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3 years ago

What is an equation for the circle with center (-1,11) and radius 5

Mathematics

2 answers:

Makovka662 [10]3 years ago

4 0

$r^{2} = (x-h)^{2} + (y-k)^{2}$

so...

$25 = (x+1)^{2} + (y-11)^{2}$

qaws [65]3 years ago

3 0

This is ur required result in attachment

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There were 78 girls trying out for dance. They were placed into groups containing d dancers. How many groups were there, in term

Lera25 [3.4K]

Step-by-step explanation:

The hole never disappears; at the end of the story, the first things that went into the hole come out of the sky and back into the city. There's no sign that the hole is gone

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3 years ago

The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.

salantis [7]

a) $X$

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

In this problem, the score on the exam is normally distributed with the following parameters:

$\mu=78.1$ (mean)

$\sigma = 10.8$ (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

$X$

And the probability for this to occur can be written as:

$p(X$

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between $z=-\infty$ and $z=Z$ , where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

$Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02$

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

$p(X$

And by looking at the z-score tables, we find that this probability is:

$p(z$

And so,

$p(X$

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

$p(X>89.1)$

First of all, we have to calculate the z-score corresponding to this value of X, which is:

$Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02$

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

$p(z>1.02) =p(z$

Because the normal distribution is symmetric.

But from part b) we know that

$p(z$

Therefore:

$p(X>89.1)=p(z>1.02)=0.1539$

Here we want to find the probability that the randomly chosen score is between 67.1 and 89.1, which can be written as

$p(67.1$

Or also as

$p(67.1$

Since the overall probability under the whole distribution must be 1.

From part b) and c) we know that:

$p(X$

$p(X>89.1)=0.1539$

Therefore, here we find immediately than:

$p(67.1$

7 0

3 years ago

A recent national survey indicated that 73 percent of respondents try to include locally grown foods in their diets. A 95 percen

lilavasa [31]

Answer:

Correct option is:

"Less than 75% of all people in the country try to include locally grown foods in their diets."

Step-by-step explanation:

A hypothesis for the proportion of people who include locally grown foods in their diets can be defined as:

H₀: The proportion of people who include locally grown foods in their diets is p.

Hₐ: The proportion of people who include locally grown foods in their diets is difference from p.

The decision rule can be based on the 95% confidence interval.

If the confidence interval consists the hypothetical value of the true proportion then the null hypothesis will be accepted or else the null hypothesis will be rejected.

The 95% confidence interval for the proportion of all people in the country who try to include locally grown foods in their diets is given as (0.70, 0.76).

This interval implies that there is 0.95 probability that the true proportion of people who include locally grown foods in their diets is between 70% and 76%.

Assuming that the claim made was supported by the 95% confidence interval.

The claim should be:

Less than 75% of all people in the country try to include locally grown foods in their diets.

Because the 95% confidence interval defines the range of the true proportion as 70% to 76%. Most of the values are below 75%.

Thus, the correct option is:

"Less than 75% of all people in the country try to include locally grown foods in their diets."

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3 years ago

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Need help please with this question

SashulF [63]

Answer:

5/8

Step-by-step explanation:

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3 years ago

How is the counting principle mathematically used?

Paha777 [63]

Answer: multiplying outcomes

Step-by-step explanation:

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3 years ago