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3 years ago

How many 8 oz glasses of water are in 3 liters?

Mathematics

2 answers:

Slav-nsk [51]3 years ago

7 0

1 liter = 33.8140226 US fluid ounces

zimovet [89]3 years ago

3 0

About 96 becuse in one liter there is four 8 oz glases so that makes 32 and if you add 32 three times you get 96

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3 years ago

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3 years ago

A sample of 47 observations is selected from a normal population. The sample mean is 30, and the population standard deviation i

erma4kov [3.2K]

Answer:

We accept the null hypothesis and reject the alternate hypothesis. There is no evidence to conclude that the population mean is greater than 29. The population mean is less than or equal to 29.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 29

Sample mean, $\bar{x}$ = 30

Sample size, n = 47

Alpha, α = 0.05

Population standard deviation, σ = 5

First, we design the null and the alternate hypothesis

$H_{0}: \mu \leq 29\\H_A: \mu > 29$

a) This is a one-tailed test because the alternate hypothesis is in greater than direction.

We use One-tailed z test to perform this hypothesis.

b) $z_{stat} > z_{critical}$ , we reject the null hypothesis and accept the alternate hypothesis and if $z_{stat} < z_{critical}$ , we accept the null hypothesis and reject the alternate hypothesis.

c) Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{30 - 29}{\frac{5}{\sqrt{47}} } = 1.37$

d) Now, $z_{critical} \text{ at 0.05 level of significance } = 1.64$

Since,

$z_{stat} < z_{critical}$

We accept the null hypothesis and reject the alternate hypothesis. There is no evidence to conclude that the population mean is greater than 29. The population mean is less than or equal to 29.

e) P-value is 0.0853

On the basis of p value we again accept the null hypothesis.

5 0

3 years ago

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$