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alina1380 [7]
3 years ago
13

Ayuda es un examen determinado raíz cuadrada entera de cada número y el residuo en cada caso º60 º86 º96 Raíz:_residuo:_ raíz:__

residuo__ raíz__residuo
Mathematics
1 answer:
konstantin123 [22]3 years ago
4 0

Answer:

Los pasos para calcular la raíz cuadrada entera y su residuo o resta se dará en la explicación;

Step-by-step explanation:

-60: Para saber exactamente cuál es su raíz cuadrada entera debemos buscar entre que cuadrados perfectos se encuentra comprendida este valor;

Tenemos que 7x7=49  y 8x8 = 64 (60 está comprendido entre  7 y 8) ,pero como la raíz de 49 se aproxima más a 60 entonces la raíz de 60 es 7 y su residuo o resta será el valor que falta para llegar a 60; así que tenemos:

\sqrt{60}=7  ; residuo:  60-49 =11

-86:    9x9 =81 ;81 se aproxima más a 86 ,entonces la raíz cuadrada entera de 86 es 9 y su residuo es 5:

\sqrt{86}=9;  residuo : 86-81= 5

-96:   9x9=81 ;  81 por tanteo se aproxima más a 91,entonces la raíz cuadrada  entera de 96 es 9 y su residuo es 15.

\sqrt{96= 9x9 =81 ; residuo :  96-81= 15.

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D

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On average, Stan drinks 3/4 of a 8 ounce glass of water in 1/2 of an hour.
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<h3>Stan drinks 12 ounces of water in 1 hour</h3>

<em><u>Solution:</u></em>

Given that,

Stan drinks 3/4 of a 8 ounce glass of water in 1/2 of an hour

Which means,

\frac{3}{4}\ of\ 8\ ounce = \frac{1}{2}\ hour

\frac{3}{4} \times 8 = 0.5\ hour\\

Therefore,

6 ounces is drank in 0.5 hour

We have to find the water he drinks in an hour

Let "x" be the water he drinks in 1 hour

Thus we get,

6 ounce = 0.5 hour

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This forms a proportion and we can solve the sum by cross multiplying

6 \times 1 = 0.5 \times x\\\\0.5x = 6\\\\x = \frac{6}{0.5}\\\\x = 12

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3 years ago
Parts being manufactured at a plant are supposed to weigh 65 grams. Suppose the distribution of weights has a Normal distributio
andrew-mc [135]

Answer:

0.64%.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 75 grams and a standard deviation of 22 grams.

This means that \mu = 75, \sigma = 22

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This means that n = 144, s = \frac{22}{\sqrt{144}} = 1.8333

More than 80 or less than 70:

Both are the same distance from the mean, so we find one probability and multiply by 2.

The probability that it is less than 70 is the pvalue of Z when X = 70. So

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