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3 years ago

9

Can someone help me

2 answers:

jenyasd209 [6]3 years ago

7 0

Answer:

y=-1/4x+2

Step-by-step explanation:

sukhopar [10]3 years ago

5 0

Answer:

Step-by-step explanation:

add all the # and see what you get

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Milo cut his cake into 2 equal pieces. He ate 2pieces of cake for his birthday.

Sauron [17]

Answer:

$1$ $or \frac{2}{2}$

Step-by-step explanation:

<h2><u>We know that Milo cut his cake into 2 different pieces, so:</u></h2>

$\frac{x}{2}$

<h2><u>He then eats two pieces of the cake. so:</u></h2>

$2$ $pieces$ $of$ $cake$ $=$ $x$

$\frac{2}{2} = 1$

8 0

3 years ago

2/3 divided by 5 I Need answers

Aleonysh [2.5K]

Answer:

Step-by-step explanation:

7.5

4 0

3 years ago

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Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0

3 years ago

What are the zeros in the function

Schach [20]

$f(x)=x^3+x^2-6x\\\\text{The zeros:}\\\\f(x)=0\to x^3+x^2-6x=0\\\\x(x^2+x-6)=0\\\\x(x^2+3x-2x-6)=0\\\\x[x(x+3)-2(x+3)]=0\\\\x(x+3)(x-2)=0\iff x=0\ \vee\ x+3=0\ \vee\ x-2=0\\\\x=0\ \vee\ x=-3\ \vee\ x=2$

Answer: -3; 0 and 2

6 0

4 years ago

Let A= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Suppose an element of A is randomly selected. What is the probability that it belongs to

gregori [183]

<h3>Answer: 3/5</h3>

============================================================

Explanation:

The set {1, 2}U{2, 6, 7, 9}U{5, 6, 7} simplifies to {1,2,5,6,7,9} after we combine everything and sort the values. We toss any duplicates.

Let B = {1,2,5,6,7,9}

There are 6 items in set B out of 10 items in set A.

The probability of landing in set B, if you pick something randomly from A, is 6/10 = 3/5

8 0

3 years ago

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