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2 years ago

11

A bag has 2 blue marbles, 3 red marbles, and 5 white marbles. Which events have a probability greater than One-fifth? Select thr

ee options.

2 answers:

ki77a [65]2 years ago

6 0

Answer:

B, D, E

Step-by-step explanation:

EDG 2020

AlladinOne [14]2 years ago

3 0

Answer:

first one and last two

Step-by-step explanation:

B)

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Malik randomly picked two numbers from 1 to 9 (Includin 1 and 9). the same number could be picked more than once. The first of t

Tju [1.3M]

Answer:

2/9 = 0.22

Step-by-step explanation:

There are two ways to pick the first number odd and less than 5: 1 and 3.

With each of these, the second number drawn can be 1, 2, 3, 4, 5, 6, 7, 8 or 9.

This makes 18 total possibilities.

Out of these, the only ways to have a sum less than 5 are 1 and 1, 1 and 2, 1 and 3, and 3 and 1. This is 4 ways out of 18:

4/18 = 2/9 = 0.22

8 0

2 years ago

Create a set of data that contains 11 test scores and satisfies each condition below:Mean: 83Median: 81Mode: 80Range: 26

Lunna [17]

64 78 80 80 80 81 81 82 83 89 90 Not ecaxt but close, it should help atleast, everything is right except mean

4 0

2 years ago

A rational function can have infinitely many x-values at which it is not continuous. True or False. If false, explain

sattari [20]

<span>It is false since the rational function is discontinuous when the denominator is zero. But the denominator is a polynomial and a polynomial has only finitely many zeros. So the discontinuity points of a rational function is finite. </span>

3 0

2 years ago

The area of ABED is 49 square units. Given AGequals9 units and ACequals10 units, what fraction of the area of ACIG is represent

Mrac [35]

Answer:

The fraction of the area of ACIG represented by the shaped region is 7/18

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the square ABED find the length side of the square

we know that

AB=BE=ED=AD

The area of s square is

$A=b^{2}$

where b is the length side of the square

we have

$A=49\ units^2$

substitute

$49=b^{2}$

$b=7\ units$

therefore

$AB=BE=ED=AD=7\ units$

step 2

Find the area of ACIG

The area of rectangle ACIG is equal to

$A=(AC)(AG)$

substitute the given values

$A=(9)(10)=90\ units^2$

step 3

Find the area of shaded rectangle DEHG

The area of rectangle DEHG is equal to

$A=(DE)(DG)$

we have $DE=7\ units$

$DG=AG-AD=9-7=2\ units$

substitute $A=(7)(2)=14\ units^2$

step 4

Find the area of shaded rectangle BCFE

The area of rectangle BCFE is equal to

$A=(EF)(CF)$

we have

$EF=AC-AB=10-7=3\ units$

$CF=BE=7\ units$

substitute

$A=(3)(7)=21\ units^2$

step 5

sum the shaded areas

$14+21=35\ units^2$

step 6

Divide the area of of the shaded region by the area of ACIG

$\frac{35}{90}$

Simplify

Divide by 5 both numerator and denominator

$\frac{7}{18}$

therefore

The fraction of the area of ACIG represented by the shaped region is 7/18

5 0

2 years ago

Write v as a linear combination of u1, u2, and u3, if possible. (If not possible, enter IMPOSSIBLE.) v = (7, −14, −3, −4), u1 =

AlexFokin [52]

Answer:

IMPOSSIBLE

Step-by-step explanation:

8 0

2 years ago