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3 years ago

What two factors of 12 that add to -7

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

3 0

12 is composite number. 12=1x12 and many more but. Try 1,2,3,4,6

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In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz

Ahat [919]

Answer:

a) There is a 18.75% probability that the first question that she gets right is the second question.

b) There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

c) There is a 10.35% probability that she gets the majority of the questions right.

Step-by-step explanation:

Each question can have two outcomes. Either it is right, or it is wrong. So, for b) and c), we use the binomial probability distribution to solve this problem.

Binomial probability distribution

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

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

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

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

And $\pi$ is the probability of X happening.

In this problem we have that:

Each question has 4 choices. So for each question, Robin has a $\frac{1}{4} = 0.25$ probability of getting ir right. So $\pi = 0.25$ . There are five questions, so $n = 5$ .

(a) What is the probability that the first question she gets right is the second question?

There is a 75% probability of getting the first question wrong and there is a 25% probability of getting the second question right. These probabilities are independent.

$P = 0.75(0.25) = 0.1875$

There is a 18.75% probability that the first question that she gets right is the second question.

(b) What is the probability that she gets exactly 1 or exactly 2 questions right?

This is: $P = P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955$

$P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637$

$P = P(X = 1) + P(X = 2) = 0.3955 + 0.2637 = 0.6592$

There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

(c) What is the probability that she gets the majority of the questions right?

That is the probability that she gets 3, 4 or 5 questions right.

$P = P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 3) = C_{5,3}.(0.25)^{3}.(0.75)^{2} = 0.0879$

$P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146$

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

$P = P(X = 3) + P(X = 4) + P(X = 5) = 0.0879 + 0.0146 + 0.001 = 0.1035$

There is a 10.35% probability that she gets the majority of the questions right.

6 0

3 years ago

Helppp me plsssssssss<br><br>

Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

$\sf \:\:\:\:\:\:\:\:\:\:x_{k}=35$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k}=50$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42$
$\sf \:\:\:\:\:\:\:\:\:\:h=5$

${\bf \:\: {By\:using\:the\: formula}} \\ \\$

$\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\$

$\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\$

${\large{\frak{\pmb{\underline{Additional\: information }}}}}$

MODE

Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

In a frequency distribution the class having maximum frequency is called the modal class.

${\bf{\underline{Formula\:for\: calculating\:mode:}}} \\$

${\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\$

Where,

$\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.$

$\small \blue{ \bigstar}$ $\sf \: f_{k}=frequency\:of\:the\:modal\:class$

$\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class$

$\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class$

$\small \purple{ \bigstar}$ $\sf \: h= width \:of\:the\:class\:interval$

7 0

3 years ago

At 3:00 AM, the temperature was 44 degrees. By 7:00 AM, the temperature had risen to 60 degrees. What was the average change in

White raven [17]

60-44=16
16/4hrs= 4deg per hour

4 0

2 years ago

Two-fifths of a number is -12

notsponge [240]

X=-39 answer

2/5x=-12
2x=-12(5)
2x=-60
X=-30

6 0

3 years ago

Write an equivalent fraction of

matrenka [14]

Answer:

<h2>here's the answer</h2>

Step-by-step explanation:

$to \: do \: so \: we \: have \: to \: multiply \: \frac{9}{6} \: by \: \frac{8}{5} \: so$

$\frac{9}{6} \times \frac{8}{5}$

$\frac{72}{30}$

6 0

2 years ago