Find the values of x and y

Which recursive formula can be used to determine the total amount of money earned for each successive hour worked based on the a

sveticcg [70]

Complete question is;

An electrician earns $110 after his first hour of working for a client. His total pay based on the number of hours worked can be represented using the sequence shown.

110, 130, 150, 170, ...

Which recursive formula can be used to determine the total amount of money earned for each successive hour worked based on the amount of money currently earned?

Answer:

The recursive formula can be expressed as; f(x + 1) = f(x) + 20

Step-by-step explanation:

Electrician earns $110 dollars after working for 1 hour.

We were told his total pay based on number of hours worked can be represented by: 110, 130, 150, 170...

This means that writing it down as a function, we can say;

f(1) = $110, f(2) = $130 e.t.c

Now,since we want to express a recursive formula to explain the question, then let's say after the 1 hour worked, he earned $110, then the hour after that, his total is $130,then the hour after that, his total is $150.

This means that he earns an additional $20 each hour.

Thus;

f(x + 1) = f(x) + 20

Where x is number of hours and f(x) is payment after x number of hours

The recursive formula can be expressed as; f(x + 1) = f(x) + 20

5 0

3 years ago

What is the value of t in the equation 3(2t + 5) = 5t + 25?

11Alexandr11 [23.1K]

8 0

3 years ago

A loss of 500 in stock market is worse than a gain of $200

wariber [46]

700 stock market is worse than gain

3 0

3 years ago

Hi pleaseuor to meeting you all and have a good day

Rainbow [258]

Answer:

And you too.

thank you.

Am glad for using this app

4 0

2 years ago

Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute in

Alborosie

Answer:

a) Null hypothesis: $p\geq 0.53$

Alternative hypothesis: $p < 0.53$

b) $z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429$

$p_v =P(Z$

c) So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

$\hat p=0.46$ estimated proportion of American families owning stocks or stock funds

$p_o=0.53$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

Part a

We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:

Null hypothesis: $p\geq 0.53$

Alternative hypothesis: $p < 0.53$

Part b

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(Z$

Part c

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .

7 0

3 years ago