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3 years ago

11

The length of a rectangle is more than three times its width. If the perimeter of the rectangle is 92 meters, find the dimension

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1 answer:

Juliette [100K]3 years ago

4 0

The dimensions are 35 m x 11 m

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4.) Determine the missing side of a rectangle with an area of 480 cm² and a length of 32 cm. Please show your work.

asambeis [7]

Answer:

w = 15 cm

Step-by-step explanation:

We want to find the width, as the area of a rectangle is length (l) * width (w).

Since we know the area and the length, we have 480 = 32w

Isolating w by dividing both sides by 32 gives us w = 15 cm

8 0

2 years ago

The measure of an angle is three times the measure of its supplementary angle. What is the measure of each angle?

madreJ [45]

Answer:

45°

135°

Step-by-step explanation:

Let the measure of one angle be x

So, measure of its supplementary angle = 3x

$\implies \: x + 3x = 180 \degree \\ \\ \implies \: 4x = 180 \degree \\ \\\implies \: x = \frac{ 180 \degree}{4} \\ \\\implies \: x = 45 \degree \\ \\ \implies \: 3x = 135 \degree$

5 0

2 years ago

%100 is shown on the following tape diagram. What percent is represented by the entire tape diagram?

maxonik [38]

100% is shown on the diagram

8 0

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

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

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

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

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

= [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].

<em>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.</em>

8 0

3 years ago

Find the distance (8,-7) (-4,-2)

Vinil7 [7]

Distance = 13 units

Point A: (8, -7)

Point B: (-4, -2)

$\sqrt{(8 + 4)^{2} + (-7 + 2)^{2} }$

$\sqrt{(12)^{2} + (-5)^{2} }$

$\sqrt{144 + 25 }$

$\sqrt{169}$

$13$

7 0

2 years ago

Read 2 more answers