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mojhsa [17]
3 years ago
14

Jesse is playing a game with a number cube with faces numbered 1 through 6. What is the probability of rolling a number that is

even and a multiple of 3?
Mathematics
1 answer:
grandymaker [24]3 years ago
7 0

Answer:

The probability of rolling a number that is even and a multiple of 3 is \frac{1}{6}

Step-by-step explanation:

The total possible  outcomes of cube = { 1, 2, 3 , 4 , 5 , 6}

Now, let E: Event of getting an even number and a multiple of 3

So, out of all the outcomes, {6) is the ONLY possible favorable outcome.

\textrm{P(E)}  =\frac{\textrm{Number of favorable outcomes }}{\textrm{Total number of outcomes}}

or, P(E) = \frac{1}{6}

Hence, the probability of rolling a number that is even and a multiple of 3 is \frac{1}{6}

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Step-by-step explanation:


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3 years ago
What is the area of the rectangle?
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The area of the rectangle is A. 6000
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Jim sells hot dogs for $2.95 each and steak sandwiches for $9.95 each out of his food cart. During a busy outdoor festival, he s
Afina-wow [57]

Step-by-step explanation:

x + y = 985

2.95x + 9.95y = 7343.75

-2.95x - 2.95y = -2905.75

7y = 4438

y= 634 steak sandwiches

x + 634 = 985

x = 351 hot dogs

7 0
3 years ago
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Let f(x) = (5)^x+1 .Evaluate f (-3) without using a calculator
rewona [7]

Answer:

f\left(-3\right)=\frac{126}{125}

Step-by-step explanation:

Given

f\left(x\right)\:=\:\left(5\right)^x+1

Putting x = -3 to find f(-3)

f\left(-3\right)\:=\:\left(5\right)^{\left(-3\right)}+1

as

5^{-3}=\frac{1}{125}       ∵  \mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

so

f\left(-3\right)=\frac{1}{125}+1

\mathrm{Convert\:element\:to\:fraction}:\quad \:1=\frac{1\cdot \:125}{125}

f\left(-3\right)=\frac{1\cdot \:\:125}{125}+\frac{1}{125}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

f\left(-3\right)=\frac{1\cdot \:\:125+1}{125}

f\left(-3\right)=\frac{126}{125}      ∵ 1\cdot \:125+1=126

Thus,

f\left(-3\right)=\frac{126}{125}

6 0
4 years ago
E-1, e-2 and e-3 are three engineering students writing their assignments at night. Each of them starts at a different time and
Bas_tet [7]

E-3 will be the first student to start writing the assignment.

It is clearly stated in the statement that the order of starting the assignment and the order of the completion of the assignment is not the same, the digit in the name of the executive is also not same. This information puts some restrictions.

E-1 can neither be the first one to start nor the first one to complete. Similarly, E-2 cannot be the second one and E-3 cannot be the third one.

It is given that the last student (the third one) to start is the first student (the first one) to complete. This can neither be E-3 nor E-1. It is E-2

The final arrangement is as follows.

                                 Start           Complete

First                           E-3              E-2

Second                      E-1               E-3

Third                          E-2              E-1

E-3 is the first one to start writing the assignment.

More similar problems are solved in the link.

brainly.com/question/28391619?referrer=searchResults

#SPJ4

8 0
2 years ago
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