All categories

3 years ago

Each student need 1 quart of water if 24 students go how many gallons is that

Mathematics

2 answers:

Ivenika [448]3 years ago

7 0

6 gallons of water to supply 24 students

Bond [772]3 years ago

3 0

You would need 6 gallons of water if there are 24 students

You might be interested in

Prove the identity. TT sin x + 2. cotx sin (-x)

saveliy_v [14]

Answer:

Hi... Do this for prove that

7 0

3 years ago

Read 2 more answers

What is the area of a triangle with vertices at (−2, 1) , (2, 1) , and (3, 4) ?

VMariaS [17]

Check the picture below

area of a triangle is 1/2 bh

3 0

3 years ago

Read 2 more answers

Help.? geometry basics homework 2: segment addition postulate

Neporo4naja [7]

The value of x is 3 given that ST = 8x + 11 and TU = 12x-1

The point that bisects a line divides the line into two equal parts

If T is the midpoint of SU, the following are true:

ST = TU
ST + TU = SU

Given the following

ST = 8x + 11

TU = 12x-1

Since ST = TU

8x+11 = 12x-1

8x - 12x = -1 - 11

-4x = -12

x = 12/4

x = 3

Hence the value of x is 3 given that ST = 8x + 11 and TU = 12x-1

Learn more here: brainly.com/question/17204733

5 0

2 years ago

Read 2 more answers

Factor the trinomial: 5x^2+41x+42

babunello [35]

Answer:

(5x + 6) (x + 7)

Step By Step Explanation:

Use the sum form to the product.

5 0

2 years ago

Read 2 more answers

The mean cost of a five pound bag of shrimp is 40 dollars with a standard deviation of 7 dollars. If a sample of 51 bags of shri

eimsori [14]

Answer:

$P(|\bar{x}-40| > 0.6)=0.5404$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 40 dollars

Standard Deviation, σ = 7 dollars

Sample size,n = 51

We are given that the distribution of cost of shrimp is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{7}{\sqrt{51}} = 0.9801$

P(sample mean would differ by true mean by more than 0.6)

$P(|\bar{x}-40| > 0.6)\\\Rightarrow = 1-P(39.4$

$P(39.4 \leq x \leq 40.6) \\\\= P(\displaystyle\frac{39.4 - 40}{0.9801} \leq z \leq \displaystyle\frac{40.6-40}{0.9801})\\\\ = P(-0.6121 \leq z \leq 0.6121)\\= P(z \leq 0.6121) - P(z < -0.6121)\\= 0.7298 - 0.2702 =0.4596$

$P(|\bar{x}|-40 > 0.6)\\= 1-0.4596=0.5404$

0.5404 is the required probability.

4 0

3 years ago