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3 years ago

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A bag contains only red and blue marbles. Yasmine takes one marble at random from the bag. The probability that she takes a red

marble is 1 in 5. Yasmine returns the marble to the bag and adds ﬁve more red marbles to the bag. The probability that she takes one red marble at random is now 1 in 3. How many red marbles were originally in the bag?

1 answer:

Daniel [21]3 years ago

8 0

The answer is gonna be 40

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Please help me almost done

Gnom [1K]

Answer:

x = -8

<XCE = 30

Step-by-step explanation:

Remark

In any question like this, there are 8 possible answers. Of the 8, 4 are acute angles and 4 are obtuse. So any pair are either equal to each other, or they are supplementary. In this case the two given angles are equal.

We have a little bit of both in this question.

Solution

3x + 174 = - 7x + 94 Alternate exterior angles. Add 7x to both sides.

3x+7x +174 = 94 Combine

10x + 174 = 94 Subtract 174 from both sides

10x = 94 - 174

10x = - 80 Divide by 10

x = - 80/10

x = - 8

===================

Check

3x + 174 = -24 + 174 = 150

-7x + 94 = -7*-8 + 94

-7x + 94 = 56 + 94 = 150

<XCE

<XCE = 180 - (3x + 174)

<XCE = 180 - (-24 + 174)

<XCE = 180 - (150)

<XCE = 30

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3 years ago

Record the product 23x79

Ber [7]

When I finished the work my product is 1,817

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3 years ago

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Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is:

$\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|$

$\Delta t_{1/2} \approx 65.788\,days$

The approximate difference in the half-lives of the isotopes is 66 days.

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Coffee costs $18.96 for 3 pounds. And

trapecia [35]

Answer:

the constant of proportionality is 6.32

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3 years ago

Given that CD¯¯¯¯¯¯¯¯ is a perpendicular bisector of AB¯¯¯¯¯¯¯¯, where D is on AB¯¯¯¯¯¯¯¯, how can you use the Pythagorean Theor

Liula [17]

Answer:

$(AD)^2 + (CD)^2 = (CA)^2$ and $(CD)^2 + (BD)^2 = (CB)^2$

Step-by-step explanation:

Given

Bisector: CD

of Line AB

Required

Apply Pythagoras Theorem

From the question, CD bisects AB and it bisects it at D.

The relationship between AB and CD is given by the attachment

Considering ACD

From the attachment, we have that:

$Hypothenuse = CA$

$Opposite = CD$

$Adjacent = AD$

By Pythagoras Theorem, we have

$(AD)^2 + (CD)^2 = (CA)^2$

Considering CBD

From the attachment, we have that:

$Hypothenuse = CB$

$Opposite = CD$

$Adjacent = BD$

By Pythagoras Theorem, we have:

$(CD)^2 + (BD)^2 = (CB)^2$

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2 years ago

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