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2 years ago

What is the result when 3/8x + 2 1/5 is subtracted from 3 1/2x - 4 1/10? Don’t put a link.

Mathematics

2 answers:

Umnica [9.8K]2 years ago

6 0

Answer:

Option 3 3 1/8x - 6 3/10

Step-by-step explanation:

(3 1/2x - 4 1/10) - (3/8x + 2 1/5)

Divide it into parts:

(3 1/2x - 3/8x) + (-4 1/10 - 2 1/5)

3 1/2x - 3/8x = 3 4/8x - 3/8x = 3 1/8x

-4 1/10 - 2 1/5 = -4 1/10 - 2 2/10 = -6 3/10

So we get:

2 1/4x - 6 3/10

Setler79 [48]2 years ago

5 0

trying to cheat on a performance matter test but the answer is the last one -3 1/8 x + 6 3/10

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chloe and jack play 4 games. the probability that chloe wins any game is 4/5. what is the probability that chloe wins for the 1s

Iteru [2.4K]

$\dfrac{1}{5}\cdot\dfrac{1}{5}\cdot\dfrac{1}{5}\cdot\dfrac{4}{5}=\dfrac{4}{625}$

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3 years ago

Read 2 more answers

What is the side length? Please help, thanks

azamat

Answer:

if Area = 169/225

then

S = √(169/225)

S = √169/√225

S = 13/15

5 0

3 years ago

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

3 years ago

Help please! Spring break hw for 7th grade advanced.

likoan [24]

I am so sorry if I'm wrong, but I think 5 is b

4 0

3 years ago

Plz answer only if correct :)) <br> (this is trig.ratios) in

slamgirl [31]

Answer:

Step-by-step explanation:

Since you want to find the variable x all you have to do is simple subtraction.

The two numbers (61 and 22). Subtract those numbers and you get 39. Now it wants you to round to the jearest hundredth but there is no hundreds place so we can round to the nearest tenth. Which would be 30! Hope it helped... :)!

bye

8 0

3 years ago