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3 years ago

Does the function ƒ(x) = 5e0.25x represent exponential growth, decay, or neither?

Mathematics

1 answer:

N76 [4]3 years ago

6 0

Step-by-step explanation:

it is exponential growth because the coefficient of x is not negative

the evidence is in the image above

the gradient increases

You might be interested in

What is the area of the figure:

Leni [432]

A 45 45 90 triangle
so a = b
a^2 + b^2 = c^2
a^2 + a^2 = c^2
2a^2 = c^2
2a^2 = 24^2
 2a^2 = 576
a^2 = 288
a = 12√2
b = a = 12√2

Area of triangle:
= 1/2 (12√2)(12√2)
= 1/2(288)
= 144

answer
Area = 144 ft^2

8 0

3 years ago

Assume that T is a linear transformation. Find the standard matrix of T. T: R^3 right arrow R^2 , T(e 1) =(1,2), and T(e2 ) =( -

irina [24]

Answer:

$A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]$

Step-by-step explanation:

Given

$T:R^3->R^2$

$T(e_1) = (1,2)$

$T(e_2) = (-4,6)$

$T(e_3) = (2,-6)$

Required

Find the standard matrix

The standard matrix (A) is given by

$Ax = T(x)$

Where

$T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]$

$Ax = T(x)$ becomes

$Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]$

The x on both sides cancel out; and, we're left with:

$A = [T(e_1)\ T(e_2)\ T(e_3)]$

Recall that:

$T(e_1) = (1,2)$

$T(e_2) = (-4,6)$

$T(e_3) = (2,-6)$

In matrix:

$(a,b)$ is represented as: $\left[\begin{array}{c}a\\b\end{array}\right]$

So:

$T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]$

$T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]$

$T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]$

Substitute the above expressions in $A = [T(e_1)\ T(e_2)\ T(e_3)]$

$A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]$

Hence, the standard of the matrix A is:

$A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]$

6 0

3 years ago

You and your family are going on a

Elina [12.6K]

The total cost, including the deposit, would be $2,750. $500*5=$2,500 $2,500+$250=$2,750

4 0

2 years ago

SunDrop Candies Inc claims that their 3 ounce bag of candies contains over 90 pieces of candy.

nlexa [21]

Answer:

a) X~N(μ;σ²)

b) [87.695 ; 89.464]

c) and d) check the explanation below

Step-by-step explanation:

Hello!

The objective of this experiment is to test if the claim of the company is correct and the 3-ounce bags of candies contain over 90 pieces of candy.

For this, a sample of 26 bags was taken and the number of candies per bag was counted.

The study variable is X: number of candy pieces in a 3 ounces bag.

Using the sample data I've made a Shapiro-Wilks test,

H₀: X has a normal distribution

H₁: X doesn't have a normal distribution

α: 0.05

p-value: 0.5675

Since the p-value is greater than the significance level, we can assume that the study variable has normal distribution:

X~N(μ;σ²)

To make this confidence interval, considering I only have sample data, I choose to use the Student t statistic.

Sample mean X[bar]= 88.58

Sample standard deviation S= 2.19

$t_{n-1;1-\alpha/2} = t_{25;0.975} = 2.060$

The formula for the interval is:

X[bar] ± $t_{n-1;1-\alpha/2}$ * (S/√n)

88.58 ± 2.060 * (2.19/√26)

[87.695 ; 89.464]

With a confidence level of 95%, you'd expect that the interval [87.69; 89.46] contains the value of the population mean of the number of candy pieces in a 3 ounces bag.

The SunDrop Candies' claim is that their 3 ounces bag contains over 90 pieces of candy. Symbolically: μ > 90

Since their claim would lead to a one-tailed hypothesis test and the calculated confidence interval is two-tailed, you cannot use it to decide whether or not the company's claim is true. But, seeing as the two bonds of the interval are below 90 you could think that the company's claim is not correct. Of course this is just an idea, you must perform a statistical test to test if the company is wrong or not and, of course, have a statistical backing of your conclusions.

I hope it helps!

6 0

3 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$

Length of curve= $s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})$

Length of curve= $s=32.66$

5 0

3 years ago