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4 years ago

9

Find the probability of winning a lottery by selecting the correct six integers where the order in which these integers are sele

cted does not matter from the positive integers not exceeding 30

1 answer:

insens350 [35]4 years ago

7 0

There are C(30, 6) = 30!/(6!*(30-6)!) = 593,775 ways to pick 6 numbers from the first 30 positive integers.

The probabilty of matching 6 randomly chosen integers is 1/593,775.

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HELP! "translate: the product of 8 and a number increased by 2 use x as the variable"

Bumek [7]

Answer:

variable is (<em> </em><em>x</em><em> </em><em>)</em>

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<em>8</em><em> </em><em>x</em>

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3 0

3 years ago

Read 2 more answers

Simplify:<br> -2x5. 4x^2

Karo-lina-s [1.5K]

Answer:

<em>Hey mate, here's ur answer:</em>

<em>---------------------------------------------------</em>

<em>-2x^5. 4x^2 </em>

<em>=(−2)*x5*4*x2 </em>

<em>=−8*x7 </em>

<u><em>=−8x7</em></u>

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<em>Hope it helps</em>

<em>#stayhomestaysafemate</em>

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3 years ago

Jamie has 8/10 of a candy bar leftover. He wants to split it into 1/3. How many 1/3 pieces can he make

anyanavicka [17]

He can make 3 pieces. Since he has 8/10 of a candy bar, it is then simplified to 4/5 of a candy bar. You need to find 1/3 of that, that is 4/15. Then how many times does 4/15 fit into 4/5, well simply do this: (4/15)x=4/5. Solving for x should give you 3.

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4 years ago

Read 2 more answers

Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam

Vlad [161]

Answer

$P(A) = 0.30$

$P(B) = 0.77$

$P(A\ n\ B) = 0.22$

$P(A\ u\ B) = 0.85$

Explanation:

Given

See attachment for proper data presentation

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

Here:

$Yes = 22$ and $No = 8$

$n(A) = Yes + No$

$n(A) = 22 + 8$

$n(A) = 30$

P(A) is then calculated as:

$P(A) = \frac{n(A)}{Sample}$

$P(A) = \frac{30}{100}$

$P(A) = 0.30$

Solving (b): P(B)

We only consider data in the Yes column.

Here:

$(1) = 22$ $(2) = 25$ and $(3) = 30$

$n(B) = (1) + (2) + (3)$

$n(B) = 22 + 25 + 30$

$n(B) = 77$

P(B) is then calculated as:

$P(B) = \frac{n(B)}{Sample}$

$P(B) = \frac{77}{100}$

$P(B) = 0.77$

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

i.e. [Supplier 1][Yes]

This is represented as: n(A n B)

$n(A\ n\ B) = 22$

The probability is then calculated as:

$P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}$

$P(A\ n\ B) = \frac{22}{100}$

$P(A\ n\ B) = 0.22$

Solving (d): P(A u B)

This is calculated as:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$

This gives:

$P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}$

Take LCM

$P(A\ u\ B) = \frac{30+77-22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

$P(A\ u\ B) = 0.85$

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3 years ago

Please help soon <br> What is m

uranmaximum [27]

The answer to your problem is 69

3 0

3 years ago