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natima [27]
3 years ago
13

I got g=(3d-2)^2-5 Simplified : g=9d-1 Did I get it correct?

Mathematics
2 answers:
SVEN [57.7K]3 years ago
4 0

Answer:

No

Step-by-step explanation:

g=(3d-2)^2-5

g=(3d-2)(3d-2)-5

<em>remember to use distributive property when squaring a value with 2 or more terms.</em>

g = 9d^2-12d+4-5

g = 9d^2-12d-1

Alex787 [66]3 years ago
4 0

Answer:

Close but no, g=3d^2-1

Step-by-step explanation:

Since d cannot be determined, we cannot square just the three, we have to square x too so 3d^2

I hope this helps u pls give a brainliest and a thx

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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
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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:

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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}} }

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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.

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