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3 years ago

Solve each equation. Check your solution. –3b + 7 = –15 + 2b

Mathematics

2 answers:

Gnoma [55]3 years ago

6 0

Exact form: b= 22/5

Decimal form: b= 4.4

Mixed number form: b= 4 2/5

aliya0001 [1]3 years ago

6 0

Answer:

Solution:

$b = \frac{22}{5?}$

Step-by-step explanation:

Rearrange terms:

$- 3b + 7 = - 15 + 2b$

$- 3b + 7 = 2b - 15$

subtract 7 from both sides of the equation:

$- 3b + 7 = 2b - 15$

simplify:

subtract the numbers

⬇️

$- 3b = 2b - 22$

subtract ⬇️ from both sides of the equation

$2b$

$- 3b = 2b - 22$

$- 3b - 2b = 2b - 22 - 2b$

simplify:

⬇️combine like terms

$- 5b = - 22$

divide both sides of the equation by the same term:

$- 5b = - 22$

$\frac{ - 5b}{ - 5?} = \frac{ - 22}{ - 5?}$

simplify:

⬇️cancel terms that are in both the numerator and the denominator

⬇️divide the numbers

$b = \frac{22}{?5}$

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Hello from MrBillDoesMath!

Answer:

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3 years ago

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3 years ago

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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:

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

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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