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3 years ago

A 110 point final exam is

Mathematics

2 answers:

statuscvo [17]3 years ago

4 0

Answer:

There are 45 2-point questions and 5 4-point questions

Step-by-step explanation:

NeX [460]3 years ago

4 0

Answer: The two point questions are 45, while the four point questions are 5.

Step-by-step explanation: We shall start by assigning letters to the unknown variables. Let the first type of question be called a and the second type of question be called b.

If the final exam is made up of 50 questions, that means

a + b = 50

Also if some questions are two points while others are four points and the total score for all of them is 110 points, then

2a + 4b = 110

(That is, we assume that a is the two point question and b is the four point question, so total score equals 110)

We now have a pair of simultaneous equations

a + b = 50 ————(1)

2a + 4b = 110 ——-(2)

From equation (1), we shall make a the subject of the equation,

hence let a = 50 - b

Substitute for the value of a in equation (2)

2a + 4b = 110

2(50 - b) + 4b = 110

100 - 2b + 4b = 110

By collecting like terms we now have

4b - 2b = 110 - 100

2b = 10

Divide both sides of the equation by 2

b = 5

Now substitute for the value of b into equation (1)

a + b = 50

a + 5 = 50

Subtract 5 from both sides of the equation

a = 45.

Therefore the two point questions are 45, while the four point questions are 5

**A quick check would show the following results ; 2(45) + 4(5) = 90 + 20 = 110.**

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2 years ago

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Advocard [28]

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2 years ago

A practice law exam has 100 questions, each with 5 possible choices. A student took the exam and received 13 out of 100.If the s

Cloud [144]

Answer:

$z=\frac{13-20}{4}=-1.75$

Assuming:

H0: $\mu \geq 20$

H1: $\mu$

$p_v = P(Z$

Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest (number of correct answers in the test), on this case we now that:

$X \sim Binom(n=100, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We need to check the conditions in order to use the normal approximation.

$np=100*0.2=20 \geq 10$

$n(1-p)=20*(1-0.2)=16 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np=100*0.2=20$

$\sigma=\sqrt{np(1-p)}=\sqrt{100*0.2(1-0.2)}=4$

So we can approximate the random variable X like this:

$X\sim N(\mu =20, \sigma=4)$

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 letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z$