What point on the graph represents ignors data

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1 answer:

Firdavs [7]3 years ago

3 0

The origin or (0,0) I need more character lol

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The image produced by an object is –10. 0 cm from a concave mirror that has a focal length of 5. 0 cm. The distance of the objec

Anon25 [30]

The distance of the object from the mirror is 3cm.

To calculate the distance of the object from the mirror, we make use of the mirror formula

<h3>Mirror formula</h3>

1/f = (1/u)+(1/v)............ Equation 1

Where:

f = Focal length of the mirror
v = image distance
u = object distance.

Using the real is positive convention,

Given:

f = 5 cm
v = -10 cm

Substitute these values into equation 1 and solve for u

1/5 = (-1/10)+(1/u)
1/5+(1/10) = 1/u
1/u = (2+1)/10
1/u = 3/10
u = 10/3
u = 3.33
u = 3 cm

Hence, The distance of the object from the mirror is 3cm.

Learn more about mirror formula here: brainly.com/question/26257987

7 0

2 years ago

the distribution of lengths of salmon from a certain river is approximately normal with standard deviation 3.5 inches. if 10 per

juin [17]

Complete question;

<em>The distribution of lengths of salmon from a certain river is approximately normal with a standard deviation of 3.5 inches. If 10 percent of salmon are longer than 30 inches, which of the following is closest to the mean of the distribution? 26 inches A 28 inches B 30 inches C 33 inches D 34 inches</em>

Option B is correct. The value that is closest to the mean of the distribution is 30inches.

The formula for calculating the z-score is expressed as:

$z=\frac{ x-\mu}{\sigma}$

Given the following parameters

$\sigma = 3.5\\x = 26in$

If 10 percent of salmon are longer than 30 inches, then:

P(x>30) = 0.1

Using the z table to get the value corresponding to the mean area 0.1.

The required z-score will be -1.285

Substitute the resulting parameters into the formula to get the mean of the distribution.

$z=\frac{x-\mu}{\sigma}\\-1.285=\frac{26-\mu}{3.5}\\-1.285 \times 3.5 = 26 - \mu\\ -4.4975 = 26 - \mu\\\mu = 26 + 4.4975\\\mu = 30.4975\\\mu \approx 30in$