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Vladimir79 [104]

3 years ago

8

The expression 15h + 216 represents the approximate amount of money a family spends at an amusement park in a day, where h is nu

mber of hours the family stays at the park. About how much can a family expect to spend if it stays 6.5 hours?
Please help!

1 answer:

ad-work [718]3 years ago

7 0

Answer: $313.50

Step-by-step explanation:

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6x9x(-5) what is the answer to this question?

9966 [12]

-270. Use the google calculator.

7 0

3 years ago

Please help i always give brainliets and thanks

pav-90 [236]

Answer:

(3)

Step-by-step explanation:

(1), (2) and (4)

can be written as Fr= p1-p2, F and r are inversely proportional

(3) r is directly proportional with F which is opposite to the given formula

hence this is incorrect

6 0

4 years ago

Jack Purchased a new iPhone for $300. It is expected to lose 25% of its value each

asambeis [7]

Answer:

94.92

Step-by-step explanation:

300*(75/100)^4=94.92

8 0

3 years ago

3. From the table below, find Prof. Xin expected value of lateness. (5 points) Lateness P(Lateness) On Time 4/5 1 Hour Late 1/10

wariber [46]

Answer:

The expected value of lateness $\frac{7}{20}$ hours.

Step-by-step explanation:

The probability distribution of lateness is as follows:

Lateness P (Lateness)

On Time 4/5

1 Hour Late 1/10

2 Hours Late 1/20

3 Hours Late 1/20

The formula of expected value of a random variable is:

$E(X)=\sum x\cdot P(X=x)$

Compute the expected value of lateness as follows:

$E(X)=\sum x\cdot P(X=x)$

$=(0\times \frac{4}{5})+(1\times \frac{1}{10})+(2\times \frac{1}{20})+(3\times \frac{1}{20})\\\\=0+\frac{1}{10}+\frac{1}{10}+\frac{3}{20}\\\\=\frac{2+2+3}{20}\\\\=\frac{7}{20}$

Thus, the expected value of lateness $\frac{7}{20}$ hours.

8 0

3 years ago

Please help solve this system of equations

stepan [7]

Make a substitution:

$\begin{cases}u=2x+y\\v=2x-y\end{cases}$

Then the system becomes

$\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}$

Simplifying the equations gives

$\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}$

which is to say,

$\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}$

$\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81$

$\implies\dfrac uv=\pm27$

$\implies u=\pm27v$

Substituting this into the new system gives

$\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1$

$\implies u=\pm27$

Then

$\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13$

(meaning two solutions are (7, 13) and (-7, -13))

8 0

3 years ago