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djyliett [7]
3 years ago
10

Alec is doing a test of 50 questions. He gets 2 marks for every question that is correct, but loses 1 mark for every question th

at is wrong. Alec answers every question and scores 67 marks. How many questions did he answer correctly?
Mathematics
1 answer:
slega [8]3 years ago
4 0

Alec got 33 and a half questions correctly because he got one mark on the 34th Question

Step-by-step explanation:

50×2=100

100-67=33 and a half

You might be interested in
A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January
Blababa [14]

Answer:

a) 27.89% probability that two have ever boycotted goods for ethical reasons

b) 41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) 41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) The expected number is 2.3 and the standard deviation is 1.33.

Step-by-step explanation:

We use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

23% of the respondents have boycotted goods for ethical reasons.

This means that p = 0.23

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

This is P(X = 2) when n = 6. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{6,2}.(0.23)^{2}.(0.77)^{4} = 0.2789

27.89% probability that two have ever boycotted goods for ethical reasons

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Either less than two have, or at least two. The sum of the probabilities of these events is decimal 1. So

P(X < 2) + P(X \geq 2) = 1

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084

P(X = 1) = C_{6,1}.(0.23)^{1}.(0.77)^{5} = 0.3735

P(X < 2) = P(X = 0) + P(X = 1) = 0.2084 + 0.3735 = 0.5819

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5819 = 0.4181

41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

Now n = 10.

P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343

P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225

P(X = 5) = C_{10,5}.(0.23)^{5}.(0.77)^{5} = 0.0439

P(X = 6) = C_{10,6}.(0.23)^{6}.(0.77)^{4} = 0.0109

P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2343 + 0.1225 + 0.0439 + 0.0109 = 0.4116

41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

E(X) = np = 10*0.23 = 2.3

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.23*0.77} = 1.33

The expected number is 2.3 and the standard deviation is 1.33.

5 0
3 years ago
What is 120÷37 please show the work and answer
Firlakuza [10]

120/37

=3.2

or =3.0

Hope this helps!

Thanks!

-Charlie

3 0
2 years ago
Find the similarity ratio of the larger triangle to the smaller one.
Ad libitum [116K]

Answer:

A

Step-by-step explanation:

The base of the larger triangle is <em>twice as large</em> as the smaller one: <em>6 v 3 --> 6 : 3 --></em> 2 : 1.

8 0
2 years ago
Pls answer<br> Subtract -37 from -53
scZoUnD [109]

Answer:

-37 subtract -53

-53 subtract -37 = -16

Step-by-step explanation:

4 0
3 years ago
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What would the range be of this function? I found the domain
Licemer1 [7]
The range would be1\leq  y
6 0
3 years ago
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