A circle is defined by the equation given below.

How many solutions exist for the system of equations in the graph?

katovenus [111]

Answer:

Two solutions exist for the system of equations in the graph.

Step-by-step explanation:

Solutions are when the two graphs intersect.

6 0

2 years ago

The FDA regulates that fresh Albacore tuna fish that is consumed is allowed to contain 0.82 ppm of mercury or less. A laboratory

yanalaym [24]

Answer:

A sample size of 183 is needed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.97}{2} = 0.015$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.015 = 0.985$ , so Z = 2.17.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Margin of error within 0.023 ppm of mercury with 97% confidence. What sample size is needed?

We have to find n for which $M = 0.023$ . We have that $\sigma = 0.143$ . So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.023 = 2.17\frac{0.143}{\sqrt{n}}$

$0.023\sqrt{n} = 2.17*0.143$

$\sqrt{n} = \frac{2.17*0.143}{0.023}$

$(\sqrt{n})^2 = (\frac{2.17*0.143}{0.023})^2$

$n = 182.02$

We have to round up(182 is not quite in the desired margin of error), so a sample size of 183 is needed.

7 0

3 years ago

Write the standard form 12 ten thousands 8 thousands 14 hundreds 7 ones

andre [41]

128,147.............................................

3 0

3 years ago

Find x<br><br> plz help!!!!!!!!

kompoz [17]

<span>'x' can be anything. There's no information here to say that it must be
one thing or another. You can't find 'x', because there is no equation.

You can probably simplify the fraction by doing some factoring, but there's
no way to tell a value for 'x'. As long as 'x' is not -4, it can be anything, and
its value determines the value of the fraction.</span>

7 0

4 years ago

Read 2 more answers

The germination rate is the rate at which plants begin to grow after the seed is planted.

chubhunter [2.5K]

Answer:

$z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59$

$p_v =P(z$

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 proportion of germinated seeds is significantly lower than 0.9 or 90%

Step-by-step explanation:

Data given and notation

n=15 represent the random sample taken

X=7 represent the number of seeds germinated

$\hat p=\frac{7}{15}=0.467$ estimated proportion of seeds germinated

$p_o=0.9$ is the value that we want to test

$\alpha$ represent the significance level

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 the true proportion of germinated seeds is less than 0.9 or 90%.:

Null hypothesis: $p\geq 0.9$

Alternative hypothesis: $p < 0.9$

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

$z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59$

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

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

$p_v =P(z$

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 proportion of germinated seeds is significantly lower than 0.9 or 90%

4 0

3 years ago