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Luba_88 [7]
2 years ago
10

Find the other endpoint of the line segment with the given endpoint and midpoint.

Mathematics
1 answer:
BARSIC [14]2 years ago
6 0

Answer:

Step-by-step explanation:

16, -11

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HELP? :))))))))))))​
erik [133]

Answer:

the answer is B

Step-by-step explanation

Cant really explain but next time put the topic for the subject so people will understand.

7 0
2 years ago
What is z^(3)+5z+3z^(2)+1 - 4 - 2z^(2) simplified
garik1379 [7]
I hope this helps you



z^3+5z+3z^2-4-2z^2



z^3+z^2+5z-4
3 0
3 years ago
1.8 + 45 tenths = ___
Ket [755]

Answer: 6.3

Step-by-step explanation:

1.8 + 45 tenths equal 6.3

8 0
3 years ago
Determine the percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain wei
yawa3891 [41]

Answer:

The result indicates that the percentage of all samples of three men that have mean brain weights within (1.24 * sampling error) of the mean is 78.50%.

Step-by-step explanation:

Note: This question is not complete. The complete question is therefore provided before answering the question as follows:

According to one study, brain weights of men are normally distributed with mean = 1.20 kg and a standard deviation = 0.14 kg.

Determine the percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.20 kg. Interpret your answer in terms of sampling error.

The explanation of the answers is now provided as follows:

Based on the Central limit theorem, it possible to say that the mean of sampling distribution (μₓ) is approximately equal to the population mean (μ) as follows:

μₓ = μ = 1.20 kg …………………………. (1)

Also, the standard deviation of the sampling distribution can be written as follows:

σₓ = (σ/√N) ……………………….. (2)

Where:

σ = population standard deviation = 0.14 kg

N = Sample size = 3

Substituting the values into equation (2), we have:

σₓ = 0.14 / √3 = 0.0808

Since we are to determine the percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.20 kg, this implies that we have:

P(1.10 ≤ x ≤ 1.30)

Therefore, 1.10 and 1.30 have to be first normalized or standardized as follows:

For 1.10 kg

z = (x - μₓ) / σₓ = (1.10 - 1.20) / 0.0808 = -1.24

For 1.30 kg

z = (x - μₓ)/σₓ = (1.30 - 1.20) / 0.0808 = 1.24

The required probability can be determined when P(1.10 ≤ x ≤ 1.30) = P(-1.24 ≤ z ≤ 1.24).

From the normal distribution table, the following can be obtained for these probabilities:

P(1.10 ≤ x ≤ 1.30) = P(-1.24 ≤ z ≤ 1.24) = P(z ≤ 1.24) - P(z ≤ -1.24) = 0.89251 - 0.10749 = 0.7850, or 78.50%

Therefore, the sampling error is equal to 0.0808 which is the standard deviation of the sampling distribution.

In terms of the sampling error, the result indicates that the percentage of all samples of three men that have mean brain weights within (1.24 * sampling error) of the mean is 78.50%.

3 0
3 years ago
Round 132,554 to the nearest tenth
777dan777 [17]
If you ment "round to the nearest tens" your answer would be 132,550.
8 0
3 years ago
Read 2 more answers
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