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2 years ago

Use the following function rule to find h(w–5). Simplify your answer. h(y)=y²–y–5 h(w–5)=

Mathematics

1 answer:

Dima020 [189]2 years ago

8 0

The value of the function is h(w - 5) = w^2 - 11w +25

<h3>How to determine the function?</h3>

The function is given as:

h(y) = y^2 - y - 5

Substitute w - 5 for y

h(w - 5) = (w - 5)^2 - (w - 5) - 5

Expand the equation

h(w - 5) = w^2 - 10w + 25 - w + 5 - 5

Evaluate the like terms

h(w - 5) = w^2 - 11w +25

Hence, the value of the function is h(w - 5) = w^2 - 11w +25

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2 years ago

The price of tomatoes went from $1.12 per pound to $1.96 per pound in 3 years. find the rate of change of the price of tomatoes.

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3 years ago

Show all work and reasoning

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Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

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$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

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3 years ago

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3 years ago

A group of 54 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic

Ket [755]

Answer:

$t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601$

P-value

First we find the degrees of freedom given by:

$df = n-1= 54-1=53$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{53}>0.9601)=0.3414$

Step-by-step explanation:

Data given and notation

$\bar X=5.25$ represent the sample mean

$s=3.98$ represent the sample standard deviation

$n=54$ sample size

$\mu_o =4.73$ represent the value that we want to test

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

z 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.

We need to conduct a hypothesis in order to check if the mean differs from 4.73, the system of hypothesis would be:

Null hypothesis: $\mu =4.73$

Alternative hypothesis: $\mu \neq 4.73$

Since we know don't the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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".

Calculate the statistic

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

$t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601$

P-value

First we find the degrees of freedom given by:

$df = n-1= 54-1=53$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{53}>0.9601)=0.3414$

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3 years ago