In how many different ways can 5 books be arranged on a shelf?

Another way to write (6+7)x

Blizzard [7]

Answer:

13x

Step-by-step explanation:

you just have to add the six and the seven together and then add the x

3 0

3 years ago

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Jimmy is trying to factor the quadratic equation $ax^2 + bx + c = 0.$ He assumes that it will factor in the form \[ax^2 + bx + c

Shkiper50 [21]

If

$ax^2+bx+c=(Ax+B)(Cx+D)$

then

$ax^2+bx+c=ACx^2+(AD+BC)x+BD$

$\implies a=AC$

If $a$ is known and Jimmy wants to find $A$ , then he has to know the value of $C$ .

6 0

3 years ago

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A coin is to be tossed as many times as necessary to turn up one head. Thus the elements c of the sample space C are H, TH, TTH,

slavikrds [6]

Answer:

Step-by-step explanation:

As stated in the question, the probability to toss a coin and turn up heads in the first try is $\frac{1}{2}$ , in the second is $\frac{1}{4}$ , in the third is $\frac{1}{8}$ and so on. Then, P(C) is given by the next sum:

$P(C)=\sum^{\infty}_{n=1}(\frac{1}{2} )^{n}=1$

This is a geometric series with factor $\frac{1}{2}$ . Then is convergent to $\frac{1}{1-\frac{1}{2}}-1=1.$ . With this we have proved that P(C)=1.

Now, observe that

$P(H)=\frac{1}{2}, P(TH)=\frac{1}{4},P(TTH)=\frac{1}{8},P(TTTH)=\frac{1}{16},P(TTTTH)=\frac{1}{32},P(TTTTTH)=\frac{1}{64}.$

Then

$P(C1)=P(H)+P(TH)+P(TTH)+P(TTTH)+P(TTTTH)=\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} =\frac{31}{32}$

$P(C2)=P(TTTTH)+P(TTTTTH)=\frac{1}{32}+\frac{1}{64} =\frac{3}{64}$

$P(C1\cap C2)=P(TTTTH)=\frac{1}{32}$

and

$P(C1\cup C2)=P(H)+P(TH)+P(TTH)+P(TTTH)+P(TTTTH)+P(TTTTTH)=\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{32} +\frac{1}{64}=\frac{63}{64}$

5 0

3 years ago

y = x^{4} - x^{3} - 28x^{2} - 20x + 48 How many possible negative real zeros, positive real zeros, and non-real zeros does this

OleMash [197]

$f(x)= x^{4} - x^{3} - 28x^{2} - 20x + 48$

There are two changes of sign, so there are 2 or 0 possible positive roots.

$f(-x)= x^{4} +x^{3} - 28x^{2}+ 20x + 48$

There are two changes of sign, so there are 2 or 0 possible negative roots.

There are 4,2 or 0 possible non-real roots.

3 0

3 years ago

In a certain Algebra 2 class of 29 students, 7 of them play basketball and 24 of them play baseball. There are 3 students who pl

antiseptic1488 [7]

Answer:

$\frac{5}{29}$

Step-by-step explanation:

Let $n(A)$ represent students playing basketball, $n(B)$ represent students playing baseball.

Then, $n(A)=7$ , $n(B)=24$

Let $n(S)$ be the total number of students. So, $n(S)=29$ .

Now,

$P(A)=\frac{n(A)}{n(S)}=\frac{7}{29}$

$P(B)=\frac{n(B)}{n(S)}=\frac{24}{29}$

3 students play neither of the sport. So, students playing either of the two sports is given as:

$n(A\cup B)=n(S)-3\\n(A\cup B)=29-3=26$

∴ $P(A\cup B)=\frac{n(A\cup B)}{n(S)}=\frac{26}{29}$

From the probability addition theorem,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Where, $P(A\cap B)$ is the probability that a student chosen randomly from the class plays both basketball and baseball.

Plug in all the values and solve for $P(A\cap B)$ . This gives,

$\frac{26}{29}=\frac{7}{29}+\frac{24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{7+24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{31}{29}+P(A\cap B)\\\\P(A\cap B=\frac{31}{29}-\frac{26}{29}\\\\P(A\cap B=\frac{31-26}{29}=\frac{5}{29}$

Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is $\frac{5}{29}$

6 0

3 years ago