All categories

3 years ago

DIVIDE 54.8 by 2.6 and round the quotient to a whole number

Mathematics

1 answer:

OlgaM077 [116]3 years ago

3 0

Answer:

Step-by-step explanation:

54.8/2.6=21.0769230769

21.0769230769 rounded would be 21

You might be interested in

A yard has a perimeter of 276 feet. If six times the length of the yard equals seventeen times the width, what are it's dimensio

Mariana [72]

I hope this helps you

8 0

3 years ago

One of the smallest pools on the ship has about 45,000 gallons of water in it.

gtnhenbr [62]

Amount of water in the pool at the end of the day is 5457798.7 gallons

Explanation:

Given:

Initial amount of water in the pool = 45,000 gallons

Increase in amount = 0.75 in per minute

Time, t = 1 day

t = 24 X 60 min

t = 1440 min

So,

Increase in amount of water in 1 day = 0.75 in X 1440

= 1080 in

Volume of 1080 in of water = (1080 in)³

Volume from cubic inch to gallon = 5453298.7 gallon

Amount of water in the pool at the end of the day = 45000 + 5453298.7 gallon

= 5457798.7 gallon

5 0

3 years ago

I need help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

SpyIntel [72]

Answer:

sorry i do not know can you put the pic bigger

Step-by-step explanation:

8 0

3 years ago

Please help, ill give 69 points <33

RUDIKE [14]

Isee s yes why that look

8 0

2 years ago

Read 2 more answers

Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From

maksim [4K]

Answer:

The 95% CI for the difference of means is:

$-155.45 \leq \mu_1-\mu_2 \leq -44.55$

Step-by-step explanation:

The question is incomplete:

"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size: 4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

$M_d=M_1-M_2=200-300=-100$

The standard deviation can be estimated as:

$\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45$

The degrees of freedom are:

$df=n_1+n_2-2=10+4-2=12$

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

$M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55$

8 0

3 years ago