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malfutka [58]
2 years ago
11

Is the point (-2, 5) a solution to the equation 2x + 5y = 0?

Mathematics
1 answer:
ratelena [41]2 years ago
5 0

Answer:

No

Step-by-step explanation:

substitute -2 for 'x' and 5 for 'y'

2(-2) + 5(5) = 0?

-4 + 25 = 21, not 0

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The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives 13.113.113, point, 1 years; th
skad [1K]

Answer:

The Probability = 0.16

Step-by-step explanation:

Empirical rule formula states that:

68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

Average = 13.1 years

Standard deviation = 1.5 years

x = 14.6 years

= 14.6 -13.1/1.5

= 1

This falls within 1 standard deviations of the mean

68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

Hence:

100 - 68%/2 = 32/2 = 16%

Converting to probability = 0.16

The probability of a meerkat living longer than 14.6 years is 0.16

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Bonds are a(n) _______________ instrument.
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Step-by-step explanation:

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Katya measured the growth of a plant seeding. The seedling grew 1/3 inch by the end of the first week and another 5/6 inch by th
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2 years ago
Given a population with a mean of muμequals=100100 and a variance of sigma squaredσ2equals=3636​, the central limit theorem appl
lakkis [162]

Answer:

a) \bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)

\mu_{\bar X}=100 \sigma^2_{\bar X}=1

b) P(\bar X >101)=1-P(\bar X

c) P(\bar X

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".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: P(A)+P(A') =1

Let X the random variable that represent the variable of interest on this case, and for this case we know the distribution for X is given by:  

X \sim N(\mu=100,\sigma=6)  

And let \bar X represent the sample mean, by the central limit theorem, the distribution for the sample mean is given by:  

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})  

a. What are the mean and variance of the sampling distribution for the sample​ means?

\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)

\mu_{\bar X}=100 \sigma^2_{\bar X}=1.2^2=1.44

b. What is the probability that x overbarxgreater than>101

First we can to find the z score for the value of 101. And in order to do this we need to apply the formula for the z score given by:  

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

If we apply this formula to our probability we got this:  

z=\frac{101-100}{\frac{6}{\sqrt{25}}}=0.833  

And we want to find this probability:

P(\bar X >101)=1-P(\bar X

On this last step we use the complement rule.  

c. What is the probability that x bar 98less than

First we can to find the z score for the value of 98.

z=\frac{98-100}{\frac{6}{\sqrt{25}}}=-1.67  

And we want to find this probability:

P(\bar X

5 0
3 years ago
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