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3 years ago

Multiple Events Help???

Mathematics

1 answer:

Alexxx [7]3 years ago

8 0

From the diagram you can see that $Pr(A)=0.5$ and $Pr(C|A)=0.4$ .

Use formula $Pr(A\cap C)=Pr(A)\cdot Pr(C|A)$ for calculating $Pr(A\cap C):$

$Pr(A\cap C)=0.5\cdot 0.4=0.2$ .

Answer: $Pr(A\cap C)=0.2$ .

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Last year the company you work for did 232,000 in business. This year you anticipate an increase of 7% how much business do you

Nataly_w [17]

7% of 232,000 is 16,240
232,000 + 16,240 is 248,240
so 248,240 is the answer

3 0

3 years ago

Read 2 more answers

2y to the power of 4 x 5y to the power of 3

Yakvenalex [24]

Answer:

(2•5y to the 7th power)

Step-by-step explanation:

Step 1: (2y to the 4th power • 5) • y to the 3rd power

Step 2: (2•5y to the 4th power) • y to the 3rd power

Step 3: 3.1 y to the 4th power multiplied by y to the 3rd power = y to the (4 + 3) power = y to the 7th power

Final answer: (2•5y to the 7th power)

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3 years ago

I can’t explain look at picture

const2013 [10]

Answer:

x = 12.48

y = 13.22

Step-by-step explanation

Im not sure if its correct

10^2 + b^2 = 16^2

100 + b^2 = 256

b^2 = √156

b = 12.48

b= 12 ( If needed to round to the nearest whole number)

a^2 + 15^2 = 20^2

a^2 + 225 = 400

a^2 = √175

a = 13.22

a = 13 ( If needed to round to nearest whole number)

3 0

3 years ago

you are charged $16.05 after tax for a meal. Assume sales tax is 7%, what was the menu price for the meal?

Anton [14]

$1.07x=16.05$

$16.05/1.07=15$

$15=x$

7 0

3 years ago

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A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago