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2 years ago

PLZ NEED HELP ASAP SHOW WORK TO PLZ IT WOULD MEAN A LOT

Mathematics

1 answer:

Marizza181 [45]2 years ago

7 0

First Chart: Perimeter

Square Portion:

Original Side Lengths: P = 4 (1 + 1 + 1 + 1 ) =4

Double Side Lengths: P = 8 (2 x 4 = 8)

Triple Side Lengths: P = 12 (4 x 3 = 12)

Quadruple Side Lengths: P = 16 (4 x 4 = 16)

Rectangle Portion:

Original Side Lengths: P = 6 (1 x 2 + 2 x 2 = 6)

Double Side Lengths: P = 12 (2 x 2 + 4 x 2 = 12)

Triple Side Lengths: P = 24 (4 x 2 + 8 x 2 = 24)

Quadruple Side Lengths: P = 48 (8 x 2 + 16 x 2 = 48)

Second Chart: Area

Square Portion:

Original Side Lengths: A = 1 (1 x 1 = 1)

Double Side Lengths: A = 4 (2 x 2 = 4)

Triple Side Lengths: A = 9 (3 x 3 = 9

Quadruple Side Lengths: A = 16 ( 4 x 4 = 16)

Rectangle Portion:

Original Side Lengths: A = 2 ( 1 x 2 = 2 )

Double Side Lengths: A = 8 ( 2 x 4 = 8)

Triple Side Lengths: A = 18 ( 3 x 6 = 18)

Quadruple Side Lengths: A = 32 (4 x 8 = 32)

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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

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.

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Answer:

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