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MA_775_DIABLO [31]
2 years ago
12

Amir has 103 days to finsih project he works 7 days a week on the project how many week and days does amir have to finish the pr

oject
Mathematics
1 answer:
aleksandr82 [10.1K]2 years ago
5 0

Answer: 14 weeks 5 days

Step-by-step explanation:

From the question, we are informed that Amir has 103 days to finish a project and that he works 7 days a week on the project.

The number of weeks and days that he has to finish the project will be:

= 103/7

= 14 5/7

= 14 weeks 5 days

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A line intersects the point (-11, 4) and has a slope of -2. What are the inputs to the point-slope formula? y - [?] = [](x-[])​
Lyrx [107]

y - 4 = -2 ( x - ( -11)) which would simplify to y - 4 = -2 ( x + 11) !!

5 0
2 years ago
Systems of equations, solve by substitution please!<br> x + 7y = -37<br> 4x - 3y = 7
Svetradugi [14.3K]
Answer:

1. x = -37 - 7y

2. x = 7+3y/4


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5 0
2 years ago
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What is the value of x?​
myrzilka [38]

Answer:

<h2>x = 20</h2><h2 />

Step-by-step explanation:

∠VRW and ∠TRS form a pair of vertical angles then they are congruent

therefore

3x = x + 40

⇔

2x = 40

⇔

x = 20

5 0
3 years ago
Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10959}{25} =438.36

Sum of squares of differences = 175413.76

S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49

Population mean, μ = 425 mg/L

Sample mean, \bar{x} = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813

Now, t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108

Since,                        

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0
3 years ago
Please Help! Will give BRAINLIEST!!!
Anestetic [448]

Answer:

about 200 people

4 0
3 years ago
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