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Katena32 [7]
3 years ago
7

Solve the equation: |2x + 5| = 11.

Mathematics
2 answers:
MrRa [10]3 years ago
7 0

Answer:

x=3 or x= -8

Step-by-step explanation:

VladimirAG [237]3 years ago
5 0

Answer:

x = -8 and x = 3

Step-by-step explanation:

Split the equation into 2 possible cases:

2x + 5 = 11

2x + 5 = -11

Then, solve these 2 equations.

2x + 5 = 11

2x = 6

x = 3

2x + 5 = -11

2x = -16

x = -8

So, the 2 solutions are x = -8 and x = 3

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Reptile [31]

Answer:

Step-by-step explanation:

Equations are commonly used in scienctific discoveries to calculate power, work, distance, and time. It is also common to use the circumference and area equations to help with construction and design. Overall, math is a very important part of our lives, and it is also important to learn about it.

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The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit
Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

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

By the Central Limit Theorem

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

Z = \frac{160 - 159}{1.68}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257

X = 150

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

Z = \frac{158 - 159}{1.68}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

7 0
3 years ago
Which is the simplified form of p⁰? <br>O p <br>O 1/p <br>O 0 <br>O 1​
tatuchka [14]

Answer:

It is p = 1

Step-by-step explanation:

p⁰=1 because it is basically saying and when simplifying fractional exponents we can flip the bottom exponent up to the top if we change the sign, leaving us with : (p⁵-p⁵)=p⁰

6 0
3 years ago
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