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lisabon 2012 [21]
3 years ago
15

A. -4m - 12 - 5m = 13​

Mathematics
1 answer:
vladimir2022 [97]3 years ago
6 0

Answer:

Hope this helps!

Step-by-step explanation:

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1. The degree of 5x³+4x²+7x degree of the quadratic Polynomial is
poizon [28]

Answer:

3

Step-by-step explanation:

Degree of the polynomial is the highest degree of the polynomial. Here highest degree is 3.

3 0
3 years ago
HELP ME ANSWER #5 PLEASEEE!!!!!!
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ABF AND BFC. PLS MARK ME BRAINLIEST. THANK YOUUUU
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3 years ago
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The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability
amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Also, important to remember that the standard deviation is the square root of the variance.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

Z = \frac{X - \mu}{\sigma}

By the Central limit theorem

Z = \frac{X - \mu}{s}

Z = \frac{70.5 - 69}{1.97565}

Z = 0.76

Z = 0.76 has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

8 0
3 years ago
of the vehicles on the road 1/2 are cars and 1/8 are motorcycles what fraction of the vehicles are not cars or motorcycles?
viktelen [127]
1/2 of 8 is 4 so there would be 4/8 motorcycles + 1/8 cars would be 5/8 and the remainder is 3/8 so 3/8 is not cars or motorcycles
5 0
3 years ago
You are in a traffic jam that is 7/8 mile long. The cars are in one lane bumper to bumper. The average length of a car is 0.003
KengaRu [80]

Answer:

Step-by-step explanation:

The average length of each car is

0.003 mile. The length of each car includes the bumper. The cars are in one lane. This means all the cars can be assumed to form a long line of cars that is uniform in length. This length is divided into equal lengths if cars.

The length of the traffic jam which is also the length of the line of is 7/8 = 0.875 mile long.

The number of cars would be in the that would be in the traffic jam will be length of the traffic jam/ length of each car. It becomes

0.875/0.003 = 292 cars

7 0
4 years ago
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