What is x/6-3=15 find for x

Akimi4 [234]

Answer:

Cells are the basic building blocks of all living things. The human body is composed of trillions of cells. They provide structure for the body, take in nutrients from food, convert those nutrients into energy, and carry out specialized functions.

Step-by-step explanation:

I hope that helps you out.

6 0

3 years ago

Read 2 more answers

29. New grass seeds grow rapidly. A grass seed has grown to 0.3 millimeters tall. Tomorrow it will be 0.9 millimeters

Cerrena [4.2K]

Answer:

The answer is a(n)=12+(n-1)11; 166 millimeter

3 0

3 years ago

Under his cell phone plan, Ian pays a flat cost of $59.50 per month and $5 per gigabyte. He wants to keep his bill under $80 per

inessss [21]

You would start by saying his budget is $80 so you start with 80= he pays an initial fee of $59.50 every month plus $5 per gigabyte so you would set it up like
80=59.50+5x
X being how many gigabytes he uses then you solve 80-59.50 is 20.5 divided by 5 gives you 4.1 but you round down cause you don’t want to pass the limit so therefor the most gigabytes he can use is 4

3 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

Has anybody gotten a 100% on their FLVS Math segment exam?

Flura [38]

Answer:

i have

Step-by-step explanation:

3 0

2 years ago