Ask question

Ask question

All categories

3 years ago

14

If f(x)=x+3 and g(x)= x-1 , what is (f g)(26)

1 answer:

lesantik [10]3 years ago

8 0

Answer:

28

Step-by-step explanation:

fog(x) =f{x-1}

=x-1+3

X+2

fog(x)=x+2

fog(26)=26+2

=28

You might be interested in

Solve the equation: -2/5 n = -30<br> A. n=-75<br> B. n=-12<br> C. n=12<br> D. n=75

OLEGan [10]

Assuming your equation is: -2/5n = -30

-2/5n = -30 <- We get rid of the "/5n" by multiplying each side by 5n
-2 = -30 * 5n
-2 = -150n
-1 = -75n
75n = 1

But this doesn't match up with any of your answers, so if you wouldn't mind giving me the equation again, i'll work it out

6 0

3 years ago

Read 2 more answers

A boat maker wanted to build a canoe 6 ft long and 2 1/2 wide but decided that those dimensions were too small. The boat maker w

Sati [7]

The ratio of length to width would change slightly, before adding on to the dimensions, the ratio is 6 : 2.5 with the added dimensions, the ratio would change to 8 : 4.5 with simple math you can see that 2.5 is less then half of 6 while 4.5 is more then half of 8

6 0

3 years ago

Read 2 more answers

Give me ans please......……………………

jeka57 [31]

hope it helps

it says I can't send if it has less than 20 characters why?

7 0

3 years ago

A television and DVD player cost a total of 1132. The cost of the television is three times the cost of the DVD player. What is

lubasha [3.4K]

Let x rep cost of dvd in $
so the cost of tv is 3x
and x + 3x = 1132
4x=1132
x=1132/4
x= 283 and 3x=3(283)=849

7 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