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3 years ago

8

Each boy in a class has 7 pockets and each girl in that class has 11 pockets. How many boys and how many girls are there in this

class if there are 147 pockets and 17 heads?

2 answers:

jeyben [28]3 years ago

8 0

Answer: There are 10 boys and 7 girls in the class.

Step-by-step explanation:

Let x be the number of boys and y be the number of girls then according to the question we have the following equations :-

$x+y=17...............................(1)\\\\7x+11y=147...........................(2)$

The value of y from (1) will be

$y=17-x$ , substitute this in equation (2), we get

$7x+11(17-x)=147\\\\\Rightarrow\ 7x+187x-11x=147\\\\\Rightarrow\ 187-4x=147\\\\\Rightarrow\ -4x=147-187\\\\\Rightarrow\ -4x=-40\\\\\Rightarrow\ x=10$

∴ $y=17-10=7$

Hence, there are 10 boys and 7 girls in the class.

Aloiza [94]3 years ago

6 0

10 boys equals 70 pockets
7 girls equals 77 pockets

that equals 147 pockets and 17 heads

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Slope is Zero 0

Step-by-step explanation:

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3 years ago

If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal

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Answer:

a

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

b

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Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 800$

The variance is $var(x) = 1600 \ kg$

The range consider is $x_1 = 778 \ kg \ x_2 = 834 \ kg$

The value consider in second question is $x = 900 \ kg$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{var (x)}$

substituting value

$\sigma = \sqrt{1600}$

$\sigma = 40$

The percentage of a cucumber give the crop amount between 778 and 834 kg is mathematically represented as

$P(x_1 < X < x_2 ) = P( \frac{x_1 - \mu }{\sigma} < \frac{X - \mu }{ \sigma } < \frac{x_2 - \mu }{\sigma } )$

Generally $\frac{X - \mu }{ \sigma } = Z (standardized \ value \ of \ X)$

So

$P(x_1 < X < x_2 ) = P( \frac{778 - 800 }{40} < Z< \frac{834 - 800 }{40 } )$

$P(x_1 < X < x_2 ) = P(z_2 < 0.85) - P(z_1 < -0.55)$

From the z-table the value for $P(z_1 < 0.85) = 0.80234$

and $P(z_1 < -0.55) = 0.29116$

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$P(x_1 < X < x_2 ) = 0.80234 - 0.29116$

$P(x_1 < X < x_2 ) = 0.51118$

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$P(x_1 < X < x_2 ) = 51.1 \%$

The probability of cucumber give the crop exceed 900 kg is mathematically represented as

$P(X > x ) = P(\frac{X - \mu }{\sigma } > \frac{x - \mu }{\sigma } )$

substituting values

$P(X > x ) = P( \frac{X - \mu }{\sigma } >\frac{900 - 800 }{40 } )$

$P(X > x ) = P(Z >2.5 )$

From the z-table the value for $P(Z > 2.5 ) = 0.0062097$

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Step-by-step explanation:

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Read 2 more answers