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3 years ago

What is the area of the composite figure? Card F

Mathematics

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

total area is bottom rectangle area+ upper rectangle area + two triangles.

whole area.

(24x19) + (13x9) +(24x13) =885

area of trapezoid

whole area - bottom rectangle

885-(24x19) = 429

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2 years ago

Lesson Posttest

stepladder [879]

Answer:

Step-by-step explanation:

Probability = number of fruit type/total number of fruit. Total number of fruit = 5 + 9 + 5 = 19.

The probability of drawing an apple is P(apple) = number of apples/total number of fruit = 5/19.

The probability of drawing a peach is P(peach) = number of peaches/total number of fruit = 9/19

The probability of drawing an apple is P(orange) = number of oranges/total number of fruit = 5/19

The probability of drawing either an apple, peach or orange at the first draw of fruit from the bag is

P(apple or peach or orange) = P(apple) + P(peach) + P(orange)

= 5/19 + 9/19 + 5/19

= (5 + 9 + 5)/19

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3 0

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

lorasvet [3.4K]

Answer:

The percentage is

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

The probability is $P(Z > 2.5 ) = 0.0062097$

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

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

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

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

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

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

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

The percentage is

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

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 $P(Z > 2.5 ) = 0.0062097$

7 0

3 years ago