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2 years ago

ΔEFG ~ ΔLMN. Find LM.

Mathematics

2 answers:

bezimeni [28]2 years ago

7 0

The solution is 2.5. The problem can be solved using the proportion:

Vika [28.1K]2 years ago

6 0

EFG ~ LMN, if EF ~ LM, then whatever the number is for EF times whatever the others are multiplied by, you will get LM
for example:

FG (10) ~ MN (20)

if EF is (5), then LM is (10) (for you are multiplying by two.

Another example:

if FG is (100) ~ MN will be (200) (again for you are multiplying by two.

hope this helps

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Kyle challenges you to a game. Each player rolls a standard number cube twice. If the sum of the results is divisible by 3, you

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Answer:

The probability of you winning is 1/3

No, it is not a fair game

Step-by-step explanation:

The first thing we need to have here is the sample space. This refers to the set of all possible results that can occur from the rolling.

Please check attachment for this

Kindly note that the total number of possible outcomes is 36.

And in the attachment, sums which are divisible by 3 are circled.

The number of circles we can count is 12

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2 years ago

A wheat farmer cuts down the stalks of wheat and gathers them in 200 piles. The 200 gathered piles will be put on a truck. The t

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Answer:

Check the explanation

Step-by-step explanation:

We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.

We know, the sample mean is an unbiased estimator of the population mean. Therefore,

$\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5$

where \mu is the mean weight of grain for all the 200 piles.

Hence, the total grain weight of the population is

$\widehat{Y}=Y_1+Y_2+...+Y_{200}$

= $200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs$

2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

S= $\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486$

Then, the standard error of $\widehat{Y} is$

$\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:$ =83.785

Hence, a 95% bound on the error of estimates is

$[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]$

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

$\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45$

The sample ratio is

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Ket [755]

The equation used for this is

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2 years ago