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Arisa [49]
3 years ago
5

Nadia is asked to find a number such that the difference between five times the number and three less than twice the number is e

qual to eighteen. Her work and answer are shown below. 5n−(3−2n)5n−3+2n7n−37nn=====181818213 Is Nadia correct? If not, identify and correct her error.
Mathematics
2 answers:
kolbaska11 [484]3 years ago
6 0

Answer:

Nadia is incorrect, please check explanations for the correction to her work

Step-by-step explanation:

Here, we want to validate the wok of Nadia

Let the number be n

5 times the number is 5 * n = 5n

Difference between 5n and 3 less than twice the number

3 less than twice the number is 2n-3

So it should be 2n-3 ;

The solution is thus;

5n - (2n-3) = 18

5n - 2n + 3 = 18

3n + 3 = 18

3n = 18-3

3n = 15

n = 15/3

n = 5

Nataliya [291]3 years ago
4 0

Answer:

Nadia is incorrect; she did not translate the given information into an equation properly. The first equation should be 5n−(2n−3)=18.

Step-by-step explanation:

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Braden biked 10 8/9 miles on Monday. On Tuesday, he biked 3/ 4 as far as Monday. How many total miles did he bike over the two d
torisob [31]

Answer:

19 1/18 miles.

Step-by-step explanation:

For Tuesday's ride we need to multiply 10 8/9 by 3/4.

10 8/9 = (9*10+8) / 9 =  98/9 miles.

98 / 9  * 3/4

= 98/3 * 1/4

49/6

= 8 1/6 miles.

Total miles over 2 days = 10 8/9 + 8 1/6

= 18 + 8/9 + 1/6

= 18 +  48/54 + 9/54

= 18 57/54

= 19 3/54

= 19 1/18 miles.

5 0
3 years ago
A manufacturing company produces steel housings for electrical equipment. The main component part of the housing is a steel trou
Mrrafil [7]

Answer:

Check the explanation

Step-by-step explanation:

(a)

H0: Population mean = 8.46

H1: Population mean is not equal to 8.46

The test statistic (t) is given by the following expression -

{t=\frac{\overline{x}-\mu_0}{s/\sqrt{n}}}

We have calculated the sample mean (8.421) and sample standard deviation (0.0461). Therefore, the test statistic (t) will be -

t = (8.421 - 8.46)/(0.0461/sqrt(49)) = -5.92

The left-sided critical value from t-distribution (two-tailed) is -2.01. Thus the test statistic falls in the rejection region and we reject the null hypothesis at 95% confidence level and conclude that the population mean is different from 8.46 inches.

(b)

There are the following four assumptions for a single sample t-test.

The observations are in ratio scale.

The observations have been taken in such a manner that every single observation is independent and uncorrelated from the others.

There should not be any significant outliers in the sample observations.

The sample data should be approximately normal.

(c)

The first assumption is true as the data is in inches. The second assumption cannot be tested now. It will depend upon the situation and study design which we assume as correct in this case. The third assumption is checked by plotting a box plot and finding the outliers. The final assumption can be checked using the Anderson-Darling normality test.

Kindly check the graphical table in the attached images below.

(d)

The data is normal (the p-value of Anderson-Darling test in > 0.05). However, two points in the Box plot are outliers though not grossly different from the entire sample data set. So, we conclude that the assumptions are valid in this case for conducting the t-test.

7 0
3 years ago
Simplify the expression 4(2x - 3y) ASAP
uranmaximum [27]

Answer:

8x - 12y

Step-by-step explanation:

4(2x-3y)

Distribute/multiply the 4 to everything in the parentheses

4 * 2x = 8x      4 * -3y = -12y

8x - 12y

4 0
3 years ago
Read 2 more answers
Can someone help me with this please?
photoshop1234 [79]
Your answer is B ...
7 0
2 years ago
Suppose that the functions r and a are defined for all real numbers x as follows. r(x)=2x-1 S(x)=5x write the expressions for (r
NeTakaya

\boxed{(r-s)(x)=-3x-1} \\ \\ \boxed{(r\cdot s)(x)=10x^2-5x} \\ \\ \boxed{(r+s)(-2)=-15}

<h2>Explanation:</h2>

In this exercise, we have the following functions:

r(x)=2x-1 \\ \\ s(x)=5x

And they are defined for all real numbers x. So we have to write the following expressions:

First expression:

(r-s)(x)

That is, we subtract s(x) from r(x):

(r-s)(x)=2x-1-5x \\ \\ Combine \ like \ terms: \\ \\ (r-s)(x)=(2x-5x)-1 \\ \\ \boxed{(r-s)(x)=-3x-1}

Second expression:

(r\cdot s)(x)

That is, we get the product of s(x) and r(x):

(r\cdot s)(x)=(2x-1)(5x) \\ \\ By \ distributive \ property: \\ \\ (r\cdot s)(x)=(2x)(5x)-(1)(5x) \\ \\ \boxed{(r\cdot s)(x)=10x^2-5x}

Third expression:

Here we need to evaluate:

(r+s)(-2)

First of all, we find the sum of functions r(x) and s(x):

(r+s)(x)=2x-1+5x \\ \\ Combine \ like \ terms: \\ \\ (r+s)(x)=(2x+5x)-1 \\ \\ (r+s)(x)=7x-1

Finally, substituting x = -2:

(r+s)(-2)=7(-2)-1 \\ \\ (r+s)(-2)=-14-1 \\ \\ \boxed{(r+s)(-2)=-15}

<h2>Learn more: </h2>

Parabola: brainly.com/question/12178203

#LearnWithBrainly

5 0
3 years ago
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