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3 years ago

Slove for x. 2x+34=64

Mathematics

2 answers:

Romashka [77]3 years ago

6 0

Answer:

$\boxed{x = 15}$

Step-by-step explanation:

$= > 2x + 34 = 64 \\ \\ = > 2x = 64 - 34 \\ \\ = > 2x = 30 \\ \\ = > x = \frac{30}{2} \\ \\ = > x = 15$

blagie [28]3 years ago

5 0

Step-by-step explanation:

2x +34 =64

2x=30

X=30/2

X=15

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The length of an instant message conversation is normally distributed with a mean of 5 minutes and a standard deviation of .7 mi

Charra [1.4K]

The <em><u>correct answer</u></em> is:

0.0764

Explanation:

We will use a z-score to answer this. The formula for a z-score is

$z=\frac{X-\mu}{\sigma}$ , where μ is the mean and σ is the standard deviation. Using our information in this problem, we have

$z=\frac{6-5}{0.7}=\frac{1}{0.7}=1.43$

Using a z-table, we look up the z-score 1.43. The area under the curve to the left of, or less than, this value is 0.9236. This means the probability that the time is greater than this is 1-0.9236 = 0.0764.

7 0

3 years ago

Read 2 more answers

Since the area under the normal curve within two standard deviations of the mean is 0.95, the area under the normal curve that c

alukav5142 [94]

Answer:

$X \sim N (\mu ,\sigma)$

And for this case we know this condition:

$P(\mu-2\sigma$

By the complement rule we know that:

$P(X< \mu -2\sigma \cup X>\mu +2\sigma) = 1-0.95=0.05$

But since the distribution is symmetrical we know that:

$P(X\mu +2\sigma) = 0.025$

So then the statement for this case is FALSE.

b. False

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

For this case if we define the random variable of interest X and we know that this random variable follows a normal distribution:

$X \sim N (\mu ,\sigma)$

And for this case we know this condition:

$P(\mu-2\sigma$

By the complement rule we know that:

$P(X< \mu -2\sigma \cup X>\mu +2\sigma) = 1-0.95=0.05$

But since the distribution is symmetrical we know that:

$P(X\mu +2\sigma) = 0.025$

So then the statement for this case is FALSE.

b. False

5 0