Set up the equation, relating the new ratio of salt

I need help on this I will give brainlist help it will mean a lot this is my last math problem help me!

Natalka [10]

Answer:

the second one

Step-by-step explanation:

i did it in my head and double checked :)

5 0

3 years ago

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Write whether the scenario represents a linear function or an exponential function? Explain why.

Ad libitum [116K]

The scenario represents a linear function. The rate is at a constant increase therefore it is linear.

Linear because it’s a constant rate

Since it’s doubled, and doesn’t go at a constant rate, it is a exponential function

Exponential since it increases by a multiplicative rate. It’s not constant

8 0

2 years ago

When Ryan is serving at a restaurant, there is a 0.75 probability that each party will order drinks with their meal. During one

Triss [41]

Answer:

0.82 = 82% probability that at least one party will not order drinks

Step-by-step explanation:

For each party, there are only two possible outcomes. Either they will order drinks with their meal, or they will not. The probability of a party ordering drinks with their meal is independent of other parties. So the binomial probability distribution is used 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.

When Ryan is serving at a restaurant, there is a 0.75 probability that each party will order drinks with their meal.

This means that $p = 0.75$

During one hour, Ryan served 6 parties. Assuming that each party is equally likely to order drinks, what is the probability that at least one party will not order drinks?

6 parties, so n = 6.

Either all parties will order drinks, or at least one will not. The sum of the probabilities of these events is decimal 1. So

$P(X = 6) + P(X < 6) = 1$

We want P(X < 6). So

$P(X < 6) = 1 - P(X = 6)$

In which

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

$P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.18$

$P(X < 6) = 1 - P(X = 6) = 1 - 0.18 = 0.82$

0.82 = 82% probability that at least one party will not order drinks

5 0

3 years ago

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D. Evaluate each of the following. (5 points)

Harrizon [31]

Problem 1

f(x) = x + 1/x

f(1) = 1 + 1/1

f(1) = 1 + 1

f(1) = 2

Answer: 2

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

Problem 2

(f o f)(1) = f( f(1) )

(f o f)(1) = f( 2 )

(f o f)(1) = 2 + 1/2

(f o f)(1) = 4/2 + 1/2

(f o f)(1) = 5/2

Answer: 5/2

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

Problem 3

(f o f o f)(1) = f( f(f(1)) )

(f o f o f)(1) = f( (f o f)(1) )

(f o f o f)(1) = f( 5/2 )

(f o f o f)(1) = 5/2 + 1/(5/2)

(f o f o f)(1) = 5/2 + 2/5

(f o f o f)(1) = 25/10 + 4/10

(f o f o f)(1) = 29/10

Answer: 29/10

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

Problem 4

(f o f o f o f)(1) = f( f(f(f(1)) )

(f o f o f o f)(1) = f( (f o f o f)(1) )

(f o f o f o f)(1) = f( 29/10 )

(f o f o f o f)(1) = 29/10 + 1/(29/10)

(f o f o f o f)(1) = 29/10 + 10/29

(f o f o f o f)(1) = 841/290 + 100/290

(f o f o f o f)(1) = 941/290

Answer: 941/290

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Problem 5

(f o f o f o f o f)(1) = f( f(f(f(f(1)))) )

(f o f o f o f o f)(1) = f( (f o f o f o f)(1) )

(f o f o f o f o f)(1) = f( 941/290 )

(f o f o f o f o f)(1) = 941/290 + 1/(941/290)

(f o f o f o f o f)(1) = 941/290 + 290/941

(f o f o f o f o f)(1) = 885481/272890 + 84100/272890

(f o f o f o f o f)(1) = 969581/272890

Answer: 969581/272890

4 0

3 years ago

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Question 16 (Essay Worth 7 points)<br><br> Verify the identity.<br><br> tan (x + π/2) = -cot x

Rus_ich [418]

Step-by-step explanation:

We know that tan=sin/cos, so tan(x+π/2)=

$\frac{sin(x+pi/2)}{cos(x+pi/2)}$

Then, we know that sin(u+v)=sin(u)cos(v)+cos(u)sin(v),

so our equation is then

$\frac{sin(x)cos(\pi/2)+cos(x)sin(\pi/2)}{cos(x+\pi/2)} = \frac{cos(x)}{cos(x+\pi/2) }$

Then, cos(u+v)=cos(u)cos(v)-sin(u)sin(v), so our expression is then

$\frac{cos(x)}{cos(x)cos(\pi/2)-sin(x)sin(\pi/2)} = \frac{cos(x)}{-sin(x)} = -cot(x)$

6 0

3 years ago