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3 years ago

○a 25°○b 31°○c 23°○d 59°

Mathematics

2 answers:

Kamila [148]3 years ago

7 0

Answer:

Step-by-step explanation:

∠ BDC + ∠ ADC = ∠ ADB , substitute values

- 6x + 53 - 9x + 22 = 90

- 15x + 75 = 90 ( subtract 75 from both sides )

- 15x = 15 ( divide both sides by - 15 )

x = - 1

Then

∠ ADC = - 9x + 22 = - 9(- 1) + 22 = 9 + 22 = 31° → b

Vaselesa [24]3 years ago

5 0

Answer:

i think it would be B

Step-by-step explanation:

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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.

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

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

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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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○a 25°○b 31°○c 23°○d 59°​

○a 25°○b 31°○c 23°○d 59°