Find the x value to 3x+9=6 and 3x+9=-6

Write the expression as a single fraction:<br> please help, giving brainliest :)

nignag [31]

Answer:

$\frac{4a^{2} + 15b}{10ab}$

Step-by-step explanation:

find the least common denominator for both fractions then combine them to find:

$\frac{4a^{2} + 15b}{10ab}$

4 0

3 years ago

The median of the list of numbers is 16, 20, 36, 18, 22, 19, 22

Monica [59]

It's 20. Since it is median you will have to find the middle number. So in the sequence in order: 16, 18, 19, 20, 22, 22, 36. Next you will just find which number is in the middle. Also, tip: When doing this, if you have two medians such as 22 and 23. You will go to the number in between those, which would be 22.5.

5 0

3 years ago

2 triangles are shown. The second triangle is rotated 90 degrees to the right. A point is between both triangles.

Semmy [17]

Answer:

Its center of rotation

Step-by-step explanation:

got it on edge.

4 0

3 years ago

Read 2 more answers

Add the following lengths: 4' 9" + 7" = <br>

lina2011 [118]

Answer:

Your answer to this question is 16

8 0

3 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