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Aleksandr-060686 [28]
3 years ago
7

Marta’s fruit stand sold many oranges on Thursday and twice as many on Friday. All together, on both days, she sold 108. Which e

quation represents the problem? 108 – 2x = x 2x – x = 108 x + 108 = 2x x + 2x = 108
Mathematics
2 answers:
emmasim [6.3K]3 years ago
7 0

Answer:

D is the Answer on Edgen.

Step-by-step explanation:

108 – 2x = x

2x – x = 108

x + 108 = 2x

x + 2x = 108

lianna [129]3 years ago
4 0

Answer:

<u>The equations that represent correctly the problem are A. 108 - 2x = x and D. x + 2x = 108</u>

Step-by-step explanation:

Oranges sold on Thursday = x

Oranges sold on Friday = 2x

Oranges sold on Thursday and Friday = 108

Formula for calculating the value of x :

x + 2x = 108

<u>The correct answer to this question could be alternative D. x + 2x = 108, but if we take a closer look, we'll find that there are other valid options.</u>

A. 108 - 2x = x

- 2x - x = - 108 (changing the positions and putting the variables on one side and the whole number on the other)

<u>x + 2x = 108 (multiplying by - 1 at both sides and it's also correct)</u>

B. 2x - x = 108 (This is incorrect because the sales of Thursday plus the sales of Friday equal to 108, not minus)

C. x + 108 = 2x

x - 2x = - 108  (This is also incorrect because the sales of Thursday plus the sales of Friday equal to 108, not minus)

<u>Now, let's find the solution</u>

x + 2x = 108

3x = 108

X = 36 (Oranges sold on Thursday)

2x = 72 (Oranges sold on Friday)

<u>After doing all this explanations, we found that equations A and D represent the problem. </u>

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In 2000, the circulation of a local newspaper was 3,250. In 2001, its circulation was 3,640. In 2002, the circulation was 4,100.
vodka [1.7K]

Percent increase from 2000 to 2001 is 12% and from 2001 to 2002 is 12.6%. Thus period of 2001 to 2002 has higher increase percentage.

<u>Solution:</u>

Given that,

Total circulation of local newspaper in 2000 = 3250

Total circulation of local newspaper in 2001 = 3640

Total circulation of local newspaper in 2002 = 4100

<em><u>Finding percent of increase in the newspaper’s circulation from 2000 to 2001:</u></em>

\text { Percent increase from } 2000 \text { to } 2001=\frac{\text { change in circulation from 2001 - 2000 }}{\text { circulation in } 2000} \times 100

\begin{array}{l}{=\frac{3640-3250}{3250} \times 100} \\\\ {=\frac{390}{3250} \times 100=12 \%}\end{array}

<em><u>Finding percent of increase in the newspaper’s circulation from 2001 to 2002:</u></em>

\begin{array}{l}{\text { Percent increase from } 2001 \text { to } 2002=\frac{\text { change in circulation }}{\text {circulation in } 2001} \times 100} \\\\ {=\frac{4100-3640}{3640} \times 100} \\\\ {=\frac{460}{3640} \times 100=12.6 \%}\end{array}

As we can see 12% < 12.6%

So period of 2001 to 2002 has higher increase percentage.

7 0
3 years ago
A study was conducted to determine whether magnets were effective in treating pain. The values represent measurements of pain us
amm1812

Answer:

Given:

Sham: n= 20,     x=0.44,   s=1.24,

Magnet:n= 20,  x =0.49,   s= 0.95

For Sham:

Sample size, n = 20

Sample mean = 0.44

Standard deviation = 1.24

For Magnet:

Sample size = 20

Sample mean = 0.49

Standard deviation = 0.95

The null and alternative hypotheses:

H0: s1²=s2²

H1: s1² ≠ s2²

a) To find the test statistics, use the formula:

\frac{s1^2}{s2^2}

\frac{1.24^2}{0.95^2} = \frac{1.5376}{0.9025} = 1.7037

Test statistics = 1.7037

b) P-value:

Sham: degrees of freedom = n - 1 = 20 - 1 = 19

Magnet: degrees of freedom = n - 1 = 20 - 1 = 19

The critical values:

[Za/2, df1, df2)], [(1 - Za/2), df1, df2]

f[0.05/2, 19, 19], f[(1 - 0.05/2), 19, 19]

f[0.025, 19, 19], f[0.975, 19, 19]

(2.526, 0.3958)

The rejection region:

Reject H0, if  F < 0.3958 or if F > 2.526

c) Conclusion:

Since the critical values of test statistic is between (0.3958 < 1.7037 < 2.526), we fail to reject null hypothesis H0.

There is insufficient evidence to to support the claim that those given a sham treatment have reductions that vary more than those treated with magnets

3 0
2 years ago
How many solutions can a single variable linear equation contain?
dimaraw [331]
No solution is  the correct answer i  think
6 0
3 years ago
I need help on this, if anyone knows how to do this can you help me out pls
ASHA 777 [7]
<h3>Answer:  (3, 0)</h3>

=================================================

Explanation:

Let's isolate x in the first equation.

x-2y = 3

x = 3+2y

Then we'll plug this into the second equation

Afterwards, solve for y.

2x + 4y = 6

2(3+2y)+4y = 6

6+4y+4y = 6

8y+6 = 6

8y = 6-6

8y = 0

y = 0/8

y = 0

Use this to find x.

x = 3+2y

x = 3+2(0)

x = 3

The solution is therefore (x,y) = (3, 0)

If you were to graph both lines, then they would intersect at the location (3,0).

-----------------------------

Checking the answer:

Plug x = 3 and y = 0 into the first equation.

x-2y = 3

3-2(0) = 3

3 - 0 = 3

3 = 3 that works

Repeat for the other equation

2x+4y = 6

2(3) + 4(0) = 6

6 + 0 = 6

6 = 6 that works as well

Both equations are true when (x,y) = (3,0).

The solution is confirmed.

6 0
2 years ago
Please help ill mark brainliest
ella [17]

<u>Answer</u>:

Equation  :    y = 6

<u>Explanation</u>:

It is a straight line, cuts y axis and does not meet x axis.

Equation of the line is y = 6

3 0
2 years ago
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