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pickupchik [31]
3 years ago
10

An object is 3.6 m from the pinhole. Its image is 4.2 cm from the opposite side of the pinhole. The height of the image is 0.8 c

m. What is the height of the object?
Mathematics
1 answer:
madreJ [45]3 years ago
5 0

Answer:

Height of the object is 68.6 cm.

Step-by-step explanation:

The height of the object can be determined by:

\frac{object distance from pinhole}{image distance from pinhole} = \frac{object height}{image height}

From the given question;

object distance from pinhole = 3.6 m = 360 cm

image distance from pinhole = 4.2 cm

height of image = 0.8 cm

So that;

\frac{360}{4.2} = \frac{object height}{0.8}

⇒ object height = \frac{360*0.8}{4.2}

                           = \frac{288}{4.2}

                           = 68.571

object height = 68.6 cm

Thus, the height of the object is 68.6 cm.

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The net of a triangular prism and its dimensions are shown below. What is the total surface area of the prism? a) 54 m squared b
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2 years ago
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How can I solve this?!
Nutka1998 [239]

A

evaluate f(5) and f(2)

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8 0
3 years ago
In rectangle PQRS, PR = 18x – 28 and QS = x + 380. Find the value of x and the length of each diagonal.
chubhunter [2.5K]

Answer:

x = 24, PR = 404, QS = 404

Step-by-step explanation:

PR and QS are both diagonals of the rectangle. The diagonals of a rectangle bisect each other and are of equal length. Thus, we can say:

PR = QS

18x - 28 = x + 380

Solving for x, we get:

18x - x = 380 + 28

17x = 408

x = 408/17

x = 24

Thus, PR = 18(24) - 28 = 404  and  QS = 24 + 380 = 404

Third choice is right.

7 0
3 years ago
A test is used to assess readiness for college. In a recent​ year, the mean test score was 20.3 and the standard deviation was 4
Vlada [557]

Answer:

\mu = 20.3

\sigma = 4.9

And we can find the limits in order to consider values as significantly low and high like this:

Low\leq \mu -2 \sigma= 20.3- 2*4.9 = 10.5

High\geq \mu +2 \sigma= 20.3+ 2*4.9 = 30.1

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

For this case we can consider a value to be significantly low if we have that the z  score is lower or equal to - 2 and we can consider a value to be significantly high if its z score is  higher tor equal to 2.

For this case we have the mean and the deviation given:

\mu = 20.3

\sigma = 4.9

And we can find the limits in order to consider values as significantly low and high like this:

Low \leq \mu -2 \sigma= 20.3- 2*4.9 = 10.5

High\geq \mu +2 \sigma= 20.3+ 2*4.9 = 30.1

6 0
3 years ago
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