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Irina-Kira [14]

3 years ago

5

There are 9 red gum balls, 5 green gum balls, 8 yellow gum balls, and 8 blue gum balls in a machine. find p(green, then yellow).

1 answer:

creativ13 [48]3 years ago

5 0

The correct answer is choice D. To find this you will find the probability of each event happening and then multiply them together.

Green: 5/30
Yellow: 8/29 (you subtract one from the denominator because you already picked a green the first time)

5/30 x 8/29 = 40/870

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3 years ago

If DM = 35, what is the value of r?

Gennadij [26K]

Answer:

$r = 11$

Step-by-step explanation:

Given

$DM = 35$

$DG = r + 5$

$GM = 3r - 14$

Required

Find r

From the question, we understand that G is a point between D and M:

This implies that:

$DM = DG + GM$

Substitute values for DM, DG and GM

$35 = r + 5 + 3r - 14$

Collect Like Terms

$r + 3r = 35 - 5 + 14$

$4r = 44$

Solve for r

$r = 44/4$

$r = 11$

3 0

3 years ago

A worker at a landscape design center uses a machine to fill bags with potting soil. Assume that the quantity put in each bag fo

defon

Answer:

a) The expected value is given by:

$E(X) =\frac{a+b}{2}= \frac{10+12}{2}= 11$

The variance is given by:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(12-10)^2}{12}= 0.3333$

And the standard deviation is just the square root of the variance and we got:

$Sd(x) = \sqrt{0.3333}= 0.5774$

b) $P(X$

And we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a}, a \leq x \leq b$

And using this function we got:

$P(X$

c) $P(X>10.5)$

And using the complement rule and the cumulative distribution function we got:

$P(X>10.5)= 1-P(X$

Step-by-step explanation:

For this case we define the random variable X=quantity put in each bag , and we know that the distribution for X is given by:

$X \sim Unif (a= 10, b =12)$

Part a

The expected value is given by:

$E(X) =\frac{a+b}{2}= \frac{10+12}{2}= 11$

The variance is given by:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(12-10)^2}{12}= 0.3333$

And the standard deviation is just the square root of the variance and we got:

$Sd(x) = \sqrt{0.3333}= 0.5774$

Part b

For this case we want this probability:

$P(X$

And we can use the cumulative distribution function given by:

$F(x) = \frac{x-a}{b-a}, a \leq x \leq b$

And using this function we got:

$P(X$

Part c

For this case we want this probability:

$P(X>10.5)$

And using the complement rule and the cumulative distribution function we got:

$P(X>10.5)= 1-P(X$

3 0

3 years ago