What is the answer to 16.3-2.666

Which example models a negative integer A.Students in a classroom B. The age of a parent C. Money lost from a pocket D. the heig

Rufina [12.5K]

Answer:

C- Money lost from a pocket

Step-by-step explanation:

It would be C because something is decreasing ( now a negative) from what they did have.

4 0

2 years ago

Anthony goes to the gym for __30____ minutes on Monday. Every day he ___increases____his gym time by _____5 minutes_______. If h

Stolb23 [73]

Geometric sequence is a sequence of numbers that have a common ratio between them.

6 0

2 years ago

Round 0.051866 to 2 significant figure

kobusy [5.1K]

Answer:

your answer would be 0.052

5 0

2 years ago

The can of corn shown is a right circular cylinder with a height of 11 cm. The volume of the can is 486 cubic centimeters. What

hammer [34]

SOLUTION

Since the can is a cylinder, we will use the formula for volume of a cylinder to solve the problem.

Volume of a cylinder is given as

$\begin{gathered} V=\pi r^2h \\ where\text{ r = radius = ?} \\ h=height=11cm \\ V=volume=486cm^3 \end{gathered}$

Applying we have

$\begin{gathered} V=\pi r^2h \\ 486=3.14\times r^2\times11 \\ 486=34.54r^2 \\ r^2=\frac{486}{34.54} \\ r=\sqrt{14.07064} \\ r=3.75108 \end{gathered}$

Hence the answer is 3.8, the second option is the answer

6 0

1 year ago

According to Inc, 79% of job seekers used social media in their job search in 2018. Many believe this number is inflated by the

Fynjy0 [20]

Answer:

Null hypothesis: $p\leq 0.79$

Alternative hypothesis: $p > 0.79$

$z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786$

$p_v =P(z>2.786)=0.00267$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.

Step-by-step explanation:

1) Data given and notation

n=750 represent the random sample taken

X=314 represent the respondents that use social media in their job search.

$\hat p=\frac{314}{370}=0.849$ estimated proportion of respondents that use social media in their job search

$p_o=0.79$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is hgiher than 0.79.:

Null hypothesis: $p\leq 0.79$

Alternative hypothesis: $p > 0.79$

When we conduct a proportion test we need to use the z statisticc, 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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786$

4) 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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>2.786)=0.00267$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.

7 0

3 years ago