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2 years ago

PLSSS HELP IF YOU TURLY KNOW THISS

Mathematics

2 answers:

aleksandrvk [35]2 years ago

3 0

Answer:

2 inches

Step-by-step explanation:

Lyrx [107]2 years ago

3 0

Answer:

2 inches

Step-by-step explanation:

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Most dermatologists recommend using sunscreens that have a Sun Protection Factor (SPF) of at least 30. One of the authors wanted

Basile [38]

Answer: We are 95% confident that the mean SPF level of sunscreen used by all Cal Poly students is between 29.0813 and 40.0187.

Step-by-step explanation:

Given : Significance level : $\alpha: 1-0.95=0.05$

Sample size : n= 52 , which is a large sample (n>30), so we use z-test.

By using z-value table,

Critical value: $z_{\alpha/2}=1.96$

Sample mean : $\overline{x}= 34.55$

Standard deviation : $\sigma= 20.12$

The confidence interval for population means is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e. $34.55\pm(1.96)\dfrac{ 20.12}{\sqrt{52}}$

$=34.55\pm5.46867829455\\\\\approx34.55\pm5.4687\\\\=(34.55-5.4687, 34.55+5.4687)=(29.0813,\ 40.0187)$

Now, the 95% confidence interval for the ppopulation mean = . (29.0813, 40.0187)

Hence, We are 95% confident that the mean SPF level of sunscreen used by all Cal Poly students is between 29.0813 and 40.0187.

7 0

3 years ago

Please answer quickly I only have 5 mins lol. The ratio of students polled in 6th grade who prefer lemonade to Iced tea is 8:4,

zhannawk [14.2K]

Answer:

Let total students who prefer lemonade is 2a and who prefer ice tea is a.

Students in 6th grade=39

2a+a=39

3a=39

a=39/3

a=13

students who prefer lemonate is 13

and those who prefer ice tea is 2a=26

4 0

2 years ago

a city bus completes its entire route 3 times each hour how long will it take the bus to complete the entire route 10 times

SOVA2 [1]

Answer: It will take 3.33 hours to complete it's entire route 10 times.

Step-by-step explanation: Time taken to complete it's entire route 3 times= 1 hour

∴Time taken to complete it's entire route 10times = 10/3

= 3.33 hours

3 0

3 years ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [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 given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$