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AlladinOne [14]

3 years ago

11

A bag contains 7 pears and 3 apples. If you select 2 pieces of fruit without looking, how many ways can you get 2 pears.

1 answer:

enot [183]3 years ago

6 0

Answer:

There are 2 ways, since we'd chosen 2 fruits

PROBABILITY:

( $\frac{2}{10}$ = $\frac{1}{5}$ ) or $\frac{1}{10}$

METHOD:

In the 2 fruits we took both of them might be pears either maybe one of them can be a pear. The same is possible with the other fruit.

<em>PLEASE DO MARK MY ANSWER AS THE BRAINLIEST..</em>

<em><HOPE IT HELPS></em>

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Jessica spent S28 on fruit at the grocery store. She spent a total of $40 at the store. What percentage of the total did she spe

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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