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Jet001 [13]
2 years ago
12

What measures do you need to find the area of a trapezoid?

Mathematics
1 answer:
Fantom [35]2 years ago
8 0
Perimeter and length
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Is selecting the winning ticket in a lottery a binomial or hypergeometric distribution?
Paraphin [41]

Answer:

hypergeometric

Step-by-step explanation:

Meeting ID: 849 7736 5835

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7 0
2 years ago
Find the slope of the graph.
professor190 [17]

Answer:

Slope= m = Δy/Δx = (y₂-y₁)/(x₂-x₁) = rise / run  

Step-by-step explanation:

Hope that helps .  Just plug in your values if you are given two points

for example: (3,4) = (x₂,y₂)    (1,2) = (x₁,y₁)  

Just substitute in the equation above

(4-2)/(3-1) = 2/2 =1  

good luck

6 0
2 years ago
Of 575 broiler chickens purchased from various kinds of food stores in different regions of a country and tested for types of ba
chubhunter [2.5K]

Answer:

a) 0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

b) We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

c) For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p represent the real population proportion of interest  

\hat p =0.68 represent the estimated proportion

n=575 is the sample size required (variable of interest)  

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

Part a

The confidence interval would be given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58  

And replacing into the confidence interval formula we got:  

0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

Part b

We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

Part c

For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

4 0
3 years ago
(3+5)x7-[10+(45-15)\5]
Gelneren [198K]
Use\ PEMDAS\\\\(3+5)\times7-[10+\frac{(45-15)}{5}]=(8)\times7-(10+\frac{30}{5})\\\\=56-(10+6)=56-16=\boxed{\boxed{40}}
7 0
2 years ago
Rory earned an 84% on his test. He answered 21 questions correctly. How many total questions were on the
Anna35 [415]

Answer:

21 is 84% to 25, so there would be 25 total questions on the test

4 0
2 years ago
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