Please help due tomorrow

Find the perimeter of the kite

Vilka [71]

the perimeter of the kite:

20+20+33+33=106mm

4 0

2 years ago

Read 2 more answers

3. In the figure below, E is on AD and C is on BD. The dimensions given are in meters. What is the length, in meters, of CD?

damaskus [11]

Answer:

C. 3,960

Step-by-step explanation:

∆ADB ~ ∆EDC. Therefore,

$\frac{AB}{EC} = \frac{DB}{DC}$

$\frac{2,000}{1,200} = \frac{2,640 + x}{x}$ (substitution)

Solve for x

$\frac{5}{3} = \frac{2,640 + x}{x}$

Cross multiply

$5*x = 3(2,640 + x)$

$5x = 7,920 + 3x$

5x - 3x = 7,920

2x = 7,920

x = 7,920/2

x = 3,960 meters

7 0

2 years ago

Tom is throwing darts at a target. In his last 30 throws, Tom has hit the target 20 times. You want to know the estimated probab

Musya8 [376]

Each scenario can be used to simulate probability, and there are 3 correct scenarios and 2 incorrect scenarios in the list of options

<h3>How to categorize the simulations?</h3>

From the question, we have the following parameters:

Number of throws = 30
Number of hits = 20

This means that the probability of hit is:

P(Hit) = 20/30

Simplify

P(Hit) = 2/3

Using the complement rule,

P(Miss) = 1/3

The above means that the simulation that represents the situation must have the following parameters:

P(Success) = 2/3
P(Failure) = 1/3
Number of experiments = 3

Using the above highlights, the correct scenarios are:

Rolling a die three times with numbers 1 to 4 representing a hit
Spinner a spinner of 3 equal sections three times with two sections representing hit
Spinner a spinner of 6 equal sections three times with four sections representing hit

Read more about probability at:

brainly.com/question/25870256

#SPJ1

6 0

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:

$p_v = P(Z$

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

Please choose an answer below.<br> A) 3.4 <br> B) 36.0 <br> C) 38.6 <br> D) 41.2

schepotkina [342]

Answer:

It would be D.

Step-by-step explanation:

Hope this helps!!!

6 0

3 years ago