Question 9 of 10

Any help would be appreciated. Thank you

Anestetic [448]

14-3^2 x (-2)
=14-9 x (-2)
=14- (-18)
=32

5 0

3 years ago

Write an expression for the sequence of operations described below.

Hatshy [7]

The expression for the sequence of operations described below is d/3 + 5

<h3>Expressions and equations:</h3>

Let the unknown value be "d"

d divided by 3 is expressed as d/3

Adding 5 to d/3 to get the expression below;

g(d) = d/3 + 5

Hence the expression for the sequence of operations described below is d/3 + 5

Learn more on expressions here: brainly.com/question/29114

3 0

3 years ago

A major credit card company has determined that customers charge between $100 and $1100 per month. If the monthly amount charged

Alexxx [7]

Answer:

a) $E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600$

b) First we need to calculate the variance given by this formula:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33$

And the deviation would be:

$Sd(X) = \sqrt{83333.33}= 288.675$

c) $P(600 < X< 889) = P(X$

And using the cdf we got:

$P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289$

d) $P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578$

Step-by-step explanation:

For this case we define the random variable X who represent the customers charge, and the distribution for X on this case is:

$X \sim Unif (a= 100, b=1100)$

Part a

For this case the average is given by the expected value and we can use the following formula:

$E(X) = \frac{a+b}{2} = \frac{100+1100}{2}=600$

Part b

First we need to calculate the variance given by this formula:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(1100-100)^2}{12} = 83333.33$

And the deviation would be:

$Sd(X) = \sqrt{83333.33}= 288.675$

Part c

For this case we want to find the percent between 600 and 889, so we can use the cumulative distribution function given by:

$F(x) = \frac{X -100}{1100-100}= \frac{x-100}{1000}, 100 \leq X \leq 1100$

And we can find this probability:

$P(600 < X< 889) = P(X$

And using the cdf we got:

$P(600 < X< 889)= F(889) -F(600) = \frac{889-100}{1000} -\frac{600-100}{1000}= 0.789- 0.5= 0.289$

Part d

$P(311 < X< 889) = P(X$

And using the cdf we got:

$P(311 < X< 889)= F(889) -F(311) = \frac{889-100}{1000} -\frac{311-100}{1000}= 0.789- 0.211= 0.578$

6 0

3 years ago

Solve the following system of equations and show all work. y = x^2 + 3 y = x + 5 (10 points)

svetoff [14.1K]

Answer:

(- 1, 4 ) and (2, 7 )

Step-by-step explanation:

Given the 2 equations

y = x² + 3 → (1)

y = x + 5 → (2)

Substitute y = x² + 3 into (2)

x² + 3 = x + 5 ← subtract x + 5 from both sides

x² - x - 2 = 0 ← in standard form

(x - 2)(x + 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 2 = 0 ⇒ x = 2

x + 1 = 0 ⇒ x = - 1

Substitute these values into (2) for corresponding values of y

x = 2 : y = 2 + 5 = 7 ⇒ (2, 7 )

x = - 1 : y = - 1 + 5 = 4 ⇒ (- 1, 4 )

4 0

4 years ago

Round the numbers to the highest place value.

Kaylis [27]

Answer:

$2398 = 2000$

$56254 = 60000$

$71563 = 70000$

$33021 = 30000$

$9645 = 10000$

Step-by-step explanation:

Given

$2398$

$56254$

$71563$

$33021$

$9645$

Required

Round up to highest place value

$2398$

Highest place value here is thousand.

So:

$2398 = 2000$ , when approximated

$56254$

Highest place value here is ten thousand.

So:

$56254 = 60000$ , when approximated

$71563$

Highest place value here is ten thousand.

So:

$71563 = 70000$ when approximated

$33021$

Highest place value here is ten thousand.

So:

$33021 = 30000$ when approximated

$9645$

Highest place value here is ten thousand.

So:

$9645 = 10000$ when approximated

4 0

3 years ago