All categories

3 years ago

3 gal = pt please I'm in 5th grade

1 answer:

7 0

24 pints
since there are 8 pints in a gallon you can do 3 x 8 and get 24 cause you are converting a bigger unit into a smaller unit

You might be interested in

To find the value of x in the equation 9/4x+2=20, you first get the term with x by itself on one side of the equation. How can y

Scorpion4ik [409]

Answer:

Step-by-step explanation:

9/4x + 2 = 20

To get x by it's self you do the same on both sides of the = to cancel out each side.

9/4x + 2 = 20

(Multiply 9 on both sides)

4x + 2 = 20*9

(Now subtract 2 from both sides)

4x = 20*9 - 2

And finally divide both sides by 4

x = (20*9 - 2)/4

Then solve for x as normal.

3 0

3 years ago

450 is 75% of what number. show ur work

alex41 [277]

Let N is the number

75% x N = 450
75/100 x N = 450
75N = 450 x 100
N = (450 x 100)/75
N = 600

So, the number is 600.

3 0

3 years ago

An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "

FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

$\large X_{sentences}$ the mean of the “sentences” group

$\large S_{sentences}$ the standard deviation of the “sentences” group

$\large X_{intentional}$ the mean of the “intentional” group

$\large S_{intentional}$ the standard deviation of the “intentional” group

Then, we can calculate by using the computer

$\large X_{sentences}=28.75$

$\large S_{sentences}=3.53553$

$\large X_{intentional}=31.625$

$\large S_{intentional}=1.40788$

$\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875$

The standard error of the difference (of the means) for a sample of size 8 is calculated with the formula

$\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}$

So, the standard error of the difference is

$\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546$

In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of 95% for the difference of the means corresponding to a level of significance of 0.05 (5%).

If this interval contains the zero, we can say there is no significant difference.

Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples.

We get the $\large t_{0.05}$ which is a value of t such that the area under the Student's t distribution outside the interval $\large [-t_{0.05}, +t_{0.05}]$ is less than 0.05.