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3 years ago

HELP!!!! Graduation depends on you guys today

1 answer:

3 0

There are the combinations that result in a total less than 7 and at least one die showing a 3:

[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36

There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:

1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36

Therefore, the required probability is: 5/36

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What is the sign of f on the interval -2

TEA [102]

Answer:

$f$ is sometimes positive and sometimes negative.

Step-by-step explanation:

$f(x)=(x-3)(x+2)(x+4)(x+4)(x-1)(2x-9)$

Take $x=-1\in(-2,\frac{9}{2})\ as-2$

$f(-1)=(-1-3)(-1+2)(-1+4)(-1-1)(-2-9)\\\\=(-4)(1)(3)(-2)(-11)\\\\=-264\\\\f(-1)$

Take $x=2\in(-2,\frac{9}{2})\ as\ -2$

$f(2)=(2,-3)(2+2)(2+4)(2-1)(2\times2-9)\\\\=(-1)(4)(6)(1)(-5)\\\\=120\\\\f(2)>0$

Hence $f(x)$ for $x=-1$ and $f(x)>0$ for $x=2$

So $f$ is sometimes positive and sometimes negative in the interval.

3 0

3 years ago

PLEASE HELP!!!! WILL MARK BRAINLIEST!!!!

gogolik [260]

Angle F would be 93°

3 0

2 years ago

Read 2 more answers

Peaches are being sold for $2 per pound. If x represents the number of pounds of peaches bought and y represents the total cost

Mama L [17]

2X=Y

2 because it is two dollars per pound.

8 0

3 years ago

Read 2 more answers

What is the 79th term of this arithmetic sequence? 3, 9, 15, 21,

hodyreva [135]

6 times 79 = 474 so the 79th term is 474

5 0

3 years ago

a. Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability th

hodyreva [135]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

3 or less than 3 tires among the 10 tires should be blemished.

There are some possibilities.

1000 tires are blemished, (5000 - 1000) = 4000 tires are not blemished.

From the 5000 tires, 10 tires can be chosen in $^{5000}C_{10}$ ways.

Possibility 1: <u>3 tires are blemished.</u>

Total 10 tires can be chosen with 3 blemished tires in $^{1000}C_3 \times {4000}C_7$ ways.

Possibility 2: <u>2 tires are blemished.</u>

Total 10 tires can be chosen with 2 blemished tires in $^{1000}C_2 \times ^{4000}C_8$ ways.

Possibility 3: <u>1 tires are blemished.</u>

Total 10 tires can be chosen with 1 blemished tires in $^{1000}C_1 \times^{4000}C_9$ ways.

Possibility 4: <u>No tires are blemished.</u>

Total 10 tires can be chosen with no blemished tires in $^{1000}C_0 \times^{4000}C_{10}$ ways.

Hence, the required probability is $\frac{^{1000}C_1 \times^{4000}C_9 + ^{1000}C_0 \times^{4000}C_{10} + ^{1000}C_2 \times^{4000}C_8 + ^{1000}C_3 \times^{4000}C_7}{^{5000}C_{10} }$ .

6 0

3 years ago