of 1 1/4 yards, a width of 2 1/2

Veseljchak [2.6K]

The volume of rectangular prism is the product of its length, width and height.
So the volume V of prism with length $1 \frac{1}{4}$ yards, width $2 \frac{1}{2}$ yards and height of 4 yards will be:

$V=1 \frac{1}{4} * 2 \frac{1}{2} *4 \\ \\ 
V= \frac{5}{4}* \frac{5}{2}*4 \\ \\ 
V= \frac{25}{2} \\ \\ 
V=12 \frac{1}{2}$

Thus the volume of prism as a mixed number in simplest form will be $12 \frac{1}{2}$ cubic yards

4 0

2 years ago

The data below represent the number of days absent per year in a population of three employees of a small company:

love history [14]

The sampling distribution is illustrated below.

<h3>How to illustrate the sampling?</h3>

Based on the information given, the population is illustrated as 1, 4, 7. The sample without replacement will be:

<u>Sample Mean</u>

1 and 4. 2.5

1 and 7. 4

4 and 7. 5.5

The population mean will be:

= (1 + 4 + 7)/3

= 4

The sample mean is an unbiased estimator of the population mean.

The sampling with replacement illustrates that the mean will be:

= 36/9 = 4

Also, when n = 3, the number of samples will be:

= 3³ = 27

Learn more about sampling on:

brainly.com/question/17831271

#SPJ1

8 0

1 year ago

Michmoo Computer Company sells computers and computer parts by mail. The company assures its customers that products are mailed

Anastaziya [24]

Answer: (60.858, 69.142)

Step-by-step explanation:

The formula to find the confidence interval for mean :

$\overline{x}\pm z_c\dfrac{\sigma}{\sqrt{n}}$ , where $\overline{x}$ is the sample mean , $\sigma$ is the population standard deviation , n is the sample size and $z_c$ is the two-tailed test value for z.

Let x represents the time taken to mail products for all orders received at the office of this company.

As per given , we have

Confidence level : 95%

n= 62

sample mean : $\overline{x}=65$ hours

Population standard deviation : $\sigma=18$ hours

z-value for 93% confidence interval: $z_c=1.8119$ [using z-value table]

Now, 93% confidence the mean time taken to mail products for all orders received at the office of this company :-

$65\pm (1.8119)\dfrac{18}{\sqrt{62}}\\\\ 65\pm4.142\\\\=(65-4.142,\ 65+4.142)\\\\= (60.858,\ 69.142)$

Thus , 93% confidence the mean time taken to mail products for all orders received at the office of this company : (60.858, 69.142)

8 0

3 years ago

Help? pls? (10 points) (image down below to make it easier to understand.)

dusya [7]

This is middle school??? AinT no way

7 0

2 years ago

Read 2 more answers

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the

kodGreya [7K]

Answer:

p=0.7103 (4-game series)
p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters n=4, p=0.6:

$P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\ i=4,5,6,7$

#We find probabilities for the different values of i:

$P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659$

Hence, probability of the stronger team winning overall is:

$=P(X=4)+P(X=5)+P(X=6)+P(X=7)\\\\=0.7103$

#Define Y as the random variable for winning 2/3 games.:

$P(Y=2)={1\choose 1}0.6^2=0.3600\\\\P(Y=3)={2\choose3}0.6^20.4=0.2880\\\\P(win)=0.2880+0.3600=0.6480$

Hence, probability of the stronger team winning in 2 out 3 game series is 0.6480

The stronger team has a higher chance of winning in a 4-game series(0.7103>0.6480)

8 0

2 years ago

What is the solution f/4 = 12/5​

What is the solution f/4 = 12/5