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1 year ago

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The probability of flowering plants bearing flowers in January is given the table. If a plant has flowered in January, what is t

he probability that it is a peony?
A. 70%
B. 73%
C. 76%
D. Insufficient data

2 answers:

ololo11 [35]1 year ago

7 0

Answer:

Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

Step-by-step explanation:

We know that:

Probability =Total favourable outcomes/ Total possible outcomes

Probability of flowering plants bearing flowers in January is given in the table. It means that out of 100 peony 70 peony flowers in January.

There are five kinds of flowers.

We have to select peony and find it's Probability.

As all flowers are independent, their flowering phenomenon does not depend on each other.

So,Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

Lisa [10]1 year ago

3 0

Answer:

70%

Step-by-step explanation:

Given : The probability of flowering plants bearing flowers in January is given the table.

To Find : If a plant has flowered in January, what is the probability that it is a peony?

Solution :

Probability of event = $\frac{\text{Favorable event}}{\text{Total events}}$

The probability of bearing peony in January is given the table i.e. 70%

Hence If a plant has flowered in January,the probability that it is a peony is 70%

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Read 2 more answers

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$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }$

Step-by-step explanation:

Given

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9}$

Required

Find the products

From laws of indices;

$\sqrt[m]{a} * \sqrt[m]{b} = \sqrt[m]{a*b}$

So;

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16x^7 * 12x^9}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16* x^7 * 12 * x^9}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^7 * x^9}$

From laws of indices

$a^m * a^n = a^(m+n); So,$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{7+9}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{16}}$

Expand 16 * 12

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4*4*4*3* x^{16}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4^3 *3* x^{16}}$

From laws of imdices

$a^{\frac{1}{m}} = \sqrt[m]{a}$

So;

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^3 *3* x^{16}})^{\frac{1}{3}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^{3*{\frac{1}{3}}} *3^{{\frac{1}{3}}}* x^{16*{\frac{1}{3}}}})$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{\frac{16}{3}}}$

Divide 16 by 3 (Write as ,mixed number)

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5\frac{1}{3}}}$

Split mixed numbers

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5+\frac{1}{3}}}$

Apply law of indices

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5}*{x ^\frac{1}{3}}}$

Reorder

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 * x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}$

Apply law of indices

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3} *\sqrt[3]{x} }$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3*x} }$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3x} }$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }$

6 0

1 year ago