(Geometry Question) For Tangent A, would it be 1.05 or 0.9524?

When Aria commutes to work, the amount of time it takes her to arrive is normally distributed with a mean of 28 minutes and a st

MrRa [10]

Answer:

Her commute would be between 32 and 35 minutes 33 times.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 28, \sigma = 4.5$

Proportion of days in which the commute is between 32 and 35 minutes:

This is the pvalue of Z when X = 35 subtracted by the pvalue of Z when X = 32.

X = 35

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{35 - 28}{4.5}$

$Z = 1.555$

$Z = 1.555$ has a pvalue of 0.94.

X = 32

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{32 - 28}{4.5}$

$Z = 0.89$

$Z = 0.89$ has a pvalue of 0.8133.

0.94 - 0.8133 = 0.1267

Out of 262 days:

Each day, 0.1267 probability

0.1267*262 = 33

Her commute would be between 32 and 35 minutes 33 times.

6 0

4 years ago

How does the graph of g(x) = 5x – 3 compare to the graph of f(x) = 5*?

Karo-lina-s [1.5K]

Answer:

the graph is shifted down 3 units

Step-by-step explanation:

5x-3

meaning the original graph of 5x going down three units

3 0

3 years ago

Convert 2 liters to gallons.

Iteru [2.4K]

2 liters to gallons is 0.53 Gallons.

5 0

3 years ago

Find the particular solution of the differential equation? /=5^3+9^2, when =1, =8

kipiarov [429]

Answer:

$\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Solving Differentials - Integrals

Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Step-by-step explanation:

*Note:

Ignore the Integration Constant C on the left hand side of the differential equation when integrating.

Step 1: Define

$\displaystyle \frac{ds}{dt} = 5t^3 + \frac{9}{t^2}$

t = 1

s = 8

Step 2: Integrate

[Derivative] Rewrite [Leibniz's Notation]: $\displaystyle ds = (5t^3 + \frac{9}{t^2})dt$
[Equality Property] Integrate both sides: $\displaystyle \int {} \, ds = \int {(5t^3 + \frac{9}{t^2})} \, dt$
[Left Integral] Reverse Power Rule: $\displaystyle s = \int {(5t^3 + \frac{9}{t^2})} \, dt$
[Right Integral] Rewrite [Integration Property - Addition]: $\displaystyle s = \int {5t^3} \, dt + \int {\frac{9}{t^2}} \, dt$
[Right Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle s = 5\int {t^3} \, dt + 9\int {\frac{1}{t^2}} \, dt$
[Right Integrals] Rewrite [Exponential Rule - Rewrite]: $\displaystyle s = 5\int {t^3} \, dt + 9\int {t^{-2}} \, dt$
[Right Integrals] Reverse Power Rule: $\displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{t^{-1}}{-1}) + C$
[Right Integrals] Rewrite [Exponential Rule - Rewrite]: $\displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{1}{t}) + C$
Multiply: $\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} + C$

Step 3: Solve

Substitute in variables: $\displaystyle 8 = \frac{5(1)^4}{4} + \frac{9}{1} + C$
Evaluate exponents: $\displaystyle 8 = \frac{5}{4} + \frac{9}{1} + C$
Divide: $\displaystyle 8 = \frac{5}{4} + 9 + C$
Add: $\displaystyle 8 = \frac{41}{4} + C$
[Subtraction Property of Equality] Isolate C: $\displaystyle \frac{-9}{4} = C$
Rewrite: $\displaystyle C = \frac{-9}{4}$

Particular Solution: $\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}$

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

Unit: Differentials Equations and Slope Fields

Book: College Calculus 10e

3 0

3 years ago

16,900,000 in scientific notation

stich3 [128]

16,900,000 in scientific notation is 16.9 times 10^7.

When you move the decimal point to the 1, it goes 7 spaces to the left, so the exponent will be a positive 7.

6 0

3 years ago