All categories

3 years ago

PLS HELP!

Mathematics

1 answer:

Elden [556K]3 years ago

5 0

Answer:

f^-1(x) = 4x-3

Step-by-step explanation:

f(x) = (x+3)/4

y = (x+3)/4

Exchange x and y

x = (y+3)/4

Solve for y

4x = y+3

Subtract 3

4x-3 = y

The inverse

f^-1(x) = 4x-3

You might be interested in

An online furniture store sells chairs for $200 each and tables for $600 each. Every day, the store can ship a maximum of 32 pie

wolverine [178]

Answer:

If the minimum 13 chair were sold , then the minimum number of Table sold is 14

Step-by-step explanation:

Given as :

The cost of each chair = $ 200

The cost of each table = $ 600

The total number of furniture sold per day = 32

The minimum amount of selling per day = $ 12000

Let the total number of chair = C

The total number of table = T

So , according to question

C + T = 32 .......1

200 C + 600 T = 12000 .......2

Solving eq 1 and 2

200 C + 600 T = 12000

200 × ( C + T ) = 32 × 200

I.e 200 C + 200 T = 6400

or, ( 200 C + 600 T ) - ( 200 C + 200 T ) = 12,000 - 6400

Or, 400 T = 5600

∴ T = $\frac{5600}{400}$

I.e T = 14

Put The value of T in eq 2

So, 200 C + 600 × 14 = 12000

or , 200 C + 8400 = 12,000

Or, 200 C = 12000 - 8400

Or, 200 C = 3600

∴ C = $\frac{3600}{200}$

I.e C = 18

The number of chair sold is 18

If the number of chair sold is 13 ,

Then the min number of table sold = 200 × 18 + 600 T = 12000

i.e 600 T = 12000 - 3600

or, 600 T = 8400

∴ T = $\frac{8400}{600}$

I.e T = 14

Hence if the minimum 13 chair were sold , then the minimum number of Table sold is 14 . Answer

5 0

3 years ago

How do I do this ? I have 120 degrees but I need to solve for x

Darina [25.2K]

Since the legs are the same length, we know that the remaining two angles must be the same.

120+2x=180
2x=60
x=30

Final answer: x=30 degrees

3 0

3 years ago

How many solutions does the system of equations have?

lora16 [44]

As 27 will always be equal to 27, you can place whatever you want for x and y and the equation would still be satisfied, as it turns out that it doesn't depend on the x or y value.
Therefore, the system has infinite solutions.

Be safe!

7 0

4 years ago

Read 2 more answers

OMG PLS HELP WITH THIS IM PANICKING OMG I GOT A F IN MATH MY MOM JUST YELLED AT ME IM CRYING PLS HELP WITH THS-

scoundrel [369]

Sorry about that, here's the answers.

1. expanded form

2. regroup

3. place value

4. A, C, D. The strategy I used was place value and regrouping. I chose these by using place value to help me understand how to regroup and what each option would sum to.

5. D. Ten thousands place

5 0

3 years ago

Read 2 more answers

Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$