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2 years ago

What is the product of seven and eight decreased by “x” if x =15?

Mathematics

2 answers:

jekas [21]2 years ago

7 0

Answer:

Step-by-step explanation:

product is another word for +

decreased is another word for -

____________________________

(7 + 8) - x

(7+8) - 15

15 - 15

= 0

yKpoI14uk [10]2 years ago

3 0

Answer:

Step-by-step explanation:

Vocab:

Product: Multiplication

Decrease: Subtraction

(7*8)-x

Substitute:

(7*8)-15

Multiply:

(56)-15

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Y=-1\4z+5 please grqph

svet-max [94.6K]

Answer

Attached the graph

Step by step explanation

Y = -1/4z + 5

Let's form the table values

Here z is the independent variable and y is the dependent values.

Let's take z = -1, 0, 1, 2 and find the corresponding y-values

z y

-1 5.25

0 5

1 4.75

2 4.5

Now let's plot the points and draw the graph.

Here is the graph.

5 0

2 years ago

Which number is an irrational number? A. B. C. D.

Ivahew [28]

There’s nothing so let’s just go with C

6 0

2 years ago

Solve these compound interest problems and round your answer to the nearest 100th.

MatroZZZ [7]

Answer:

$2,226.96

Step-by-step explanation:

You are going to want to use the compound interest formula, which is shown below.

$A=P(1+\frac{r}{n} )^{nt}$

P = initial balance

r = interest rate

n = number of times compounded annually

t = time

First, change 10% into its decimal form:

10% -> $\frac{10}{100}$ -> 0.1

Now lets plug in the values into the equation:

$A=500(1+\frac{0.1}{12})^{15(12)}$

$A=2,226.96$

The final amount after 15 years is $2,226.96

4 0

2 years ago

"the life length x (in years) of a dvd player" is exponentially distributed with mean 5 years. what is the probability that a mo

loris [4]

The exponential distribution is:
$\lambda e^{- \lambda x}$
where $\lambda = \frac{1}{mean}=\frac{1}{5}$

The probability we want is how likely will a dvd player last more than 8 years, given it has already lasted 5 years
To find this, you use conditional probability.
$P(A|B) = \frac{P(A and B)}{P(B)}$
where A is P(x>8) and B is P(x>5)
To find these probabilities, integrate over the distribution:
$P(x \ \textgreater \ N) = \int_N^{\infty} \frac{1}{5} e^{-x/5} dx = e^{-N/5}$

Sub into conditional probability formula:

$P(x \ \textgreater \ 8 | x\ \textgreater \ 5) = \frac{P(x\ \textgreater \ 8)}{P(x\ \textgreater \ 5)} = \frac{e^{-8/5}}{e^{-5/5}} = e^{-3/5} = 0.549$

Final Answer: Given a dvd player is more than 5 years old, the probability that it will last another 3 more years is about 54.9%

3 0

2 years ago

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

2 years ago