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Alborosie
3 years ago
15

Pls help me with thissssss

Mathematics
2 answers:
sergejj [24]3 years ago
7 0
Divide the two numerators and get 3.25 multiply 13 hours by that and the answer to the first part is 42.25
nydimaria [60]3 years ago
4 0
Yes that’s very true
You might be interested in
Find the most common ratio 72,12,2, 1/3,1/18
liraira [26]

Answer:

1/6

Step-by-step explanation:

To find the common ratio, you compare a few pairs of consecutive terms, by dividing an element by its predecessor.

12 / 72 = 1/6

2 / 12 = 1 / 6

1/3 / 2 = 1 / 6

The ratio is constant... so that's your common ratio to go from one term to the next.  

To go from one term to the next, you have to multiply by 1/6.

7 0
2 years ago
The mean per capita income is 16,127 dollars per annum with a variance of 682,276. What is the probability that the sample mean
MakcuM [25]

Answer:

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Step-by-step explanation:

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

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, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

The standard deviation is the square root of the variance. So

\mu = 16127, \sigma = \sqrt{682276} = 826, n = 476, s = \frac{826}{\sqrt{476}} = 37.86

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Either it differs by 104 or less dollars, or it differs by more than 104 dollars. The sum of the probabilities of these events is 100. I am going to find the probability that it differs by 104 or less dollars first.

Probability that it differs by 104 or less dollars first.

pvalue of Z when X = 16127 + 104 = 16231 subtracted by the pvalue of Z when X = 16127 - 104 = 16023. So

X = 16231

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

By the Central Limit Theorem

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

Z = \frac{16231 - 16127}{37.86}

Z = 2.75

Z = 2.75 has a pvalue of 0.9970

X = 16023

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

Z = \frac{16023 - 16127}{37.86}

Z = -2.75

Z = -2.75 has a pvalue of 0.0030

0.9970 - 0.0030 = 0.9940

99.40% probability that it differs by 104 or less.

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

p + 99.40 = 100

p = 0.60

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

7 0
3 years ago
Simplify Simplify 6 - 23 + (-9 + 5) · 2.
Vlad [161]

Answer:

The answer is −42

7 0
2 years ago
What is the probability of picking an ace of spades in a deck of cards (That has 52 cards in it) write your answer as a fraction
Jet001 [13]

The answer to the problem is 4 over 52

4/52




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3 0
2 years ago
Read 2 more answers
Jackson bought 3 shirts costing $15.98, $19, and $24.50.How much money did he spend altogether?
marusya05 [52]

Answer:

Jackson spent $59.48 altogether

Step-by-step explanation:

15.98+19= 34.98

34.98+24.50= 59.48

6 0
2 years ago
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