Please solve this help

The solution to the equation x3 = 45

Darina [25.2K]

Answer:
x=15

Because 3x15=45

Hope that helped!

6 0

3 years ago

Read 2 more answers

Find the area of an equilateral triangle if a side is 8 times the square root of 3

irga5000 [103]

The area would be about 83.14.

You can find this by starting with the formula for the area of an equilateral triangle:

A = $\frac{\sqrt{3}}{4} S^{2}$

In this equation, A is the area and S is the side. So we'll just plug in the value.

A = $\frac{\sqrt{3}}{4} S^{2}$

A = $\frac{\sqrt{3}}{4} (8\sqrt{3})^{2}$

A = $\frac{\sqrt{3}}{4} 24$

A = 83.14

4 0

3 years ago

Select the example of an inequality. o 14y = 2 0 32 - 4 044 = 15 4y < 64

faltersainse [42]

The last one is the answer

4 0

3 years ago

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

A scale model of a rectangular building lot measures 7 feet by 5 feet. If the actual house will be built using a scale factor of

eimsori [14]

Answer:

The area of the actual building lot is $5,040\ ft^2$

Step-by-step explanation:

Scaling

The scaled rectangular building measures 7 feet by 5 feet. We know both magnitudes were scaled by a factor of 12, thus the real dimensions of the house to be built are 7*12=84 feet, and 5*12=60 feet.

The area of the actual building lot is calculated by:

$A=84\ ft*60\ ft=5,040\ ft^2$

The area of the actual building lot is $5,040\ ft^2$

4 0

3 years ago