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3 years ago

What is the area of the trapezoid?

Mathematics

2 answers:

iris [78.8K]3 years ago

5 0

Answer:

Step-by-step explanation:

AlladinOne [14]3 years ago

3 0

Answer:

C) 84 cm2

Step-by-step explanation:

$A=\frac{a+b}{2} h\\\\A=\frac{8+16}{2} *7\\\\A=\frac{24}{2} *7\\\\A=12*7\\A=84$

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Kelsey uses the similarity statement ARST - ARUV to describe the triangles below.

SpyIntel [72]

8.0 is the best answer

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3 years ago

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The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

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4 years ago

Point (2,3) is on a circle with the center (3,5). What is the radius of the circle?

ruslelena [56]

Can you add the picture to the question?

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3 years ago

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HELP!! 70 points will be given! Picture included

kkurt [141]

Answer: a) -13/16

Step-by-step explanation: Start by setting equations equal and rearrange X^3 - x^2 + 1 = 0. Visual inspection of graph shows x between -1 and -1/2. Start with x = 3/4 plug in and calculate: just a little too small. Try going halfway towards -1: x =-7/8 Plug in and the answer is very far from 0. Go halfway back towards -3/4: -13/16 and the equality is very close.

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4 years ago

The total number of people at a football game was 5600. Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

rewona [7]

Answer:

1100 field-side tickets and 4500 end-zone tickets.

Step-by-step explanation:

Let x represent number of field side tickets and y represent number of end-zone tickets.

We have been given that the total number of people at a football game was 5600. We can represent this information in an equation as:

$x+y=5600...(1)$