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3 years ago

If x=3 units, y=11 units, and h=7 what is the area of the trapizoid

Mathematics

1 answer:

Elanso [62]3 years ago

5 0

Answer:

area of trapizoid=49 unit^2

Step-by-step explanation:

area of trapizoid=0.5(x+y)h

where, x=parallel side=3units

y=parallel side =11 units

h=distance between two parallel sides

area of trapizoid=0.5(x+y)h..............(1)

put x=3 units,y=11 units and h=7 in equ (1)

area of trapizoid=0.5(3+11)7

area of trapizoid= $0.5\times 14\times 7$

area of trapizoid=49 unit^2 answer

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3 years ago

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P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

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So, a 98% confidence interval for the population mean, $\mu$ is;

P(-2.426 < $t_3_9$ < 2.426) = 0.98 {As the critical value of z at 1% level

of significance are -2.426 & 2.426}

P(-2.426 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.426) = 0.98

P( $-2.426 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $23.92-2.426 \times {\frac{6.72}{\sqrt{40} } }$ , $23.92+2.426 \times {\frac{6.72}{\sqrt{40} } }$ ]

= [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

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3 years ago