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3 years ago

Is this function linear, quadratic, or exponential?

Mathematics

1 answer:

svetoff [14.1K]3 years ago

7 0

Answer:

Step-by-step explanation:

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Barrett works at an ice cream shop. The function f(x) represents the amount of money in dollars Barrett earns per gallon of ice

Paha777 [63]

Answer:

$f(g(x))=6x^3} +4$

Step-by-step explanation:

Given:

$f(x)=2x^2+4\\\\g(x)=\sqrt{3x^3}$

Required:

f(g(x))=?

Solution:

let f(g(x))=f(X), where X=g(x)

so $f(X)=2X^2+4$

put X= $g(x)=\sqrt{3x^3}$ , we get

$f(g(x))=2(\sqrt{3x^3} )^2+4$

$f(g(x))=2(\sqrt{3x^3} )^2+4\\\\f(g(x))=2({3x^3} )+4\\\\f(g(x))=6x^3} +4$

5 0

3 years ago

A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard devi

Luden [163]

Answer:

Null hypothesis: $\mu \leq 3.3$

Alternative hypothesis: $\mu > 3.3$

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$

Since is a one right tailed test the p value would be:

$p_v =P(z>1.768)=0.039$

Step-by-step explanation:

Data given and notation

$\bar X=3.4$ represent the sample mean

$\sigma=0.4$ represent the population deviation for the sample

$n=50$ sample size

$\mu_o =3.3$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:

Null hypothesis: $\mu \leq 3.3$

Alternative hypothesis: $\mu > 3.3$

Compute the test statistic

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$

P value

Since is a one right tailed test the p value would be:

$p_v =P(z>1.768)=0.039$

6 0

3 years ago

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

$P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)$

Calculation the value from standard normal z table, we have,

$P(x < 5) =0.7287= 72.87\%$

b) P( port handles 3 or more million tons of cargo per week)

$P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)$

Calculating the value from the standard normal table we have,

$1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%$

c)P( port handles between 3 million and 4 million tons of cargo per week)

$P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%$