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3 years ago

What is the value of i 20+1

Mathematics

2 answers:

zmey [24]3 years ago

6 0

Answer:

<h2>21 no sure what the "i" means</h2>

Step-by-step explanation:

Ipatiy [6.2K]3 years ago

6 0

I’m pretty sure its 21

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All rectangles are paralloelograms

Anni [7]

Answer:

not all rectangles are paralloelograms.

Step-by-step explanation:

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3 years ago

Which of the following is equivalent to 1 cup

kipiarov [429]

1 cup=16 tablespoons

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3 years ago

Read 2 more answers

Malia buys 2 loaves of bread priced at $2.46 per loaf and 2 pound of cheese priced at

aleksandrvk [35]

She spent $19.88 on cheese and bread

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2 years ago

15 points to whoever answers correct

Pepsi [2]

<u>Answer</u>

C. 4 + 6

D. 3.15151515.. + 4.6

E. √64 × √25

<u>Explanation</u>

A rational number is a number that can be expressed as a quotient or a fraction. That is it can be written as a/b.

π×6 = 6π ⇒ can not be written as a rational number.

√3 + √5 ⇒ the sum of two irrational numbers can not be rational.

4 + 6 = 10. Ten is a rational number. it can be written as 10/1

3.15151515.. + 4.6 = 104/33 + 23/5

= 1279/165

√64 × √25 = 8×5=40 This can be written as 40/1

5 0

4 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