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1 year ago

Write the expression for the Area of the polygon BORT MO 6 A= O A = 6m + mp O A = mp + 1/2(6m) O A = mp + 6p

1 answer:

7 0

We are asked to determine the area of the given figure. To do that we will divide the figure into a square and a triangle. The area of the square is the product of its dimension, therefore, the area is:

$A_{sq}=mp$

The area of the triangle is given by the following formula:

$A_t=\frac{bh}{2}$

Where "b" is the base and "h" is its height. Replacing the values:

$A_t=\frac{(6)(m)}{2}=\frac{1}{2}6m$

The total area is the sum of both areas:

$A=A_{sq}+A_t$

replacing the areas:

$A=mp+\frac{1}{2}6m$

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IVE BEEN ASKING FOR HOURS!!!!!! PLEASE HELP ME ANYONE!!!!! Juliet coaches softball and soccer. she coaches both sports for 15 ho

IrinaK [193]

Answer:

x + y = 15

y = 3x + 1.5

x + (3x + 1.5) = 15

Step-by-step explanation:

Let :

Soft ball practice = x

Soccer practice = y

y = 3x + 1.5

Total practice time :

x + y = 15 - - - (1)

x + 3x + 1.5 = 15

4x + 1.5 = 15

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2 years ago

A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio

yaroslaw [1]

Answer:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

$p_v = P(Z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ we see that $p_v$ 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 chips that fail in the first 1000 hours of their use is not significantly less than 0.48.

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

$\hat p=0.45$ estimated proportion of chips that fail in the first 1000 hours of their use

$\mu_0 =0.48$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 si less then 0.48:

Null hypothesis: $p\geq 0.48$

Alternative hypothesis: $p < 0.48$

When we conduct a proportion test we need to use the z statistic, 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 is significantly different from a hypothesized value .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

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

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

$p_v = P(Z$

6 0

3 years ago

Which applies the power of a product rule to simplify (5 t) cubed?

Elena L [17]

Answer:

(5 t ) cubed = 5 cubed . t cubed = 125 t cubed applies the power of a product rule to simplify (5 t) cubed ⇒ 3rd answer

Step-by-step explanation:

Let us revise some rules of exponents

$a^{m}$ × $a^{n}$ = $a^{m+n}$