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klio [65]
2 years ago
7

PLS HELP marking brainliest 20-27 show ur work

Mathematics
1 answer:
Ivanshal [37]2 years ago
8 0

Answer:

-7?

Step-by-step explanation:

I mean, is the question 20 - 27 =? Or it's something else? Lol.

20 - 27 = -7, it's very straightforward cannot write it in more detailed way. When we substract a number from another and the first one is smaller the result is a negative number. Imagine someone sells up to 3 apples, you ask him to give you 5. He gives you all 3 of his so he has 0 but you force him to go and get another 2 for you. He lost another 2, so he has less than 0 now. He has -2 apples: 3 - 5 = -2.

Now, I don't know if this is the question or these are set of exercises you must solve and your picture just didn't upload. Let me know if you need help with anything else.

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The least common multiple of 18 and 42 is 126. Sorry I don't know how to explain it.
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Answer:

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Step-by-step explanation:

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2 years ago
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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:

a)

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

b)

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

c)

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

d)

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

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Answer:

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Step-by-step explanation:

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