What is one sixth of 72?

A hardcover book has a mass of about 1 kilogram. What could have a mass of less than 1 kilogram? Select all that apply.

insens350 [35]

A key, pencil, and toothbrush

4 0

3 years ago

SOLVE Janet piano teacher charged $12.50 for a 30-minute lesson if Janet has a 4 lesson per month how much will she owe her pian

miskamm [114]

You would multiply 12.50 by 4 because thats how much each lesson is and you go 4 times a month. You then multiply this by 12 because there are 12 months in a year. The answer comes out to $600 a year. Hope this helps. XD

3 0

4 years ago

Use the Statistical Applet: P ‑Value for a Test of One Proportion to answer the question.You have taken a sample of ????=1000 in

umka21 [38]

Answer:

$z=\frac{0.785 -0.7}{\sqrt{\frac{0.7(1-0.7)}{750}}}=5.08$

$p_v =P(z>5.08)=1.88x10^{-7}$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ 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 proportion of interest is highr than 0.67 or 67% .

Step-by-step explanation:

1) Data given and notation

n=20 represent the random sample taken

X=15 represent the people who eat breakfast

$\hat p=\frac{15}{20}=0.75$ estimated proportion of people who eat breakfast

$p_o=0.67$ 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 higher than 67%:

Null hypothesis: $p\leq 0.67$

Alternative hypothesis: $p > 0.67$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.785 -0.7}{\sqrt{\frac{0.7(1-0.7)}{750}}}=5.08$

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 next step would be calculate the p value for this test.

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

$p_v =P(z>5.08)=1.88x10^{-7}$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ 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 proportion of interest is highr than 0.67 or 67% .

6 0

3 years ago

A population of squirrels on an island has a carrying capacity of 350 individuals. If the maximum rate of increase is 1.0 per in

TiliK225 [7]

Answer:

The population growth rate is 59 squirrels per year.

Step-by-step explanation:

Given:

A population of squirrels on an island has a carrying capacity of 350 individuals. If the maximum rate of increase is 1.0 per individual per year and the population size is 275.

Now, to find the population growth rate.

Carrying capacity ( $K$ ) = 350.

Maximum rate ( $r_{max}$ ) = 1.0 per individual per year.

Population size ( $N$ ) = 275.

Now, to get the population growth rate we put formula:

$\frac{dN}{dT}= r_{max}\frac{K-N}{K}N$

$\frac{dN}{dT}=1.0\times \frac{350-275}{350}\times 275\\\\\frac{dN}{dT}=1.0\times \frac{75}{350}\times 275\\\\\frac{dN}{dT}=1.0\times 0.214\times 275\\\\\frac{dN}{dT}=58.85.$

So, rounding to nearest whole number of the population growth rate is 59 squirrels per year.

Therefore, the population growth rate is 59 squirrels per year.

6 0

3 years ago

Read the excerpt from Chapter 3 of Wheels of Change . In 1881, as women in the United States continued to struggle with hoops an

elena-s [515]

Step-by-step explanation:

Chapter 3 of Wheels of Change .

But the bicycle craze had a more lasting impact on women’s clothing than just the use of bloomers. Thanks in large part to cycling, the innovation of rational dress reformers were starting to take effect by the end of the 1890s. Corsets were on their way out, dresses were getting shorter, and women no longer wore the heavy, bulky undergarments that gave them round, unnatural shapes. These changes went a long way toward unburdening women and setting the stage for them to be healthier and more active in the coming century.

Based on the excerpt, women in the 1890s became more independent and their lives improved as a result of

a rise in the popularity of bicycles.

a rise in the popularity of corsets.

cycling becoming less socially acceptable.

clothing becoming more restrictive.

7 0

2 years ago