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Thepotemich [5.8K]
3 years ago
5

Ax-c=b solve for x *show work

Mathematics
1 answer:
Makovka662 [10]3 years ago
4 0
ax-c=b\ \ \ |add\ c\ to\ both\ sides\\\\ax=b+c\ \ \ \ |divide\ both\ sides\ by\ a\\\\\boxed{x=\frac{b+c}{a}}
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What is a power?????????????
AlladinOne [14]

Answer:

The power (or exponent) of a number says how many times to use the number in a multiplication.

It is written as a small number to the right and above the base number.

In this example the little "2" says to use 8 two times in a multiplication:

8^{2} = 8 × 8 = 64

Step-by-step explanation:

8 0
3 years ago
(8X-20)° (3X+5)°how do you solve this​
kirza4 [7]

Answer:5

Step-by-step explanation:

Simplifying

8x + -20 = 3x + 5

Reorder the terms:

-20 + 8x = 3x + 5

Reorder the terms:

-20 + 8x = 5 + 3x

Solving

-20 + 8x = 5 + 3x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-3x' to each side of the equation.

-20 + 8x + -3x = 5 + 3x + -3x

Combine like terms: 8x + -3x = 5x

-20 + 5x = 5 + 3x + -3x

Combine like terms: 3x + -3x = 0

-20 + 5x = 5 + 0

-20 + 5x = 5

Add '20' to each side of the equation.

-20 + 20 + 5x = 5 + 20

Combine like terms: -20 + 20 = 0

0 + 5x = 5 + 20

5x = 5 + 20

Combine like terms: 5 + 20 = 25

5x = 25

Divide each side by '5'.

x = 5

Simplifying

x = 5

6 0
3 years ago
The Scholastic Aptitude Test (SAT) is a standardized test for college admissions in the U.S. Scores on the SAT can range from 60
kodGreya [7K]

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

n_1 represent the sample after the preparation

n_2 represent the sample without preparation  

\bar x_1 =678 represent the mean sample after preparation

\bar x_2 =1837 represent the mean sample without preparation

s_1 =197 represent the sample deviation after preparation

s_2 =328 represent the sample deviation without preparation

\alpha=0.1 represent the significance level

Confidence =90% or 0.90

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})} (1)  

The point of estimate for \mu_1 -\mu_2

The appropiate degrees of freedom are df=n_1+ n_2 -2

Since the Confidence is 0.90 or 90%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,df)  

The standard error is given by the following formula:  

SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}  

After replace in the formula for the confidence interval we got this:

3.0 < \mu_1 -\mu_2

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

5 0
3 years ago
PLEASE HELP ASAP WILL MARK BRAINLIEST
laiz [17]

Step-by-step explanation:

here you go

mark me brainliest

8 0
3 years ago
Aaron wants to buy a bicycle that costs $128. So far he has saved $56 the equation of 8 + 56 equal 128 can be used to find the a
lutik1710 [3]

Answer:

Aaron still needs to save $9

Step-by-step explanation:

Assuming the equation is 8a + 56 = 128

8a + 56 = 128

8a = 128 - 56 = 72

8a = 72

a = 72/8 = $9

7 0
4 years ago
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