(10x2 - 6x + 8)(2x + 2)

A triangle has 2 sides that are 9 inches and the bottom is 6 inches what is the height?

Elanso [62]

The height of the isosceles triangle is 8.49 inches.

<h3>How to find the height of the triangle?</h3>

Here we have a triangle such that two of the sides measure 9 inches, and the base measures 6 inches.

So this is an isosceles triangle.

We can divide the isosceles triangle into two smaller right triangles, such that the side that measures 9 inches is the hypotenuse, the base is 3 inches, and the height of the isosceles triangle is the other cathetus.

By Pythagorean's theorem, we can write:

(9in)^2 = (3 in)^2 + h^2

Where h is the height that we are trying to find.

Solving that for h we get:

h = √( (9 in)^2 - (3in)^2) = 8.49 inches.

We conclude that the height of the isosceles triangle is 8.49 inches.

If you want to learn more about triangles:

brainly.com/question/2217700

#SPJ1

8 0

2 years ago

Find m∠Y if m∠Y is six more than three times its supplement.

Svetlanka [38]

To answer this item, we let x be the measure of angle Y. If we say supplement, it is understood that when we sum up the measures of the two angles (angle Y and its supplement) the answer should be 180. Therefore the other angle is 180-x. Going back to the condition given in this item, we will be able to generate an equation,
x = 3(180 - x)
The value of x from the equation is 135. Thus, m∠Y is 135°.

3 0

2 years ago

Read 2 more answers

PLS HELP ME I NEED ANSWER FAST

vaieri [72.5K]

Answer:

it is a.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

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:

$p_v = P(Z$

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.

6 0

3 years ago

A study of 35 golfers showed that their average score on a particular course was 92. The sample standard deviation was 5. We wan

svlad2 [7]

Answer:

$t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322$

Step-by-step explanation:

We have a sample of size n = 35, $\bar{x} = 92$ and $s = 5$ . As we want to construct a 95% confidence interval to estimate the value of the true population mean $\mu$ , we should use the pivotal quantity given by $T = \frac{\bar{X}-\mu}{S/\sqrt{n}}$ which comes from the t distribution with n - 1 = 35 - 1 = 34 degrees of freedom. Besides we should use the t-value $t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322$ , i.e., regarding the t-distribution with 34 degrees of freedom, we have an area of 0.025 above 2.0322 and below the probability density function. The rationale behind this is that $P(-2.0322\leq T\leq 2.0322) = 0.95$ because of the simmetry of the t distribution.

7 0

2 years ago