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andriy [413]
3 years ago
7

Jameer paints one wall of a house in 3 days. june paints two walls of a house in 5 days. if the two of them work together to pai

nt a whole house that has 10 walls, how long will they take? (we assume that each wall takes the same amount of time to paint for any fixed individual.)
Mathematics
1 answer:
grandymaker [24]3 years ago
7 0

Let the number of days be x.

Jameer paints 1 wall in 3 days. In 1 day, he paints 1/3 of a wall. In x days, he paints (1/3)x walls.

June paints 2 walls in 5 days. In 1 day, she paints 2/5 of a wall. In x days, she paints (2/5)x walls.

Working together, in x days they paint (1/3)x + (2/5)x walls.

We want to know the number of days they will take to paint 10 walls, so we let the expression equal 10.

(1/3)x + (2/5)x = 10

(5/15)x + (6/15)x = 10

(11/15)x = 10

x = 10 * 15/11

x = 150/11

It will take 150/11 days which is the same as 13.6363... days or approximately 13.6 days.

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Answer:

The inequality sign remains same while multiply or divide both sides by positive numbers.

The inequality sign changes while multiply or divide both sides by negative numbers.

Step-by-step explanation:

The given inequality is - 8 < 2.

Now, if we multiply 2 in both sides then - 16 < 4

Again, if we divide by 2 into both sides then  - 4 < 1

Therefore, the inequality sign remains the same while multiply or divide both sides by positive numbers.

Now, if we multiply -2 in both sides then 16 > -4

And, if we divide -2 into both sides then 4 > -1

Therefore, the inequality sign changes while multiply or divide both sides by negative numbers. (Answer)

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1 + 1 = 2

Step-by-step explanation:

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Natalia is making gift bags for her son's birthday party. Each bag costs $3.24 to make and she needs to make 15 bags. How much w
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Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus
quester [9]
Hi! 

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20;  </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510

To calculate the degrees of freedom you need to use the following equation:

df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.



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3 years ago
Variation equation and determine the quantity indicated.
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Answer:

Step-by-step explanation:

Direct variation has the form y=kx, given k=7 and x=20

y=7(20)=140

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