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Sever21 [200]
3 years ago
14

What is 2466 divided by 3

Mathematics
2 answers:
Romashka-Z-Leto [24]3 years ago
7 0
2466/3 is equal to 822. 

NemiM [27]3 years ago
5 0
The answer is 822




Hope this helped :)
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Elena L [17]

Answer:

  • D. Points D, C and A

Step-by-step explanation:

Points D, C and A are collinear as DC, CA and DA have same slope of 1/3.

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3 years ago
Suppose F and G are continuously∗ differentiable functions defined on [a, b] such that F0(x) = G0(x) for all x ∈ [a, b]. Using t
Orlov [11]

Answer:

The proof is detailed below.

Step-by-step explanation:

We will first prove that if H(x) is a differentiable function in [a,b] such that H'(x)=0 for all x∈[a, b] then H is constant. For this, take, x,y∈[a, b] with x<y. By the Mean Value Theorem, there exists some c∈(x,y) such that H(y)-H(x)=H'(c)(x-y). But H'(c)=0, thus H(y)-H(x)=0, that is, H(x)=H(y). Then H is a constant function, as it takes the same value in any two different points x,y.

Now for this exercise, consider H(x)=F(x)-G(x). Using differentiation rules, we have that H'(x)=(F-G)(x)'=F'(x)-G'(x)=0. Applying the previous result, F-G is a constant function, that is, there exists some constant C such that (F-G)(x)=C.

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4 years ago
Student scores on exams given by a certain instructor have mean 74 and standard deviation 14. This instructor is about to give t
Lapatulllka [165]

Answer:

P(\bar X >80)=P(Z>2.143)=1-P(z

Step-by-step explanation:

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".

Let X the random variable that represent the Student scores on exams given by a certain instructor, we know that X have the following distribution:

X \sim N(\mu=74, \sigma=14)

The sampling distribution for the sample mean is given by:

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

The deduction is explained below we have this:

E(\bar X)= E(\sum_{i=1}^{n}\frac{x_i}{n})= \sum_{i=1}^n \frac{E(x_i)}{n}= \frac{n\mu}{n}=\mu

Var(\bar X)=Var(\sum_{i=1}^{n}\frac{x_i}{n})= \frac{1}{n^2}\sum_{i=1}^n Var(x_i)

Since the variance for each individual observation is Var(x_i)=\sigma^2 then:

Var(\bar X)=\frac{n \sigma^2}{n^2}=\frac{\sigma}{n}

And then for this special case:

\bar X \sim N(74,\frac{14}{\sqrt{25}}=2.8)

We are interested on this probability:

P(\bar X >80)

And we have already found the probability distribution for the sample mean on part a. So on this case we can use the z score formula given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

Applying this we have the following result:

P(\bar X >80)=P(Z>\frac{80-74}{\frac{14}{\sqrt{25}}})=P(Z>2.143)

And using the normal standard distribution, Excel or a calculator we find this:

P(Z>2.143)=1-P(z

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liraira [26]

Answer:

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Step-by-step explanation:

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I WILL GIVE BRAINLIESTTT
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Step-by-step explanation:it’s a

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