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2 years ago

If x > 3, Which of these numbers could be a possible number for x?

Mathematics

2 answers:

fgiga [73]2 years ago

7 0

Answer:

Step-by-step explanation:

the Awnser is D because x is greater than three and 4 is the only number listed that is larger than 3

TEA [102]2 years ago

4 0

Answer:

D. 4

Step-by-step explanation:

x is greater than 3, which means x has to be a higher number. the only option here that is higher than 3 is 4.

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Amalia used 8 lemons to make a picture of lemonade that served six people. How many lemons would she need to make enough lemonad

11111nata11111 [884]

28 lemons is the answer to your question

8 0

3 years ago

As a marketing manager, you are tasked with selecting a website to place your advertisement. The following sampled data shows th

olasank [31]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

Website 1 : 10357, 10537, 10767, 10561, 10544, 10581, 10602, 10665, 10335, 10419, 10737, 10410, 10485, 10601, 10458, 10472, 10435, 10375, 10436, 10510, 10345, 10559, 10520, 10425, 10351, 10465, 10491, 10671, 10366, 10440, 10618, 10606, 10406, 10538, 10449, 10462

Mean, xbar = ΣX/ n ; n = sample size = 36

Xbar = 377999 / 36 = 10499.9722

Standard deviation, s = √[(x - xbar)² / (n-1]

Using calculator :

Standard deviation (Website 1 :), s = 110.239865

Website 2 : 11067, 11029, 10888, 10789, 10914, 10663, 10787, 11140, 11042, 11074, 10868, 10853, 10900, 11088, 10991, 10928, 10959, 11126, 11033, 11114, 11150, 11155, 11027, 10900, 11015, 11123, 10953, 11181, 10855, 10731, 10971, 10770, 11070, 11122, 11018, 10903

Mean, xbar = ΣX/ n ; n = sample size = 36

Xbar = 395197 / 36 = 10977.6944

Standard deviation, s = √[(x - xbar)² / (n-1]

Using calculator :

Standard deviation (Website 2), s = 132.617995

2.)

Yes, the viewership between the two websites are different with the second website has a higher mean viewership with a mean of 10977.6944.

3.)

The probability of 12000 views per month on each website :

Probability = Mean viewership per month / required viewership

Website 1 :

P(12000) = 10499.9722 / 12000 = 0.8749

Website 2 :

P(12000) = 10977.6944 / 12000 = 0.9148

4.)

More consistent website :

We use the standard deviation value, the higher the standard deviation, the higher the variability :

Website 1 should be more consistent has it has a Lower standard deviation score, hence, should show lower variability than website 2.

5.)

Website suitable for advertisement should be one with higher viewership per month in other to reach a larger audience. Hence, website 2 should be recommended for advertisement.

5 0

2 years ago

I need help on #2 if 1 meter is 1000 kilometers what would the answer be

anastassius [24]

120 times because 300 divided by 25

4 0

3 years ago

Read 2 more answers

PLEASE HELP ME GUYS OR I WONT PASS this calculus!!!!

KonstantinChe [14]

Answer:

b. $\displaystyle \frac{1}{2}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions
Function Notation
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle H(x) = \sqrt[3]{F(x)}$

Step 2: Differentiate

Rewrite function [Exponential Rule - Root Rewrite]: $\displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}$
Chain Rule: $\displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)]$
Basic Power Rule: $\displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)$
Simplify: $\displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}$
Substitute in function values: $\displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}$
Exponents: $\displaystyle H'(5) = \frac{6}{3(4)}$
Multiply: $\displaystyle H'(5) = \frac{6}{12}$
Simplify: $\displaystyle H'(5) = \frac{1}{2}$