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Cloud [144]
3 years ago
14

How much do subtract from 37/6 to make 6

Mathematics
1 answer:
choli [55]3 years ago
4 0
37÷6
=6.1666666667
- 0.1666666667
=6
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I need help finding this answer to this inequality -10[9-2x]-x≤2x-5
Ad libitum [116K]

Answer:

x≤5

Step-by-step explanation:

-10(9-2x)-x≤2x-5

-90+20x-x≤2x-5

19x-2x≤90-5

17x≤85

x≤85/17

x≤5

8 0
3 years ago
Un jardinier a récolté deux quintaux trois quarts de pommes . Il vend un quintal un quart à un voisin ; il vend aussi 7/10 de qu
Flauer [41]

Answer:

Step-by-step explanation:

Notez que: un quintal =

une unité de poids égale à 100 kg

Un jardinier a récolté deux quintaux et trois quarts de pommes.

Le nombre total de pommes récoltées = 200 kg + 3/4 du quintal de pommes

= 200 kg + 0,381 kg

= 200,381 kg

Il vend un quintal un quart à un voisin

Le montant qu'il a vendu à un voisin = 100 + 1/4 quintal de pommes

100 + 0,127

= 100,127 kg

Il vend également 7/10 d'un quintal sur le marché du village

= 7/10 × 100 kg

= 70 kg de pommes restantes

2/5 d'un quintal à un pâtissier.

2/5 × 100 kg

= 40 kg de pommes restantes

À l'aide d'une fraction, exprimez la quantité de pomme que le jardinier a gardée pour lui.

La quantité que le jardinier lui a réservée =

200 3/4 kg - (100 1/4 kg + 70 kg + 40 kg)

200,381 kg - (210,127 kg)

5 0
3 years ago
What is the solution to the inequality below x^2>81
dem82 [27]

Answer:

x>9

Step-by-step explanation:

4 0
4 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

To solve this question, we need 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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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4 years ago
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3 years ago
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