All categories

3 years ago

Which algebraic equation models the English sentence, a number subtracted from 8 is equal to 4?

Mathematics

1 answer:

torisob [31]3 years ago

5 0

I believe it is x-8=4 if I understand this correctly. Otherwise it's x-4=8.

You might be interested in

Simplify: - 3(6 - 2) + | - 4 | ÷ 2 The solution

Korolek [52]

Answer: -10

Step-by-step explanation:

7 0

3 years ago

Answer correctly

DochEvi [55]

Answer: d

Step-by-step explanation: can i have brainlist please :D

6 0

3 years ago

Please help with Linear Function Models for 20 points, Thank you.

Bumek [7]

Y=mx+b
m=slope
b=yintercept

input points
when x=1
y=1
1=1m+b

and
x=2
y=-2
-2=2m+b
multilpy first equaiton by -1 and add to othe requation

-1=-m-b
-2=2m+b +
-3=m+0b

-3=m
y=-3x+b
1=-3(1)+b
1=-3+b
add 3
4=b
y=-3x+4

first blank is -3
second is 4

4 0

3 years ago

Read 2 more answers

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.

That value can be obtained either by using a table or the computer and is found to be

$\large t_{0.05}=2.365$

Now we can compute our confidence interval

$\large (X_{sentences}-X_{intentional}) \pm t_{0.05}*(standard \;error)=-2.875\pm 2.365*1.34546$

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.

6 0

4 years ago

Dennis_Churaev [7]

Answer:

C. Approximately normal

Step-by-step explanation:

We know that when we have, np≥10, and n(1−p)≥10, and both of these are true, where n is the sample size, and p is the sample of proportions, the sampling proportions of sample distributions, will be normal in shape.

Expected successes : np=125(0.88)=110≥10

Expected failures : n(1−p)=125(1−0.88)=15≥10

7 0

2 years ago