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2 years ago

How many pints are in 192 FL oz

Mathematics

2 answers:

Bond [772]2 years ago

7 0

In 192 FL oz there are 12 us liquid pints

Vikki [24]2 years ago

6 0

12 pints. Hope I helped

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Out of 3000 students in a School of Public Health, 1500 students defined themselves regular dietary supplement users. If you tak

attashe74 [19]

Answer:

The standard error of the proportion is 0.112.

Step-by-step explanation:

The standard error of a proportion is given by the following formula:

$S = \sqrt{\frac{p(1-p)}{n}}$

In which p is the probability of a success and n is the length of the sample.

In this problem

A success is a student defining themselves regular dietary supplement users. Our of 3000 students, 1500 do. So $p = \frac{1500}{3000} = 0.5$ .

We take a sample of 20 of them, so $n = 20$ .

$S = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5*0.5}{20}} = 0.112$

8 0

2 years ago

The cost of 5 pens is $140.50. What is the cost of 1 pen?

nydimaria [60]

Answer:

28.1

Step-by-step explanation:

140.50/5

These are some expensive pens

5 0

2 years ago

Read 2 more answers

PLEASE HELP I GIVE THANKS

ololo11 [35]

The answer would be B because it crosses the X axis closer to the 0,0 than the original line

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2 years ago

Can -5 be the base of an exponential function?

sleet_krkn [62]

No, because negative base sometimes results in non-real or imaginary value.

5 0

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

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0

2 years ago