PLEASE I REALLY NEED HELP IT'S DUE TODAY AND I CAN'T DO THIS PLEASE HELP ME!!!

Charlie, a science teacher, worked 10 months out of the year. What percent of the year did he work?

Lina20 [59]

Answer:

10/12x100%= 83.3%

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

three brothers and a sister shared a sum of money equally among themselves. if the brothers alone had shared the money, then the

Virty [35]

Answer:

1. t=240

2.t/3=(t/4)+20

Step-by-step explanation:

4 0

3 years ago

A photograph negative measures 1 7/8 inches by 2 1/2 inches. The printed pictures is to have its longer dimension be 4 inches. H

Tresset [83]

4/1 / 5/2 = 4/1 * 2/5 = 8/5

1 7/8 = 15/8

15/8 * 8/5 = 120/40 = 12/4 = 3 inches

shorter side should be 3 inches

6 0

4 years ago

. He rowed 3 1/2 miles in 1/2 hour.What is the average speed in miles per hour.

Rudiy27

Answer:

7 miles per hour

Step-by-step explanation:

3 1/2 miles = 7/2 miles

1/2 hours= 30 minutes

So,

In 30 mins, travelled 7/2 miles

In 1 min, travelled 7/2 ÷ 30= 7/2× 1/30 = (7×1)/(2×30) = 7/60 miles

In 60 mins, travelled (7×60)/60 miles = 420/60 miles = 7 miles.

Hence, the speed is 7 miles per hour.

Hope this helps :)

7 0

3 years ago

Clarinex is a drug used to treat asthma. In clinical tests of this drug, 1655 patients were treated with 5- mg doses of Clarinex

bagirrra123 [75]

Answer:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

$p_v =P(z>3.363)=0.00039$

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 interest is significantly higher than 0.012 (1.2%)

Step-by-step explanation:

Data given and notation

n=1655 represent the random sample taken

$\hat p=0.021$ estimated proportion of interest

$p_o=0.012$ 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

We need to conduct a hypothesis in order to test the claim that true proportions is higher than 0.012.:

Null hypothesis: $p \leq 0.012$

Alternative hypothesis: $p > 0.012$

When we conduct a proportion test we need to use the z statisitic, 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:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

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 right tailed test the p value would be:

$p_v =P(z>3.363)=0.00039$

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 interest is significantly higher than 0.012 (1.2%)

3 0

3 years ago