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Goryan [66]
3 years ago
13

In a family with 8 ​children, excluding multiple​ births, what is the probability of having 7 boys and 1​ girl, in any​ order? a

ssume that a boy is as likely as a girl at each birth
Mathematics
1 answer:
Vadim26 [7]3 years ago
5 0
There is a 7/8 chance there will be boys and a 1/8 chance there will be a girl 

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Solve for x. Leave your answer in the simplest radical form.
MaRussiya [10]

Answer:

x = 2\sqrt{5}

Step-by-step explanation:

'x' can represent the hypotenuse of a right triangle with a base of 4 (11-7) and a height of 2

use Pythagorean Theorem to find 'x':

2² + 4² = x²

4 + 16 = x²

x² = 20

x = \sqrt{20}

x = \sqrt{4}·\sqrt{5}

x = 2\sqrt{5}

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3 years ago
Benjamin has a mask collection. He keeps 144 of the masks on his wall, which is 48% of his entire collection. What is the total
Harman [31]

Answer:

213

Step-by-step explanation:

convert percent to decimal 48%=0.48

144x0.48= 69.12

estimate 69.12 we got 69

144+69=213

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3 years ago
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What is the solution set of fx x>-5) u {xx<5)?
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Huh.........

Explanation:

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3 years ago
1000 centimeters=_______meters?
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<em>1000 Centimeters = </em><em>10 Meters</em>

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3 years ago
A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 6%
Georgia [21]

Answer:

a

\% E =  0.9 \%

b

\%E_1 = 10.1 \%

Step-by-step explanation:

From the question we are told that

The probability that an employees suffered lost-time accidents last year is P(e) =  0.06

The probability that an employees suffered lost-time accident during the current year is

P(c) =  0.05

The probability that an employee will suffer lost time during the current year given that the employee suffered lost time last year is

P(c | e) =  0.15

Generally the probability that an employee will experience lost time in both year is mathematically represented as

P(c \ n \ e) =  P(e) *  P(c \ |\ e)

=> P(c \ n \ e) =  0.06*   0.15

=> P(c \ n \ e) = 0.009

Generally the percentage of employees that will experience lost time in both year is mathematically represented as

\% E =  P(e \ n \ c ) * 100

=> \% E =  0.009 * 100

=> \% E =  0.9 \%

Generally the probability that an employee will experience at least one lost time accident over the two-year period is mathematically represented as

P(e \ u \ c) =  P(e) + P(c) - P(e \ n \  c)

=> P(e \ u \ c) =  0.06 + 0.05 - 0.009

=> P(e \ u \ c) =  0.101

Generally the percentage of the employees who will suffer at least one lost-time accident over the two-year period is mathematically represented as

\%E_1 = P(e \ u \ c) *  100

=> \%E_1 = 0.101*  100

=> \%E_1 = 10.1 \%

7 0
3 years ago
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