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Monica [59]
3 years ago
7

Please Help if you can! There are 2 red gumdrops and 3 green gumdrops in a small jar. Also, 3 pieces of butterscotch candy and 2

pieces of cinnamon candy are in another jar. If Craig draws one piece of candy from each jar without looking, what's the probability that he will get a red gumdrop and a piece of butterscotch candy? The Answer choices are 6/25, 1/2, 4/25, or 1/5! Thank you!
Mathematics
1 answer:
attashe74 [19]3 years ago
5 0
I think it would be 6/25 because you have 2/5 chance and a 3/5 chance,
2 × 3 = 6
--- --- ---
5 × 5 = 25

I hope this helps ♡ chyna
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Klio2033 [76]

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The tallest column of the line plot

Step-by-step explanation:

A line plot shows the frequency of data, in short it shows how many times something has occurred or the number of times data occurs in a group of data.

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The bar graph there does not show the frequency of the different scores. This shows the relationship between the test number and the percentage of the scores obtained. It does not tell you the individual scores obtained.

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3 years ago
Need help ASAP ? <br>3/4×20
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3/4 x 20

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3 years ago
What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)
Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\

Now to solve the integral on the right hand side we shall use Integration by parts Taking 7t^{3} as first function thus we have

\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\

Again repeating the same procedure we get

=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\

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\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\

Now solving this integral we have

\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}

Thus we have

L[7t^{3}]=\frac{7\times 3\times 2}{s^4}

where s is any complex parameter

5 0
2 years ago
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