Solve the system of equations

The original cost of an item is $68, but you have to pay $76.16.

Sonja [21]

Answer:

12%

Step-by-step explanation:

calculator

7 0

3 years ago

Find the range of the function defined by each equation: y =- x

blagie [28]

Range is all the values of y, in this case, y can be any number, so the range is all real numbers.

4 0

3 years ago

How many pennies could you have if:

Gelneren [198K]

Answer:

The number of pennies owned is 7 pennies

Step-by-step explanation:

The given parameters are;

The number of pennies left over when we break the pennies into groups of 2s = 1 penny

The number of pennies left over when we break the pennies into groups of 3s = 1 penny

Let the number of pennies owned = c

What we are given are as follows;

2 × a = c - 1

3 × b = c - 1

2 × a = 3 × b

a/b = 3/2

Therefore, if we multiply 2 by 3, and 3 by 2 we get 6

2 × 3 = 6, similarly 3 × 2 = 6

If we put 6 = c - 1, we get;

c = 6 + 1 = 7

c = 7

The number of pennies owned = 7 pennies.

4 0

3 years ago

Please help! I don’t have much to offer except the crown icon thing and 15 points.

Goshia [24]

<h2><u>SOLUTION</u><u>:</u></h2>

x/9>12

x>12×9

x>108

This is your answer

8 0

3 years ago

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

So

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

So

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0

3 years ago