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1 year ago

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You select 35 marbles from a bag. The results are as follows: 4 blue marbles, 9 green marbles, 7 red marbles, 7 white marbles, a

nd 8 yellow marbles. Find P(green). Write the probability as a fraction in simplest form, a decimal, and a percent.

1 answer:

jarptica [38.1K]1 year ago

4 0

You just take the number of the color of interest (green) and divide by total marbles

9/35 = 0.257 ~~ 25.7%

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Will give brainliest

klemol [59]

Your question is calling the figure ABCD, but the image is showing PQRS

Using the letters that you wrote:

Because the figure was only rotated and not scaled , the sides with the same letters would still be the same

So AB would be the same as A'B' ( Length of AB = Length of A′B)

and

BC = B'C'

CD = C'D'

AD = A'D'

Or by the picture:

PQ = P'Q'

QR = Q'R'

RS = R'S'

PS = P'S'

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2 years ago

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Trapezoid QRST has two right angles. A 5-in.altitude can be drawn dividing QRDT into a rectangle and an isosceles right triangle

Vika [28.1K]

Answer:

The area of the trapezoid is 57.5 square inches

Step-by-step explanation:

we know that

The trapezoid QRST can be divided into a rectangle QRDT and an isosceles right triangle RSD

see the attached figure to better understand the problem

step 1

The area of rectangle is given by the formula

$A=((QR)(RD)$

we have

$RD=5\ in$ ----> altitude

$QR=9\ in$

substitute

$A=((9)(5)=45\ in^2$

step 2

Find the area of the isosceles right triangle

The area of triangle is given by the formula

$A=\frac{1}{2}(RD)(DS)$

we have

$RD=DS=5\ in$ ---> because is an isosceles triangle

substitute

$A=\frac{1}{2}(5)(5)=12.5\ in^2$

step 3

Adds the areas

$A=45+12.5=57.5\ in^2$

3 0

3 years ago

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ValentinkaMS [17]

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2 years ago

Lizzie rolls two dice. What is the probability that the sum of the dice is:

zhenek [66]

Answer:

$A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\$

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A) = \dfrac{12}{36} = \dfrac{1}{3}$

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)} Count 1

Possible cases for sum 3: {(1,2), (2,1)} Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)} Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)} Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

$P(B) = \dfrac{15}{36} = \dfrac{5}{12}$

C. Divisible by 3 and less than 7:

$P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}$

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

$P(A \cap B) = \dfrac{7}{\text{36}}$

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What is the probability of the arrow landing on a 3 and then on an odd number?

matrenka [14]

Answer

go to your math teacher and ask him to explain more

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3 years ago