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3 years ago

What is the asymptote of the graph shown? y=

Mathematics

1 answer:

hichkok12 [17]3 years ago

3 0

Answer:

y=0

Step-by-step explanation:

you see the graph stop at the y-axis

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Write an equation that could be used to find how many miles a hybrid suv can travel in the city on 20 gallons

vagabundo [1.1K]

Answer:

m=36×20

Step-by-step explanation:

A hybrid SUV uses 36 miles per gallon in the city.

The equation that could be used to find how many miles a hybrid SUV can travel in the city on 20 gallons of gas is m=36×20.

8 0

3 years ago

(GIVING BRAINLIEST AND 15 POINTS FOR ANSWER) Solve five and three sixths minus two and one third.

a_sh-v [17]

Answer:

three and three eighteenths

5 3/6 = 33/6

2 1/3 = 7/3 = 14/6

33/6 - 14/6 = 19/6 = 3 1/6 or 3 3/18

5 0

2 years ago

The measure of angle FGJ is 25°.

ivanzaharov [21]

Answer:

50+25=75+x=75x

Step-by-step explanation:

75x nw czy dobrzee

4 0

2 years ago

Find the least common denominator of<br> 7<br> and<br> X+4 x-4

Effectus [21]

(X+4)(X-4) is the LCD because you just have to look at the denominators and see what factors or values are not already part of your LCD.

5 0

2 years ago

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